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1
—1
~: + tl.h)).
Fig. 7
Thus, by by (39·1), (391), the the functional equation
/ ~ tl.k_ ~) 1  k (h + tl.h)
...... (39"3) (393)
must be be satisfied satisfied identically. identically. now assume assume that that 4, (u) isis continuous continuous and differentiable at at uu == 0 If we now
r
M:! o
çb'(h)= cp'(h)=1~h2'
by integration, integration, that and hence, hence, by ¢(h)=! log~~~.
(394) . ..... (39"4)
§§ 3942]
21
ANGLE AND DISTANCE DISTANCE
Direct substitution in (393) (39"3) then thenverifies verifies that thatthat thatequation equationisissatisfied satisfied identically. 40. IfIfz1, z 1 ,z2 z2 are are two two arbitrary arbitrary points points of ofthe thenonEuclideari nonEuclidean plane, plane, the nonEuclidean motion (§ 35) — z1 zz~ y=1 1ZtZ
transforms z 1 into into the — z,z.). z 2 into thepoint point(z2—z1)/(1 (z2 z1)/(linto the the point point 0, and z2
rotation about point can can now now be be transformed transformed into By a rotation about 0 this last point
IIz2—ziI/I z.z,l/l1z,z21, by (39•4), (39"4), so that, by 1 D (z,' z.) = ! log
I~ =;' z"i ~ I~· =~~~ ;} Nj
Z2
..,2
N}
<1, IIzil z, I < 1, I1z21 z2l <<1.1.
• •• • •
•<(401) 40"1)
J
with the the real real axis 41. In the case case when the orthogonal orthogonal circle coincides coincides with the nonEuclidean nonEuclidean distance distance between between two two points z1, z., zz2 in the the upper half2 in follows: the Möbius' Mobius' transformation plane can be deduced as follows: transformation ...... (411) (41"1)
z1 — Z
wjI == 1, the point zplane into the the circle cirde I w transforms the real axis of the zplane z1 into the point pointw1 w1 = 0, and the the point point z2 z2 into the point point z1
In this this way way we we obtain obtain
(z22 — — zz2).  z1)/(zlw 2== (z 2).

ki—z.I w)11 lz,z:l+lz.z.!.} ])(z.,""2 ~ o g  _       , lz,z.l1 z,z.l
I (zt) > 0, )l(z2)>0. I (z,) > 0.
...... (412) ( 41"2)
J
42. The triangle theorem. It is easy to show by direct calculation from (40"1) that, that, for for any three of the the nonEuclidean nonEuclidean plane, plane, points of D (::1 , ::,) ~ D (z1 , z2 ) + D (z2 , z3 ), •••••• (421) (42·1) where the "ign Pign of equality holds if and and only only ififz;2 lies on the nonEuclidean segment whichjoins joins; thethe following z1 and;. and z3But followingproof proofof of (421) (42"1) is is much much segment which • But more instructive. All the axioms axioms which which Euclid postulates at at the thebeginning beginningofofhis hiswork, work, excepting only only the the parallel postulate, hold excepting hold also also in in the thenonEuclidean nonEuclidean plane. Hence, since the the parallel postulate isis not not used used in in the plane. Hence, since parallel postulate the earlier earlier theorems, we we can can apply apply the the first sixteen theorems, sixteen propositions propositions of Euclid's first first book without without change change to to our figures; iii book in particular, theorems theorems concerning concerning the congruence oftriangles, triangles,the the theorem theoremthat that the the greatest greatest side side of of a congruence of triangle is opposite to the greatest the triangle triangle theorem theorem greatest angle, and, lastly, the expressed (42"1). expressed by (421).
22
NON.EUCLrDEAN NONEUCLIDEAN GEOMETRY
(CHAP. II II [CHAP.
43. 43. NonEuclidean NonEuclidean length length of of a curve. curve. The conception conception of of the the length length of aa curve curve can can at at once once be beextended extended in in the the sense of We suppose of this this geometry geometry to to curves curves in in the the nonEuclidean nonEuclidean plane. plane. We the curve curve is is given given in in terms terms of of aaparameter. parameter. On On the the curve curve that the finite set setof ofpoints points arranged arranged in in order order so so that that the theparameter parameteralways always take aa finite increases (oralways alwaysdecreases) decreases)from from one pointtotothe thenext. next. Let these increases (or one point these points in order order be be taken taken as asthe thecorners cornersofofaapolygon polygonwhose whose sides sides are are nonEuclidean straight anddefine define the the length lengthof of the thepolygon polygon as as the the straightlines, lines, and sum sum of the lengths lengths of its sides; sides; the the upper upper limit limit (when (when finite) finite) of the lengths of of all such such inscribed inscribed polygons polygons now defines lengths defines the the length of the given curve. Replacing z,, z1 , z2 z2 in (401) (40·1) by zero, by z,; z + az, we obtain, as az tends to zero, I azl
D (z, z + az) ...... I= I :"I , and itit follows that the follows that the nonEuclidean nonEuclidean length length of of an an arc arc of of the the curve curve which which complex function expressed by is given given by the complex function zz (t) is expressed by the integral
<«U _ at
r~o Iz'(t)I I
dt i—jz2(tfl · h J, 1lz"(t)l
The analogous formula formula for for the case in which the orthogonal orthogonal circle is the 2); this real axis is obtained in in aa similar similar way way from from (41 (41"2); this formula formula is is
I z' I ;1 [, dt. (z—z) z t, (zz) ut, 2
curvature. 44. Geodesic Geodesic curvature. It has been been shown shown by P. Finder Finsler(10) (10) that, that,with with the the·most most general general metric, metric, It has the geodesic curvature of aa curve curve Cat C at a point geodesic curvature point PP on onititmay maybe beobtained obtained follows: consider consider any any family family of of curves curves which which contains contains C, and conas follows: construct that touches Cat P; struct thatgeodesic geodtsic line of the metric which which touches P;let let1?R be be any any tangent geodesic, and lets let sdenote denotethe theextremal extremaldistance distancePPR point of this tangent geodesic, and R and 9 the the angle of intersection* at II and() R of the geodesic geodesic PPR R with with that that curve curve of of the family passes through through R; family which which passes curvatureatat PP is then the geodesic geodesic curvature k= urn lim~ ....... (441) (44·1) lc= $._o ....... o S$
It can can be be shown shown that that the number k so It is independent independent of the the particuparticuobtained is lar choice choice of of the the family family of of curves.
Fig. 8 in the metric. * It is is assumed assumed that that the the measure me~tsure of of angles angles is is suitably suitably defined defined in
§§ 43—45] 4345]
23
ROTATIONS ROTATIONS AND TRANSLATIONS
Applying the above above definition definition of of curvature curvature to toour ournonEuclidean nonEuclidean Applying geometry, curvature is invariant invariant for for nonEuclidean nonEuclidean motions. geometry, we we see see that that curvature Further, Further, if if the the orthogonal orthogonal circle circle is is lzl = 11 and if, if, according according to to§§39, 39, t/J' (0) = 1, the curvature curvature of of a curve curve at at the the centre centreof ofthis thiscircle circle is is the the same same as its ordinary ordinary Euclidean Euclidean curvature, curvature, since since the nonEuclidean nonEuclidean straight lines through ordinary straight value of lines through this this point point are are ordinary straight lines lines and and the value angles is is the same in both geometries. curvature of angles geometries. The nonEuclidean nonEuclidean curvature' point Zo z0may therefore be calculated by writing a curve z (t) at a a point ZZo
(
• )
'(t)= 1 z ()' ...... 442 t z0 (t)atat'=0. = 0. But this and then determining the ordinary curvature of of' (t) is known known to be — lc = ""f:r'    (f." . . ..••• (44:3) (443) 2i 21 <'t)i , consequently, from we have, have, on on replacing replacing z0 by z, the final and consequently, from {44"2), (442), we formula kic = ¥z'z' (zz' zz') + (1 zz) (z'z" z'z"). . ..... (44•4) (44"4) 2i (z'z')i (z't')1 45. NonEuclidean motions. It was shown shown in in§ thatevery every nonEuclidean nonEuclidean motion motion can can be obtained obtained It was § 32 that by two successive successiveinversions inversionswith withrespect respecttotononEuclidean nonEuclideanstraight straight lines; lines; there are are three three essentially essentially distinct distinct types types ofofnonEuclidean nonEuclidean motions, motions, corresponding to different relative dispositions of these these two straight straight lines. the inversions inversions are with respect to two Suppose Suppose first first that that the two nonEuclidean nonEuclidean straight lines whichintersect intersectatat aa point F, lines which P,where where they they thake :tilake an angle !8 with one o~e another; the themotion motionisis in in this thiscase case a nonEuclidean nonEuclidean rotation P, the theangle angleof ofrotation rotation being being 9. 9. By Bykeeping keeping PP fixed fixed and and letting letting about F, ()O vary we obtain aa oneparameter oneparameter group group of of nonEuclidean nonEuclidean rotations. vary we P1be toP with Let P with respect 1 be the point inverse toP to the orthogonal orthogonal circle; circle ; then then every every circle circle of the elliptic pencil points pencil which which has has the points P and and F1 P 1 as as limiting limiting points is is transformed this group, into itself by every operation of this so that the so the circles circles of this this pencil, pencil, in in so so far far as they lie inside the the nonEuclidean nonEuclidean plane, are nonEuclidean nonEuclidean circles circles with with common common nonEuclidean centreP.P. If C nonEuclidean centre C is is any any one one of these circles, on C are at the circles, all points on Care Fig. Fig. 99 same nonEuclidean nonEuclidean distance distance from fromP; F; this distance is the the nonEuclidean nonEuclidean radius radius of ofC. C.
cC
24

(CHAP. n it [ca&p.
NONEUCIJDEAN GEOMETRY NONEUCLIDEAN
Suppose, secondly, secondly, that the thetwo two inversions inversions to to which which the irotion notion 46. Suppose, isreducible reducibleare are with with respect respect to two nonEuclidean straight straight lines i!l lines which which intersect nor nor are are parallel parallel in in the sense neither intersect sense of Lobatschewsky. Lobatschewsky. The conjugate pencil, conjugate pencil, consisting consisting of all circles circles which which are transformed transformed into themselves themselves by the motion, is hyperbolic; hyperbolic; its points M1, M 1 , M, lie on the orits common points thogonal c.arcle.Among Amongthe thecircles circlesof of this thogonakircle. pencil there there is is exactly exactly one one nonEuclidnonEuclidean straight line. line. It It intersects intersects the the nonnonEuclidean respect Euclidean straight lines with respect M2 to which the inversions inversions are are performed performed F1, P,. P2. If at the points P~o Ifthe the nonEuclidnonEucliddistance P1P2 P 1P 1 is denoted by by 4h, !k, ean distance the motion motion can be be regarded as a translation of the plane plane through through distance distance hIt Fig. 10 along nonEuclidean straight Fig. IO along the nouEuclidean straight line M P1P2M2. the nonEuclidean nonEudidean straight straight line line ~Ms M1M fixed ~P By keeping keeping the fixed 1 P,M,. By and letting letting hIt vary varywe we obtain obtain aa oneparameter oneparameter group group of of motions, motions, such that every motion motion of of the the group group transforms transforms into into itself itselfevery every circular circular arc arc that every which lies lies inside the nonEuclidean nonEuclidean plane and joins the the points pointsM1, lrL, A!2. M2• circular arcs arcs are arecalled calledilypercycles; kypercycles; a hypercycle can also be dedeThese circular fined fined as the the locus locus of ofaapoint pointwhose whose nonEuclidean nonEuclidean distance distance from from the nonEuclideanstraight straightline line~M, MM isis constant. nonEuclidean constant. Finally, if the parallel the two twosuccessive successive inversions inversions are with respect to parallel 37), whose whose common commonendpoint endpoint isis L, we nonEuclidean lines (§ 37), nonEuclidean straight lines obtain a nonEticlidean nonEuclidean motion which Grenzdrewhich isis called calledaalimitrotation limitrotation ((Grenzdrehung). kung). Given two two pairs of parallel parallelnonEuclidean nonEuclideanstraight straightlines lines LJYI, LA!,LN LN and L1M1, there isis always L1Mu LL1N1, always a nonnon1N 1, there Euclidean Euclidean motion motion which which transforms each pair into the other that the other pair, so that points L, M, Jf, N X are are transformed transformed into the N11 or or L1, L1, N1, N1, fl'/1. M1. It points L1, LlJ Ml> N that from the follows that the point pointof ofview view of nonL1 Euclidean limitrotations Euclidean geometry geometry all limitrotations are equivalent, or rather that equivalent, or that they they can can differ only sense of the rotation. differ only in in the sense For the limitrotation by For limitrotation obtained obtained by inversions with respect respect to to LM LM and and L.!I.T LV inversions with M Fig. 11 is the theonly onlyfixed fixed point. Circles Circles inside inside Fig. 11 L is 1
§§ 46—48] 4648]
PARALLEL CURVES
25
the nonEuclidean nonEuclidean plane plane which whichtouch touch the the orthogonal orthogonal circle circleatat LL are are transformed into themselves; these thesecircles circles are arecalled calledoricycles. oricycles.
47. Every Everyordinary ordinarycircle circle which which lies lies wholly wholly within the the nonEuclidean nonEuclidean nonEuclidean circle, find its nonnonplane plane isis also also aa nonEuclidean circle,and anditit isis easy easy to to find Euclidean Euclidean centre. centre. Similarly every every circle circle which which touches touches the the orthogonal circle is an oricycle, and every every circular arc whose endpoints lie lie on on the oricycle, and whose endpoints the circle orthogonal orthogonal circle circle is is a hypercycle. hypercycle. The curves curves of these these three three types types are only curves curves of of the the nonEuclidean nonEuclidean plane which which have a constant constant nonnonthe only zero By §§ 44 the curvaturecurvature· of of the theoricycle oricycleisis(disregarding (disregarding zero curvature. curvature. By 2, while that that of ofaa hypercyele hypercycle is is less less than than2,2,and andof ofaa nonnonsign) equal to 2, 2. If k denotes denotes the Euclidean circle circle greater greater than 2. the curvature curvatureand and rr the circles, we nonEuclidean nonEuclidean radius radius of of one of these circles, we obtain obtain the the relation relation e"" +e_,. k=2 e"" t:r=2coth2r.* ...... (47·1) e
48. Parallel curves. 48. curves. Consider the the aggregate Consider aggregate of nonEuclidean nonEuclidean circles circles with with given given nonnonEuclidean radius r whose whose centres centres are at the the points points of an arbitrary set A of points of the nonEuclidean nonEuclidean plane. These These circles circles cover cover a set of points B (r), whose frontier, frontier, if if itit exists, exists, contains contains all all points points of the nonEuclidean nonEuclidean from A. plane which which are are at at a distance r from If we take as as the set A aa curve and if we we let let rr If we take curve C 0 of of finite curvature, curvature, and vary while while remaining remaining less less than aasufficiently sufficiently small small upper ~pper bound, bound, we we obtain obtain a family family of parallel parallel curves curves (in (in the sense sense of the the JlOnEuclidean metric). It can metric). Ca.n be be proved proved in in just justthe thesame sameway way as as with with other other similar similar theorthogonal orthogonal trajectories trajectories problems problems of of the the calculus calculus of ofvariationst variationst that that the of such a family of equidistant curves curves are are nonEuclidean nonEuclidean straight straight lines. lines. of the thenonEuclidean nonEuclidean plane planeisissimply simply covered covered Conversely, Conversely,ifif aa portion of by a family nonEuclidean straight straight lines, the orthogonal family of nonEuclidean orthogonal trajectories trajectories of the above. Thus the the the family family are are parallel parallel curves in the sense sense defined defined above. families of parallel parallel curves are the simplest examples of families the following: following : common centre; centre; (a) nonEucliclean nonEuclidean circles circles with with aa common (b) oricycles oricycles which which touch touch the orthogonal circle circle at at aa common common point; point; (c) hypercycleshaving same endpoints endpoints M, M;,21(2 M 2 (see Fig. 10). hypercycles having the same 11"
r<"2we •* For r< we have ha.ve 2
t
14 1 4
16
k=:;.+ar45r3+ ....
t See Frank Frank and a.nd v. v. Mises. Mises. Die DieDifferentialDifferential und und Integralgleichungen der iWechanik Mechanik Integraigleichungen der und Physik. Physik. Vol. i, ch. 5. und Vol. 1,
CHAPTER CHAPTER ill HI
ELEMENTARY TRANSFORMATIONS ELEMENTARY TRANSFORMATIONS 49. The Theexponential ezponentlal function. function. 49. The function
...... (49•1) (491)
W=e"
gives rise two important important special special transformations. transformations. On On introducing introducing gives rise to two rectangular coordinates in the coordinates x, a:, yyin thezplane zplaneand andpolar polarcoordinates coordinatesp,p, 4> the wplane, wplane, i.e. writing zz = =a:+ replace ((491) 49·1) by in the z + iy and and w == pt!+, we replace two equations equations the two p=ex, p = e", çb=y. 4> = y. A horizontal strip of the iplane zplane bounded hounded by the lines yp = y, and '!J y,, y = P2, wedgeshaped region where I y,y transformed into a wedgeshaped region of of the Y22 I1< 2r, is transformed I
0
zplane plane
w— plane Fig. 12 Fig.12
wplane, the angle angle of the wplane, the the wedge wedge being being ex= =II y,— !It 1 The = Ilt/>2— t/J, II= representation is is conformal throughout the interior representation conformal throughout interior of of these these regions, regions, since neverzero. zero. since the derivative derivative of of?e"isisnever As a special special case, case, if IJ/2 — r (e.g., y1 y1 == 0, 0, 'P2 !J2 == ir), '~~"), the wedge wedge behe y,l = 7r comes a halfplane. strip, namely namely that II'Yi !J1  Y2! y2 1<<2ir, 2w, The restriction restriction on on the the width width of of the strip, may be dropped. If, for example, example, y'=O y, = 0and andY2 y2 = 21r, the strip is reprerepreI
,... , .......
/ 27T( \
\
\' ....... __ ,. /
271 27T I
/
/
zplane zplane
wplane Fig. 13 Fig.13
sented sen ted on on the the wplane wplane cut cut along along the the positive positive real real axis; axis; and if — obtainedcovers coverspart part of of the wplane IY1 y" I> 21r the wedge wedge obtained ~vplane multiply. multiply. j> 2ir
§§ 4952]
THE WEDGE AND THE THE STRIP STRIP
27
The cases when when 1!/1— '!hi is an integral multiple of 2,. are of particular for the the strip is is then transformed importance importance; for transformed into aa Riemanu Riemann surface surface with an algebraic algebraic branchpoint, and cut cut along along one one sheet. sheet. I
furnished by by (49"1) (491) 50. We Weobtain obtain the the second second special special transformation transformation furnished by considering an arbitrary arbitrary vertical vertical strip bounded bounded by by the lines lines z = = z., z = a'2. z 2 • This This strip strip isis represented represented on aa Riemann Riemann surface which covers the annular annular region region Pi P1
constant, let a'1 z 1 tend to to —  oo , constant, we let the strip x 1 <
...... (51"1) (511) gives, of course, course, on interchanging the gives, the zplane zplane and and wplane, wplane, precisely the same conformal conformal transformations transformations as as (49"1). (491). problems of of conformal conformal representation, representation, important important in the A number of problems the proofs of general general theorems, theorems, can be solved by combining the the above above transtransproofs formations with with the Mobius' formations Mobius' transformations discussed in Chapter i.I.
52. Representation Representation of of aa rectilinear rectilinear strip on 52. on aa circle. circle. The strip
28
ELEMENTARY TRANSFORMATIONS
and hence hence
[ca&P. Ill [cHAP. m
1+iz 1 + iz 2h w = .., log 1 . . .. z7r 1—2z Z7r  ZZ
...... (52"1) (521
4k dw 4h dz,r(1+z2)' dz = r (1 + z2) '
...... (52"2) '522)
gives This gives
which is positive positive when when z = 0, as the the conditions conditions of of the the problem problem required. required. 8 If the the point point zzdescribes describes the the unitcircle unitcircle zz == e; , then die 2/u dw 2ki d() — = 71" ,r cos ()0'' dO as the circumstances require. If If the the point point zz describes describes a pure imaginary as circumstances require. the = rr << 1, w describes a curve the circle circle Iz I= curve which which lies inside a finite circle and whose whose form (52·1), form isis easily easily determined; determined; for for this curve, by (521), I
2h
1 +r
r
1r
li(w)l~log
.
. ..... (52·3)
The inequality (52"3) has many many applications. applications. (523) has 53. An arbitrary wedge, wedge, whose whose angle denote by byircc, ?TOt, and whose angle we denote vertex 0, can vertex we wesuppose supposetotobe beat at w w ==0, can be transformed into a halfplane. halfplane. For, For, on on introducing introducing polar poiar coordinates z=re18 , w=pe"~> ...... {53"1) (531) in the relation relation W=Z"=e>wgz, ...... (53"2) (532) this relation takes the form form 4=cLO; p=,..., cp =OC8; ..•... (53"3) effects the the desired hence (53"2) effects desired transformatransforma 0 16 Fig. 16 tion •.. Hence, by combination with Mobius' by combination with a Mdbius' transformation,aawedge wedgecan canbebetransformed transformedinto intothe the interior interior of of a transformation, show that that the representation circle. relations (53"3) show representation of of aa wedge wedge on on circle. The relations a halfplane halfplane (or on a circular circular area) area) isis conformal conformal at points inside inside (or on at all points or on the except at at its vertex or the boundary boundary of the the wedge wedge except vertex w w ==0. 0. Two Two curves 0 at an an angle angle A X are transformed two curves which whichintersect intersectat at w w ==0 transformed into two curves which at an an angle angle A/cc A/a., so which intersect intersect at sothat that at at the origin corresponding no longer longer equal, equal, are are p,vvportional; proportional; in in these these circumstances circumstances angles, angles, though no the representation at at the theorigin originisissaid saidtotobe bequasiconformal. quasiconformal.
54. Representation 54. Representation of a circular crescent. The area area between between two two intersecting intersectingcircular circulararcs arcsororbetween betweentwo twocircles circles and this strip into aa. half* If the the wedge wedge is first first transformed into aa. strip strip (§ 49), 49), and ha.lfplane pla.ne (\152), the same function 153·2) is arrived at.
5356] §§5356]
RIEMANN SURFACES
29
contact isis transformed, transformed, by by means means of ofaaMöbius' Mobius' trans.. transhaving internal contact formation whereby common to to the thetwo twocircles circles corresponds corresponds to to formation whereby aa point P common
17 Fig. 17
or aa strip respectively. the point co, into a wedge wedge or respectively. Hence, Hence, by by the the foreforegoing, each each of ofthese these crescentshaped crescentshaped areas areas can be be represented represented conformally circle. In In the the same same way way the exterior exterior of of aa circular circular on the interior of of a circle.
Pig. 18 Fig. 18
crescent or the the exterior exterior of of two two circles circles having external contact can also be represented on the interior interior of of aa circle. circle.
55. 55. Representation Representation of ofRlemann Riemann surfaces. surfaces. is a positive If, in equation (532), (53'2), ex is positive integer n, the the equation equation effects effects on aa portion portion of a Riemann the representation representation of of the the unitcircle unitcircle Iz I <1 < 1 on Riemann surface of nn sheets, sheets, overlying overlying the circle circle Iw I <1 < 1 and and having having an an nfold nfold surface branchpoint at w w = 0. I
56. ItItisisimportant importanttotodetermine determinethe thefunction function which which represents represents the which has has the the point w0 as its its branchunitcircle on a Riemann surface surface which w0 as point § 55. We require, require, further, point but butis is otherwise otherwise of the kind described in §55. and that that w w =0 = 0 should should correspond correspond to zz = = 0 and that parallel parallel directions directions the same drawn same sense sense through through these these two two points points should should also also drawn in the correspond. · Suppose first that thatw0 w 0 is is aa positive positivenumber numberIik(/z (k << 1), and displace displace the branchpoint Riemann surfat:e branchpoint of of the the Riemann surfacetotothe the origin originby by means meansof of the •
30 30
ELRMRNTAflY TRANSFORMATIONS ELEMENTARY
Möbius' Mobius' transformation (§ 35)
11u A— w=lhu"
[CHAP. m [CHAP. III ...... (56"1} (561)
By §§55 55 the equation u = t" represents represents this new Riemann surface on on the 1
simple (scklickt)•* circle Itl < 1, and in this way t = Jtfi corresponds tow= simple (sc/dicht) tow = 0. 0. Hence the desired transformation is I
1
h,iz A" — z
t t=.., 1—h"z lk"z i.e., in terms terms of of the the original original variables, variables, 1
1
(Ik"z)" (1k,., z)" (1 —h w=h w=k 1 1 (1k"z)"z)" (1—k ..h"z)" (1 — k"z)" — k A22 (1theorem shows showsthat that The binomial binomial theorem
...... (56"2)
=
...... (56"3)
In In the thegeneral generalcase casewhen whento0 w 0== he'° ke16 the required required transformation transformation is 11
11
(1k»e••z)"(1k
e16 z}"
' e 1 / 2 16 16 (1(1 ~t»e z)"k (1 k ,.e z)"
w=h' w = k
'
1
16 •
(564)
• ••••• (56"4)
57. The Thecase case in in which which the the Riemanu Riemann surface has aa logarithmic logarithmic branchbranchhe treated in point at atto0 w 0 == kii may be in aa similar similar manner. manner. We Wefirst, first, by by means means transform this Riemaun Riemann surface into the surface already conof (56"1), transform virtue of of the thesame same paragraph, paragraph, the the transforinatransformain §§ 50; then, in virtue sidered in et transforms this surface = e' tion u = surface into the the halfplane halfplane lt (t) <<0, 0, making making A. Finally u = A, i.e. w w=0,0,correspond t = log k. k, i.e. correspond to tot= Finally the halfplane halfplane It (t) <<00 is transformed into the the circle circle IIzlI < 11 by means of the relation 1 +zlog/z +z tt= 1j— log A, = lz ' so that the so the function function producing producing the desired desired transformation transformation is seen to be 11+z +•1og"
to=
kel•
1~logh
~ Ioc" 2z
le1•
,
~logh.
...... (57"1)
2 e 1 • Ike•• lk 1 From this equation we obtain at once From once
2k log A >O. dw) = 2klogk (fdw\ 1—k2 dz •=O 1k —
2
. ..... (572) (57•2}
* Alliernau4 points of the * A Bieman.J surface is simple (schlicht) (•chlicht) if no two points ihe surfacehaveihesame surface have the same coordinate u.
§§5760] 5760]
31
EXTERIOR OF THE ELLIPSE
58. IfIfn nisismade madetotoincrease increaseindefinitely, indefinitely,the theRiemaun Riemann surface surface dealt dealt with with 58. Riemann surface surface of of§§57, having a becomes,ininthe the limit, limit, the Riemann 57, having 56 becomes, in §§56 logarithmic branchpoint. It It is logarithmic branchpoint. is therefore therefore to be be expected expected that, that,ifif4',. ~. (z) (z} denotes member of of (56"2), (562), and 4' (z)that that of (571), denotes the righthand righthand member 1/J (z) (57"1), we we shall shall haVe have ~.(z} . .•... (58"1) lim 4, (z) = 1/J(z). 'Ii (z).
..
,._
The truth truthof of(581) (58"1)can canininfact factbe bededuced deducedfrom from aa general general theorem; but verified directly, directly, as as follows. follows. In the equation equation can also be verified In 1
(Jii. )" =t= 11
put
I
I _! 6,. {log(ll.nz)lotrl1ll"•l}
••••••_(~8"2)
(1k"'z)"' urn e,.=0; lim exponent of (58"2) can can = 0; the exponent of e in (582)
1— i,.,soso that that k"= 1e,.,
.......
now be written as 1lo n{log(1ze..)log(1z+e,.z)} g og log(1e,.) · ' and when when e,. tends to zero zero this this tends tends to to the limit 1 +z 1+z log ii. logk. 1
i— z
Equation Equation (581) (58"1)now nowfollows follows at at once. once. It It can can be be shown shown in the the same same way that the the limiting limiting form form of (56"3) is (5V2). (57"2).
59. ItItcan canalso alsobebeshown shown that, that,for forall allvalues values of ofn, n, ~'.. (0) > ~·.+1 (0), ...... (59"1) and this inequality may perhapsrest restuponsorne upon somedeeper, deeper, as as yet unremarked andthisinequalitymayperhaps 1
— A,., so h"1 = property of the transformations. transformations. To To prove prove (59"1) write log li =~.so that .\,.+1
(— 2/i log /a\ ~·.. (0) = (:~1~~ k))".,,(~~'• ;~"'!)' 2A,. 1' i—/i' 'i"' —
—
the function and and the function
11'eA —
A'
A'
2A= 1 + 3! +51+ ...
A decreases. steadily decreases as A 60. 60. Representation Representation of of the the exterior ezteriorof ofan an ellipse. ellipse. We start toplane, cut along the startwith withthe theproblem problem of of representing representing the thewplane, a
u=—, w=a—
32
ELEMENTARY TRANSFORMATIONS
[CHAP. III in
transforms the cut cut wplane wplane into into the the uplane uplane cut along the negative transforms the negative real axis, and and the further transformation axis, transformation u == t2 t 2 transforms this into into the the halfhalfplane Jli (t) > 0. 0. To 'fo the the point point ww == ~ correspond correspond the the points uu = 11 and t= = 1, so that the the required required transformation transformation is is obtained obtained by by writing writing 1 ++tt 1 z=1t" =
Thus finally
w=~(z+D;
...... (60"1) (601)
the relation relation (60"1) very remarkable represents the cut w1) is very remarkable in in that itit represents plane not only on the exterior plalie ezterior but but also also on the interior interior of of the the unitcircle. unitcircle.
If in (601) 61. If (60"1) we we write
zz=re;', = re2O, where (for (for example) example) r> r 1, we obtain where
w={(r+!) sino}; ...... (611) w=~{(r+~)cos8+i(rDsin0}; (61"1) hence hence to the circle circle II z I = rr there corresponds corresponds in wplane an an ellipse ellipse in the wplane with semiaxes
rz/
1\
a=~ (r+ ;),
1\
b=~ (r;).
. ..... (6r2) (61"2)
Conversely, (61"2) determines ex, r: Conversely, ifif a, a, bare b are given, (61
r=
...... (61"3)
Thus equation (60"1) transforms transforms the the exterior exterior of the circle circle I zz I> > r> r> 11 (or the interior of the circle circle I z I < ~
1)
(6r1). (61"1). The representation representation of of the the interior of an ellipse on the interior of the The ellipse on
unitcircle cannot,on onthe the other other hand, unitcircle cannot, hand, be obtained obtained by means means of the the elementary transformations so so far employed. employed. But itit is to be that elementary be noticed that 1) represents represents the upper half the function (60"1) half of of the the ellipse ellipse (61"1) on the upper upper half halfof ofan anannular annularregion regioncut cutalong alongthe thereal realaxis; axis; this this last area, however, and therefore therefore the the semiellipse also,isis easily easily (by (by a method however, and semiellipse also, method similar to that that of 50) transformed transformed into into a rectangle. rectangle. The similar to of§§ 50) The details details of of this this calculation, which whichleads leadsto to trigonometric trigonometric functions, functions, are are left left to the reader. calculation,
62. of an 62. Representation Representation of an arbitrary arbitrarysimplyconnected simplyconnected domain on a bounded domain. domain By open connex connex (zusammen/iàngend) (zusammenkiingend) set By a domain we we understand understand an open set of points domain points of of the the complex complexplane; plane; thus thus an open open set of points is a domain
§§ 61—63] 6163]
CONNEcrriVITY
33
if and only if every every two two points points of of the the set can be joined by a continuous curve all of whose whosepoil}ts pointsbelong belongtotothe the set. set. The frontierS frontier S of of aa domain domain curve Tis defined as which are not also points T is defined as the set of limiting points of T which points of 7'. T. The The set setS8++ TT of of points points of of aadomain domain and and of of its itsfrontier frontier isis called called with the a closed closed domain domain and will will be denoted denoted by by 7'. T. In conformity conformity with is to be (see Chapter i,1, §§ 9), the point point oo is be conventions conventions already already made made (see treated like like any other other point point and and may, may, in in particular, particular, be an interior point point of a domain 7'. T. If If a oneone oneone transformation transformation isis continuous continuous at at every every point point of ofaa domain domain T, transforms T into another domain domain T'. If T is is not not identical identical with with 7', it transforms whole complex one point, and the whole complexplane, plane,itsitsfrontier frontiercontains containsatat least least one hence we we can, can, by by means means of of a Möbius' transformation, represent represent T conMobius' transformation, conformally formally on on aa domain domain T• T* which which does does uot not contain the point zz = oo.
63. The 63. The classification connectivity is classificationof of domains domainsaccording accordingtoto their their connectivity T, which which may interior important. domain 7', important. A domain may have have the the point oo as an interior to be be rnply mply connected if its frontier frontierS consists of of m distinct point, is said to S consists distinct topological incontinua. The degree degree of of connectivity connectivity is, of of course, course, a topological continua. The is to to say, say, ititisisnot notaltered alteredbybyany anycontinuous continuousoneone oneone variant, variant, that is transformation. Our main main concern concern here here isiswith withsimplyconnected simplyconnected domains, domains, whose whose frontier consists consists of of aa single single continuum. continuum. 'rhe The property property expressed expressed by by the the frontier words "simplyconnected" "simplyconnected" can can be be specified specifiedininmany manyother otherways. ways.ItIt can words proved by topological methods that that all these be proved topological methods these specifying specifying properties properties another. AAfew properties are: are: are equivalent to one another. few examples examples of such properties 7'* (a) If the domain 7'* does not contain the point then (a) If the domain T* does not contain the point oo, T• is simplyconnectedifif and and only only if the interior interior of ofevery every polygon, polygon, whose whose simplyconnected frontier belongs to T•, T*, consists frontier belongs to consists entirely of points of 7'*• T•. (b) A domain domain TT isissimplyconnected simplyconnected if and only only ifif every every curve y, y, (b) A whichjoins joinstwo twopoints pointsofofthe the frontier frontier SS and and lies within within 7', divides T7' which T, divides into at at least least two two domains. domains. (c) The (c) The same same is true true ifif every every closed closed curve within within 7' T can be be reduced reduced to aa point 7'. (In the course pointby bycontinuous continuous deformation deformation in T. course of the deformation the curve curve may may possibly possiblyhave havetoto be be taken taken through through the the deformation the point cx.) oo .) is simplyconnected if it can be represented on on the (d) AAdomain domain 7'Tis simplyconnected if circle by by means means of of aa continuous continuous oneone oneone transformation. transformation. interior of of a circle (e) The theoremmay maybeberegarded regardedas as giving giving a specifying (e) Themonodromy monodromy tkeorem specifying property for for a simplyconnected domainT• T which simplyconnected domain which does does not contain the
34
ELEMENTARY TRANSFORMATIONS
[cn.aP. [CHAP.
m
ill
point oo. This characterisation is especially especially important important in in the theory of functions. theorem states functions. The theorem states:: .A aomain T T•which wkickdoes does not not contain cuntain the tke point A &nnain point oo is simplyconnected if, w/zeneveirf(z) wlumever f (z) is an analytic analyticfunction functiunwhich wkickcan canbe becontinued continued along along every curve T"',f(z} a singkvaluedfunction(ll). curve in in T,f(z) is is a singlevalued function(11). By this this is is meant: meant: If By If an an arbitrary arbitrary functional functional element element is assigned assigned to of T, then a point ofT*, thenanalytic analyticcontinuation continuation of ofthis thiselement element along along paths paths entirely within T* T• must musteither either lead lead to tothe thesame samefunctional functional value value at any whatever path path that thatpoint point is is reached, reached, (R" must contain contain point of T*, T, bybywhatever orT* T must a singular point point of of the the function function obtained obtained by by continuation. continuation. by this definition, 1, is, by definition, Every circle, circle, for for example examplethe the unitcircle unitcircle I zz JI< <1, a simplyconnected simplyconnected domain. domain. Koebe, Koebe, in his lectures, lectures, proves proves this this as as follows: follows: A function which can can be be continued continued along along every every curve curve within within A function f(z), which the circle, circle, must must be expressible in 0 by a powerpowerthe in the the neighbourhood neighbourhood of of zz==0 series ...... (631) (63"1} f(z)=a0,a1z+a2z2+ ....
The radius radius of of convergence convergenceofofthis thisseries seriesisisatat least least unity, unity, for for ifif it The wereless lessthan than unity unity analytic analytic continuation continuation of off(z) f(z) along radius were along every every radius <1 1would within I z I< wouldnot notbe bepossible. possible. It follows follows that, for for all analytic analytic <1,1,the continuations thevalue value of ofthe thefunction function f(z) f(z) may may be be continuations within I zz I< calculatedfrom fromthe the relation relation (63"1), (631), and and the function is seen to be calculated function is seen to singlevalued. From this ititfollows follows that thataadomain domainisissimplyconnected simplyconnected if itit can can be be From with the put into into oneone oneone correspondence correspondence with the D C unitcircle Izz<1 transI < 1bybymeans meansof ofaa conformal transThus, for halfplane formation. Thus, for instance, instance, the halfplane 54), and hence also also the (§ 12), the semicircle semicircle(§ 54), quartercircle (§ 53), are simplyconnected. It can now now be be shown shownthat that the square square is is simplysimplyconnected, for foritit can can be be regarded regarded as as the sum of two every common point P two quartercircles; quartercircles; every common point of the two be joined joined to to the two quartercircles quartercircles can be middle point 00 of of the the square square by by means means of a A B middle point Fig. 19 whichlies liesentirely entirelyin in ABCM ABUM and ensegment which Fig. 19 tirely in ANUD ANODand andthe themonodromy monodromy theorem theorem applies. 64. Now Nowlet letTTbebeaasimplyconnected simplyconnected domain of the the complex complex zplane, that the such that the frontier frontier of of TT contains contains at atleast leasttwo twopoints pointsA1, .A" B1, B., and let let Consider the z. a.n interior interior point of T, T, other other than the point point oo. Consider z, be an circular arc (which may be a finite or infinite straight straight line) line) which which joins
§§ 64, 65] 651
SIMPLYCONNECTED DOMAIN
35
A, to B, and as an an interior interior point, point, and denote denote by first A1 to B1 and has has z;0 as by A A the first describing this from z0 z0 to A1, Ah frontierpoint of frontierpoint of TT which which isis met met in in describing this arc from and by B the the corresponding corresponding point of the the arc arc from from z0 z0 to B1. B, _Thus Thus all interior points pointsof ofthe thearc arcAz Az0B areinterior interiorpoints pointsof of T, T, and and its endinterior end0 B are points are frontierpoints we now now apply zplane the points frontierpoints of T. If we apply to to the zplane Möbins' transformation whereby whereby the the points points A, correspond to to the Mobius' transformation A, z0, z0 , B B correspond points 0, 1, 1, oo in the the uplane, uplane, the domain TTis transformed into intoaasimplysimplypoints 0, is transformed and uu = = oo are frontierconnected connected domain domainT,, T,, which whichisissuch such that that u = =00 and positive real interior points points of of T, T1but butall allother otherpoints points o£ of the the positive real axis are interior ofT,. points of points T1. By transformation u == vv44 we we now now obtain different By means means of of the the transformation obtain four different domains in in the the vplane vplanecorresponding correspondingtotoT,T1; let TT2 denote that one ot domains ; let denote that 2 v > 0 corresponds corresponds to straight line line them them in which whichthe the straight straight line line v> to the straight (v) >0, and is u>O*. Weshall shallprove provethat thatT2 T 2lies lies in in the the halfplane halfplane J!{ (v)>O, u> 0*. We therefore transformed by v1 W=...••• (64'1) (641) v+1 T' which into a domain T' which lies inside the circle circle Ij w T 2 were were to wI<< 1. For ifif T2 T 2 a curve joining contain pointv contain aa point v22 for which J!{ (v) ~ 0, we we coulddrawin coulddrawin T2 joining and v1 vv22 and v1 == 1; let P be bethe thefirst firstpoint pointofofintersection intersection(counting (countingfrom fromv1) 'lh) of this curve with the imaginary vaxis, vaxis, and Q Q the last intersection intersection of to P with with the the real real axis, axis, so sothat that the arc the arc from from vv1 arc PQ, PQ,which which we we 1 toP will denote denote by by Yv• y,,, lies lies either either in in the first first or or in in the the fourth fourth quadrant quadrant of of the the will vplane. all of vplane. The The relation relation uu == vv44 transforms the arc Yv into a curve rnrve y,, 'Yu all whosepoints pointsbelong belongtoto T,, T1,and andwhose whoseendpoints endpointslie lieon on the the real axis axis whose uu>> 0; by by adding adding to y, a portion portion of the axis u> u > 00 we we therefore therefore obtain obtain which lies lieswholly whollyininT,T1and andsurrounds surroundsthe thepoint point uu=0. = 0. a closed curve which But no nq such curve curve can can exist, exist, since since T1 T, isis simplyconnected simplyconnected and and does not contain either of the points uu == 0, 0, u = oo. contain domainwhose whosefrontier frontiercontains containsat at least Thus: any simplyconnected simplyconnected domain two distinct points can, tmn.iformations, be be conformally conformally repretwo can, by by simple simple transformations, sented m; a domain which which lies entirely inside the unitcircle. unitci1·cle. A multiplyconnected multiplyconnected domain domaincan can be be treated treated in A in precisely precisely the same same way provided provided that that its frontier frontier contains contains at at least leastone one continuum continuum of ofmore more way than than one one point. point. though in appearance little different different from from 65. AAtheorem theorem which, which, though appearance but little thatof § 64, that· of§ 64, is, is, on on account account of of the the value value of of the the constant constantinvolved, involved, of of the the greatest importance importance in the general general theory, theory, is is due due to to Koebel2). Koebe(l2}. * Each of of the the four four domains domains corresponding correspondingto to TT1, inparticular particular T7',, (schlicht) 1 , in 2 , is simple (•chlicht) (cf. §§56, 56, footnote).
36
ELEMENTARY TRANSFORMATIONS
III (CHAP. m
domain of the wplane not conLet T be be aa simple simple simplyconnected simplyconnected domain x;; suppose supposefurther furtherthat that a.ll all points points of the the point point w w= = oo the circle circle taining the T, but but that that w w==1 is a frontierpoint of T. T. Any I ww I<1 < 1are are points points of T, 1 is frontierpoint of Any domain T Tis is transformed by the same same function such domain 4(1 ÷ + .j2) 2 z — '651 W= •••••. (65"1) 2 2 {1 + (1 + .j2) z}
into a simple simplyconnected which lies into simplyconnecteddomain domainT' T'which lies inside inside the the unitcircle unitcircle I z I <<11 but always contains the fixed circle I z I <(1 < (1 + + J2t 4• Consider twosheeted Riemann surface surface which has Consider in in the wplane the twosheeted branchpoints at at w = 1 and w branchpoints w= = oo ; the function function I
1w=u} l—w=u2
...... (65"2)
transforms this surface surface into the the simple simple uplane, uplane, making the two points 1 correspond to w =0, and transforms anycurve curvedrawn drawninin TT and u == ± correspond to w = 0, and transforms any ±
starting starting from from w ==00 into intotwo two curves curves in in the the uplane. uplane. The 'fheaggregate aggregate of of all a.ll curves in the uplane which correspond correspondtotocurves curvesininTT and and which which start uplane which from au = + 11 fills domain which which we we will denote by T1. T1 • fills aa simplyconnected domain We shall prove that, if u0 is any complex number, the domain We prove that, if u0 is any complex number, the domain T1 T1 cannot contain both both the points ± uu,. For ifif ititdid, did, these these two two points could 0 • For to there I;., and toy,. there would correspond be joined by a curve y,. lying inside T1, wplane aa closed 7w lying lying inside inside T, in the thewplane closed curve 'Yw T, some some point pointw0 W 0 of of)'w y"' is simplyconnected corresponding to the theendpoints endpoints±±u0 u0 of of 7u. y,.. Since T Tis simplyconnected we we can can now, now, by by continuous continuous deformation, deformation,and and keeping keepingthe the curve always inside T, reduce reduce 'Yw to an arbitrarily arbitrarily small sma.ll curve curve passing passing through through a'0. w0 • But as, as, in in this thisdeformation, deformation, the theendpoints endpointsofofVu y,. remain remain fixed, fixed, we we have have clearly clearly arrived arrived at at a contradiction. 66. Equation (65·2) transforms transforms the circle circle I w w I= = 11 into aa bicircular bicircular quartic quartic curve curve whose whose equation in in polar polar coordinates coordinates is is 2 = 2 cos 28. p . ..... (66"1) 29. p2=2 This curve curve isis symmetrical symmetricalwith withrespect respecttotothe the origin; origin; and, of the the two two domains bounded by by its loops, and domains bounded loops, one one must mustlie lieentirely entirelyinside inside T1, T1 , and hence, by§ by § 65, 65, the the other must lie entirely outside outside T1. T1 • hence, The function 1u t=14u 1+u
...... (66"2)
transforms the curve curve (661) (66·1)into intothe thecurve curveCCof ofFig. Fig. 21, 21, having having a double point at t == 11;; and itittransforms which lies lies entirely entirely transformsT1 T 1 into intoaadomain domain7'2 T2 which loop of inside the curve inside curve C, C, since since the the exterior exterior of of C C corresponds correspondstoto that that loop (66"1)which which contains contains no no point point of of T1. T1 . itHtherefore thereforefollows follows that curve (661) the curve
§66] § 661
KOEBE'S THEOREM
37 37
the inner inner ioop loop of of C C lies entirely entirely inside inside 7'2. T2 • But (65'2) and (66'2) give
'J?
w) == {1.j(lwW;' ...... (663) ( 66 .3) w 1w) 2• Conand from .J2)2. from this we obtain at once once that, that, ififII w I== = I, Itt I~ (1 ++ ../2) sequently ~ lies entirely inside also, since since the the curve curve C 0 inside this this last circle; also, see further is its own respect to the unitcircle own inverse with respect unitcircle I tt I= == 1, 1, we see .J2)22 in its interior. that that 7'2 T2 must contain the the whole whole circle jt ItII== (1(1 ++ J2t interior. t== 11+
./2
u—plane up lane Fig. 20 20
theorem enunciated at The theorem a.t the thebeginning beginning of of §§65 65 is is now now obtained on on w as aa function function oft of t and then writing express was using (66'3) to express t=(1 t == (1+.,J2)2z. + ../2)2 z. . ..... (664) (66'4)
I— plane tplanc
Fig. 21 21
Since we we supposed supposedthat that T does Since does not contain the point point w w= == oo , T 7'11 does point uu === oo ; hence hence tt === — 11 lies outside outside 7'2, T 2 , and, finally, not contain the point .J2)22 lies outside T'. z=(1 + J2t lies outside
38
TRANSFORMATIONS [CRAP. ELEMENTARY TRANSFORMATIONS [CHAP. m;, §66 § 66
Remark. Let 7'Tbebea simplyconnected a simplyconnected domain domain contained contained in T and and Remark. Let and denote denote the the frontier frontier of of T7' by Yw· The containing the point w w = 0, and transformation (65·2) (652) gives transformation gives two distinct domains domains in the the uplane uplane corcorresponding to 'J"J. T. One and the other responding to Oneofofthese thesecontains contains the point u = + 1 and —1, and their boundaries be denoted denoted by yu' and y,/' the point point uu = 1, boundaries may be If u' transforms it it respectively. If ·u' is is aa point point of of 7u', Yu', the relation u" = =—  u' respectively. u' transforms The relation (66"2) (662) transforms into a point point u" of y,.". 'fhe transforms these these two domains distinct simplyconnected simplyconnected domains in the tplane. One of these, into distinct tplane. One these, say T,*, contains consists of the points lying within a contains the the point point t == 0 and consists certain continuum continuumy,'; it'; the lying outside outside aa certain the other other consists consists of the points lying Since the the domains have no continuum y/'. Since domains have no points points in in common, common, it" y/' continuum 7g".
relation t" surrounds y,'. The relation t'' = another.
f,
transforms y,' and and ye" yc" into one one
IV CHAPTER IV
SCHWARZ'S LEMMA SCHWARZ'S LEMMA &7. Schwarz's Schwarz's Theorem. Theorem. 67.
We the observation that if an We take take as as startingpoint startingpoint the observation that an arbitrary arbitrary closed domain domain T, the maximum maximum analytic function function l(z) 1(z) is regular in aa closed value f(z) I in TTis is attained S of value of ofI1/(z) attained at at one one point point at at least least of of the frontier frontierS T. The Thefamiliar familiar proof proof of of this this theorem theorem depends depends upon upon the the elementary eleme9~ fact that nonconstant function fñnction thatevery everyneighbourhood neighbourhood of of a regular regular point pointz0 Zo of a nonconstant f(z) contains points such that 11(z1) I (z} contains points~ that 1/(z.) I> l/(z0 ) I.I· I> 11(z0) nowaafunction/(z) function f(z) satisfying the following three conditions: conditions: Consider now following three (i) I1(z) (z) is is regular regular in the the circle circle Iz I <<1; 1; (ii) (ii) at atall all points points of of this this circle circle conditions (i) (i) and and (iii) (iii) it follows that 11(z)I<1; /(0) = 0. From conditions 1/(z) <1; (iii) (iii)I(O}=O. follows that "'{z) =/(z)fz ...... (671) (67"1) is also also regular regular in in Iz I<1; < 1 consequently, ; consequently, if if z0 Zo is any point point inside inside the unitcircle, and rr denotes denotesaapositive positivenumber numberlying lyingbetween between IIz0 z0 I and unity, there is is on on the the perimeter perimeter of of the the closed closed circle circle Iz I ~ r at at least least one one point point 4, (z0) z1 thatI 4, I"'(z1) (z.) I ~i II "' (Zo) j.1. But, by by hypothesis, hypothesis, 11(z1) II(~) I <<11 and z1 such that lziII ==r, I~ r, so so that thatwe we may may write write I
I"' (zo) I ~ I "' <~> I = Illz.
hence, since since rr may may be be taken arbitrarily near to unity, hence, unity,
I"' (zo) I~ 1.
. ..... (67"2) (6r2)
4' (z) is not Thus for all all points pointsinside insidethe theunitcircle unitcircle14' I"'(z) (z) II ~ 1; but but ifif"' for all interior points. points. For constant, this (z) II < 11 for this may may be be replaced by 4' I"'(z) if, 4' (z) not being constant, there were an interior point z0 such that that if, "' (z) not being constant, there were an interior point Zo such would necessarily interior points for which ) I"' (z I = there exist other interior points for 1, II' (z,)0 I On use 14' (z)I>j>1,1,and and this contradicts I"' (z) contradicts (67"2). On use of (671) (67"1} there there now now follows: THEOREM ScHWARZ's LEMMA. Ijtke If the analytic analytic/unction/(z) function 1(z) is regular THEOREM 1. SCHWARZ'S If(z)!
1/(z) I< lzl If(z)I
...... (67"3) ._ ... (6v4) (67"4)
40
SCHWABZ'S LEMMA SCHWARZ'S
[CHAP. Iv IV
(67 1) ~1,(0) (0)=/'(0), =f'(O), andourformerreasoningtherefore and our former reasoning therefore gives at By (67"1) at once once the further theorem: theorem : THEOREM 2. f(z) Tasoami 2. If 1/1(z)
satisfie,g tke Lemma, then then the conditions ecnditions qf qf &kwarz's &hwarz's Lemma, <1 orf(z) is of the form (6r4). either either If' If' (0) I< 1 or/(z) is qf form (67"4).
68. 68. Theorem Theorem of of uniqueness uniqueness for the conformal representaof simplyconnected sJ.mplyconnected domains. tion of One of of the simplest One simplest but most important important applications applications of Scpwa.rz's Lemma isis the the following. following. Let Let G be be a simplyconnected domain in the Lemma simplyconnected domain wplane, point w w == 0 in its interior; and wplane, containing, containing, we we will will suppose, suppose, the the point we assume assumethat that a function w w ==f(z) we f (z) exists exists which which represents representS G conformally on the and is is such such that that/(0) =0and and/'(0) real and and the circle circle lzl z <1, and f(O) =0 1 '(0) isisreal positive. We that thefunctionf tkefunctionf(z) (z)satisfying satisfyingthese theseconditions conditions positive. We shall prove that is necessarily unique. For if there 1(z), but there were were aa second second function function g (t), (t), distinct from from /(z), equation satisfying the same conditions, the equation I
...... (681) (68"1) =/(z) g (t) =1(z) would transform the the circle would transform circle It! It I <<11 into the circle circle Izz I<1, < 1,making making the to each each other. other. Hence centres 0, z = 0 correspond correspond to Hence both botk the functions functions centres tt ==0, ...... (682) (68"2) tt=4,(z), = ~ (z), z = yt(t), obtained by solving (681), would obtained solving (68"1), would satisfy all the the conditions conditions of ofSchwarz's Schwarz's z Lemma, and and hence, 67, both both the inequalities z Lemma, hence, by by §§67, inequalities ltiSizl, lziSitl tI It!
would holdfor z andtsatisfying t satisfying(68"2). (682). Thus ItItlI =I hold for all values of zand = Ilzz I = I(z) 4'(r) I= by Schwarz's Schwarz's Lemma, Lemma,¢ of the the form form eifz. But it and, again by 4, (z) would be of followsfrom fromthe theoriginal originalhypotheses hypothesesthat that 4,' (0)> 0, and hence follows ~· (0) > 0, hence 4, cp (z) z; I
=
that that is is to to say say
=g(cp(z))=g(z), f(z) =g(4,(z))=g(z), so so that the function/ function/(z) (z) is is unique.
69. Liouville's 69. Liouville's Theorem. A large number of important important theorems theorems follow follow readily readily on on combining combining Schwarz's Lemma Lemmawith with earlier earlier theorems. theorems. As an example example we may mention Schwarz's the theorem Liouville that aa bounded bounded integral integralfunction function isis necessarily necessarily theorem of Liouville a constant. For ifF if F (z) (z) isisan anintegral integralfunction functionsuch suchthat thatIF(z) F(z)II< .31 M for all values z, and we we write z = = Ru, of z,
F(z)—F(0) =F(z)F(O)
/( ) 1(u)— u
2JL[ 2fr1
'
§§ 68—71] 6871]
41
PICK'S THEOREM THEOREM
the function function /(u) satisfies the conditions conditions of Schwarz's Schwarz's Lemma 1(u) satisfies Lemma for for any positive value valueofofR. positive R. Hence, Hence, if lul Iu I <<1,11(u) 1 , 1/(u) I
2M
70. Invariant 10. Invariant enunciation enunciation of of Schwarz's Schwarz's Lemma. Lemma. be regular and let In the the circle circle I z I <1 < 1 let the the function function to w = 1(z) f(z) be regula.,. and I
I
<1; however,that that f(O) f(O) ==0. lf(z) I< 1; we we no no longer longer suppose, suppose, however, 0. nPnoting by zoanypoint z0 anypoint inside insidethe the unitcircle, unitcircle, transform transform the the unitcircles unitcircles in in thezplane thezplane wplane into and in the thewplane into themselves themselves by by means of the Mobius' transformations (of. (cf. §§ 35) tions t= zzo w= wf(zo) =— f(z)—f(zo) f(z)f(zo) ....... ((701) 70 .1) — 1 — to Z' 1zoz' — 1f(z 1 w — 1f(zo)f(z) 1 —f(z0) 1(z) 0)w w == w The fun.::tion w w (t) obtained by elimination of z from fromequations equations (70"1) (701) satisfies conditions of of Schwarz's Schwarz's Lemma, satisfies all three conditions Lemma, and consequently I
IIw(t)kitI. w (t) I ~ It I·
...... (10·2)
inequality can be expressed expressed by bythe thestatement statement that that the nonEuclidean This inequality (t) is is not greater distance .D (0, w (t)) of of the (0) = 0 and w distanceD w(t)) the two two points pointsww(O)=O w(t) than the the nonEuclidean nonEuclidean distance distance of the poil').ts 0, t, so that, remembering that, remembering that, as as was was proved in Chapter II, ii, nonEuclidean nonEuclidean distances distancesare are invariant invariant we obtain obtain the for the transformations transformations (70"1), we the following following theorem, theorem, which which includes includes Schwarz's Schwarz's Lemma Lemmaasasaa special specialcase caseand andwhich whichwas wasfirst first stated stated by G. G. Pi~kCl4l. PickU4. by THEOREM 3. 3. LetLet f (z) TaEOREM 1(z)bebeanananalytic analyticfunction functionwhich whichisis regular regular and denote any two such that lf(z) 11(z)I< <11 in <1, such that in the the circle circle Izt Iz I< 1, and let let z1, z~> Z 2 denote two
points p<Jints inside the unitcircle; then then either either D (f(zl), D (z,, (z1,z2) z2) ...... (70"3) (1(z1), f(zs)) 1(z2)) <
(702) with 71. Comparing Comparing (70·2) with (701) (70·1)we we obtain obtain
1/(z~ f(z
I
0)
—f(zo)f(z) ItII .I (zo)f(z)
I:£ IIIi_[ z_~z·:"'""'I
or
1./Jz~ =~(zo)
~z—zo~o
I:£< I~.((z~{(z) I; I —z(,z 1~u
42
(CHAP. [cHAP. IV Iv
SCHWARZ'S SCHWARZ'S LEMMA
and if if we we let let zz converge converge towards towards z0 Zo this gives gives
1 / ,(zo)I~" 1l/(zo) 1 If'(zo)I 1lzol2 2
Thus, since z0 is arbitrary, arbitrary, we Zo is we now have the theorem:
THEOREM 4. 1// tke conditions of Theorem TIIRhrem 3, 3, then either 1ff (z) satisfies the
If' (z) I< 1 ~ ~~j~ 12
(7r1) 000 0.. (71°1)
for /or all points z inside inside the tke unitcircle, unitcircle, or
1 1/' (z)I= 1l/(z) 1lzl2 2
(7r'2) •o•···(71°2)
for all if /or tke unitcircle; further (71°2) lwlds if ifand and only only if all points z inside the further (71 2) holds (704) (70°4)holds. koldso From (71°1) (7r1) and (7r2) that, if Iz I ~ r < 1, From (71°2)ititfollows follows at once once that,
I/' (z) I~ 1 ~r;
(713) oooooo(71"3)
hence, hence, denoting by z, z1 and z, z2 any any two two points points of of the theclosed closed circle circle jIz I ~ r, integrating along along the the straight straightline linefrom fromz1 ~ to to z2, .z,, the relation relation and integrating
L"' Jt' l'
(z) dz =f(z2) (z) dz=/(z2}
/(~)
gives at once, gives once, by the the Mean Mean Value Va.lue theorem, theorem, ThEoaui 50 .lfthefunctionf(z) satisfies 5. lftke/unctionf(z) THEoREM satisfiesthe tkeconditions conditions of ofTheorem Theorem 3, and if~, .z, z2are are any any two points of tke the closed closedcircle circleIzI two points z I ~ r< r < 1, 1, then then
I~~l /(~)/(z2) ~z2 17 1
.
o.. o(71'4) (7r4)
of equations equations(70'1) (701) and and (70'2) (702) we take the 72. As a last la.Rt application application of we take following: if we we solve solvethe the second secondofofequations equations(70'1) (701)for forf folloWing: if I (z) and take Zo =0, = 0, we we obtain obtain w+f(O) ; /(z)= w+!(O) . 0. 0(72'1) (721) 1+/(0)w 1+f (0) but, for Zo==O, the firr,t of of equations equations(70'1) (70i) becomes t=z, so becomes t=z, so that, by by 00
(702), jw (70'2), Iwl ~ Ilzl. {72°1)is is a Möbius' Mobius' transformation transformation which which transforms z Also (721) the circle circle, and and by con.::ircle Iw(U I ~ Iz I into another easily constructed circle, configure so difficulty at the sideration of sideration of the figure so obtained obtained we we arrive arrive without difficulty at the theorem: I
72—74] §§7274]
POSITIVE PosmvE REAL PARTS
43
THEOREM 6. 6. 111(z) Iff (z) is regular tile circle IzI Iz I <1, <1, then tAen regular and and 1/(.z) 11(z)I< <11 in the
1/(z) 11(z) I ~ Iz I + II (o) I 1 + 1/(0) liz I 1÷ifonki I
......(722) (72 .2>
at all at allpoints points zz inside inside the tile unitcircle. unitcircle.
Finally we deduce from from (72"2), (722), by by an an easy easy manipulation, manipulation, 1—11(z)> 1—11(0)! >1—11(0)1 1lf
lf
Theorem 33 of of§§70 may be regarded regarded as a.s aa. special special case case of aa. more more 73. Theorem =f(z) is general theorem which we will now now prove. If the the function function w w=/(z) is regular regularfor for IIz I <1, < 1, every every curve Yz in the the zplane zpla.ne which which lies entirely inside the unitcircle unitcircle corresponds corresponds to a curve curve y., in the the wplane, wpla.ne, which which will will also the unitcircle provided 1/(z) I<1 < 1for forthe thevalues values lie entirely inside inside the the unitcircle provided that that 11(z) rectifiable (in (in the ordinary sense) it considered. If the curve curve Yz sense) it of z considered. If the 7z is rectifiable defined a.s as the the upper has, by §43, §43,a.a nonEuclidean nonEuclidean length length LL (y..) which is defined ha.s, of the thenonEuclidean nonEuclidean lengths lengths of ofcertain certaininscribed inscribed curvilinear curvilinear limit of polygons; inscribed in polygons; to any anynonEuclidean nonEuclidean polygon polygon P., inscribed in 7w y.. there 70, the corresponds a curvilinear inscribed in y., and, by curvilinear polygon polygon P .. inscribed by §§70, between two two consecutive consecutive vertices vertices of F,, P ID is is never never nonEuclidean distance between greater than than that thatbetween betweenthe thetwo twocorresponding corresponding vertices vertices of P.. From follows at Yw is also rectifiable and and has ha.s aa nonnonthis itit follows at once once that that the curve y,, Eucidean Euclidean length lengthLL (y,,) (yiD)which which cannot cannot exceed exceed L L (y,). (y..). If now now the two two curves curves have have the thesame samenonEuclidean nonEuclidean length, length, correcorreelementaryarcs arcs of of7z y.. and y., must also be sponding elementary be equal, equal, so sothat that (71"2) (712) must be satisfied satisfied at all points of y,. y •. Thus, Thus, on onusing usingTheorem Theorem 44 of of §71, we obtain obtain ~he the required required theorem theorem:: we and (y..,) denote denote the nonEuclidean lengths lengths of of corresponding corresponding arcs y.. , 7w 'Yw drawn in in the the unitcircle, then tAen either L (y.,)
Pick's THEOREM. thefunction functicnw=f(z) w =1(z) isis regular regular PICK's THEOREM. IfIftile
If(z)l<1 1/(z) I< 1ininthe thecircle circle IzI<1, I z I< 1, and and if L (y.. ), L
with positive real real parts. Functions with 74:. 74. ~ctions supposethat/(z) thatf(z) is regular, that lV(z) >>00 and In the circle circle IzlzI<<1 1 suppose regular, that also Mobius' transformation also supposef(O) suppose! (0) = 1. By the Möbius' u—i 11+w +w u1 U=to=1to' U+l' 1—w u+1
u=—, w=—,
......(741) (74"1)
44
SCHWARZ'S LEMMA SCHWARZ'S
(cuAp. IV [CHAP. IV
is transformed into the uni ucircle Iw I < 1, and the halfplane halfplane lltu > 00 is transformed into u = 1, w =0 = 0 are are corresponding corresponding points. Schwarz's Schwarz's Lemma can therefore therefore be be applied applied to to the function f(z) 1
w=f(z)+1'
...... (74'2)
Consequently, the uplane, upla.ne,which which corresponds, corresponds, by by means means Consequently, the the figure figure in the =f(z),
75. Harnack's 75. Harnack's Theorem. Theorem. It will now theorem of fundamental It will nowbe be shown shownthat that aa theorem of Harnack, the fundamental which has been long recognised, conrecognised, is an almost almost obvious conimportance of which sequence of of our our last result, so that thatHarnack's Harnack'sTheorem Theorem may may without without sequence result, so Schwarz's Lemma. loss loss be replaced by Schwarz's (13, Harnack's Theorem Theoremmay maybebestated stated as as follows: follows: If If FL, Harnack's U17 (13, U2 , U 3 , ... increasing sequence of harmonic harmonic functions functions in the a monotone monotooe increasing seq'IU3nC6 of tke circle toaajinite limit at atz=0, IIzI< z I < 1, and if if U,. U.,. tends tends to finite limit z = 0,then tken U,. tends to a limit uniformly in the circle I zz I
denote
We write u,.=
U,.,
f,. (z) = u,. + i?J,.,
...... (751) (75'1)
...... (752) (75'2)
v,. isisthe the harmonic function conjugate the v,. conjugate to toU,. u,. and vanishing vanishing at at the U,. ) hypothesis, tV.. (z) == u,. ~ 0, 0, and andconsequently consequently origin, so that, by hypothesis, The functions f. (z) (z) is identically zero. The functions u,. (0) =f,. (0) > 0 unless I,. (Z)/Un (0) therefore satisfy the conditions j,.(z)/u,.(O) conditions of § §74, 74, so sothat, that, by (74'3), (743), in where where
origin,
+r 1 +t'
If. (z)l
1 +r 1+r
Un u,.
and Harnack's Harnack'~ Theorem Theorem follows follows at once from this.
§§7578] 7578]
4o
BOUNDED REAL PARTS PARTS
45
Functions with bounded real real parts. 76. Functions Suppose the the function f(z) toto satisfy Suppose function f(z) satisfythe thefollowing following conditions: conditions: (i)/(0) = 0; (ii)f(z) (ii)/(z)isisregular regularfor for IIz I< 1; (iii)there (iii)there is is a constant/i constant k such such (i)f(O) = that
I Jit/(:&) I
...... (76"1) (761)
2/i 2k 11+i(O + iw w =, log 1 .. w=—log . .
. ..... (76"2) (762)
iflzi
'(IJ
1—sw 
l11"
as a function These relations determine determine wwas function ww (z) of z, z, and and w w (z) satisfies the conditions Lemma. From From this this fact and the Schwarz's Lemma. the relation relation conditions of Schwarz's (52·3) for follows that for IIzzI <
2/t
1 +r
If/(z) I~ log ~ 1 ~. 11" r 77.
. ..... (76•3) (763)
Surfaces with with algebraic Surta.ces algebraic and and logarithmic logarithmic branchbranch· points.
(z) and Y, (z), The functions tf>,. (z) means of which, which, in §§56 and §§57, (z), by means 56 and 57, the unitcircle unitcircle was was represented represented on on the the circular circularareas areashaving having algebraic algebraic or logarithmic branchpoints, branchpoints, satisfy satisfy all conditions of Schwarz's Lemma; logarithmic Schwarz's Lemma; therefore they they must satisfy the inequalities therefore inequalities ltf>,.(z)l
4>2' (0) =
:
ft
< 1.
78. AAfurther furtherproperty propertyofofthe thefunction function used used in in§ § 57, 57, Y,(u) =
kii —
I l+ulogll log elu
I+u
,
(h<1), (k < 1),
...... (78"1) (781)
1keJulogll
follows. Consideration Consideration of of the the Möbius' may be obtained as follows. Mobius' transformation employed shows shows that, that, ififIIuI < rr < <1, u I< 1, ) 1r 1+r logk
u
and hence, in particular
l+r
I e 1_,.log Ill >8 1::r log all =p. =p. l+u
l+r
. ..... (78•3)
Consequently, by (78"1), the values (u) for for these values <Jonsequently, values assumed assumed by Y,(u) values
ScHwARz's SCHWARZ'S LEMMA
46
xv (CHAP. IV
of uu must must lie lieoutside outside a nonEuclidean circle with nonEucidean nonEuclidean of nonEucidean circle centre k, once obtain, remembering remembering that p < 1, the h, and and from from this this we we at at once inequality k+p ll/t(u)kl> 1 +kp k>(1k)p, 1+r log hh I+r
I.e. i.e.
ll/t(u)kl>(1k)e 1 r
•.
...... (784) (78"4)
79. Consider Consider now now a function function w=/(z) having the three three properties properties (i), w =f(z) having (iii) of§ of §67, 67,and andhaving havingalso alsothe thefurther furtherproperty propertythat that there is a (ii), and (iii) h has no solution real number number k, h, (0 (0<
I
*
I+r log 'I h
1/(z)kl> (1k)e 1 r
.
. ..... (792) (79"2)
Representation of ofsimple simple domains. 80. Representation Let the function function
...... (80·1) w=f(z) (801) w=/(z) represent the simple simple domain domain T, T, which which was was discussed discussed in in §§65, 65, on on the the If in (65·1) (651) interior of of the the unitcircle unitcircleIz I< <1,1,and (0) == 0. If and suppose suppose that! that/(0) w, and and if we we denote denote (1 (1 ++ J2).J2)22 by we replace zz by by—  w, by h, k, §§ 65 and 66 show that that the the function function ...... (80"2) (802) w=
}
w= (kw'f =/(z), (h — ...... (80"3) (803) k= (1 + J2)2, effects effectsaa conformal conformalrepresentati::.:. representatio. of of the the domain domain T' I" COD$idered at the the 66 which which lies lies inside inside the the unitcircle II cu end of § 66 "' I <<11 and does not contain the point = ii. Since 4) point w=k. .P(O)=O it follows follows from Schwarz's Lemma from Schwarz's Lemma that that (0) = 0 it also,by I "'I ~ r ifif Iz I ~ r, and also, by the the Theorem Theorem of § 79, that
f
1+r l+rl
lwkl>(1k)e 1 r
og
h
...... (804) (80"4}
§§ 7981] 79—81]
47
SIMPLE DOMAINS
from the thesecond secondof ofequations equations(803), (80·3), But, from h 1 — = 2 .Jh, 1k=2 Jk, hence, from from the first first equation equation of of (803), (80•3), using using the the inequalities inequalities and hence, I "' I ~,.r and and (804), (8o·4), if if II z I~ r, lf(z) I< re
1+r
2 !±!log h.
...... (80"5) (805)
1T
<21'... sothat Observing that 11++J2 = 2·41 ... < 2·7 ... <<e, e, so that Observingthat 12=241... .J2)<4, —2logh=4log(l12log k=4log(1 + J2) < 4, we we see see that (805) (80·5) includes the inequality inequality
''.
14r 41~ IT.
lf(z)l
. ..... (8o·s) (806)
we have have proved provedthat, that, in in the Thus we the conformal confonnal representation representation (801), (80"1), the domain corresponding correspondingtotothe thecircle circleIzz I ~ r <1 < 1 isisbounded boundedand afld lies lies wholly wholly 1+T . 441+r 11—t ".
within the the fixed fixedcircle circleIw IwlI~ re
On the other hand the the domain domain TT contains, contains, by byhypothesis hypothesis (see (see§§65), 65), <11 in the circle circle I w I < in it8 itsinterior, interior,and andhence, hence,by bySchwarz's Schwarz's Lemma Lemma applied to the function z=cfl(w) inverse tof(z), tof(z),the thefigure figurecorresponding corresponding to mustcontain containthe thecircle circleII w in its its interior. interior. IfIfwe wenow now apply to Izz I ~ r must wII ~ r in an arbitrary general theorem theorem:: arbitrary magnification, magnification, we obtain the following following general I
THEOREM. Let T be be a simple simple demain domain of qfthe the wplane wplane containing w in w ==00 in its interior let a be tke the dista'IUJe distance qf of tke the point paint interior but but not not containing containing w w = oo ; let
w=O from the tke frontier frontier of qf T, T, arid andletlet1(z) f(z)denote denote aafunction function which wil,ick w = 0 from represents conformallyonontke thecircle circlelzl<1. w=( Iz I< 1, making malcing w = (1correspond correspond represents TT confqrmo)Jy toz=O. to z = 0. Then,for Tken, forany anyvalue value ofzinside qf z insidethe tke circle circle Iz I < 1. \I lf(z)l 41+1•1 11z1 a~
p lZI
formulae already already established establishedcan canalso alsobe beused usedtoto obtain obtain limits for The fonnulae for —f(z,)}/(z, —z2) functionf(z). For /(Zt)}/(~ Zt) ofof thethe functionf(z). the difference difference quotient {/(zt)41~
example, by by {80•7), f(ru)fare (807),the thefunction function f(ru)/areIr'' has modulus modulus less less than Hence by by Theorem Theorem 55 of of § 71 71 inside inside the unity inside the the circle circle Iu I<< 1. Hence i.e. inside the circle I1u 1< r, i.e. the circle circle 1 < r, I zz f1< I
I+r 4 1+r
l
f(z1) f(ru,) —f(ru2) s ae 1  .. f (zt) —f(z2) f(Zt) j = jf(t"Ut)f(rus)l r(u1—u2) ZtZt r(UxU,)  1r'
IziI
l21l
in (80•7) (807)we wesee seethat that the the function functionf1(z) Byletting letting IIz I tend to zero zero in (z) 81. By of § 80 satisfies the as'. the inequalities inequalities aa~ lf{O) of§ ae'. ItItisisknown knownby by Schwarz's Schwarz's If(0) I~ I
48
rv (CHAP. IV
SCHWARZ'S LEMMA
boundfor forI/' If' (0) I; we we shall shall now Lemma that a is is the the true lower lower bound now detertrue upper upperbound boundfor forthis thissame same number, number, using using aa very very ingenious mine the true Erhard&hmidt. &kmidt.ItItis ispublished publishedhere herefor forthe thefirst firsttime time(l5). method due to Erliard (15). which follow followform formthe thebasis basisof ofthe the proof: proof: The considerations which (a) Suppose Suppose that that the thefunction function} that itit is is J (z) is not a constant and that regular within within and and on on the the frontier frontierof ofaasimplyconnected simpiyconnecteddomain domain1), D, whose whose frontier is aa regular regular curve curvey.y. Then, Then, ifj(z) if 1(z) = ua + iv, where ua and v are real,
i
uclv>O.
. ..... (81"1)
For, if account For, account is is taken takenofofthe theCauchyRiemann CauchyRiemann equations, equations, Green's Green's Theorem Theorem shows shows that that
i
uclv= fJD (u,2 +u.,/)dxdy
can be (b) The inequality inequality (81 (81"1) can be extended extended to to multiplyconnected multiplyconnected (b) The domains providedf(z) is domains provided/(z) is regular regular and and singlevalued singlevalued in the domain domain conconsidered. Let y'y' and and y" y" be be two two closed closed curves y' lies lies within within y" y" sidered. Let curves such such that that y' itself surrounds z = 0. Then Then the thefunction function log logzz isis regular regular in inthe theananand itself nular domain domain between between y' and j'y"but butititisisnot notsinglevalued. singlevalued. If however nular iv', then log p and dtfr are singlevalued z= pd1{1 and log log z = = log pp + itfr, singlevalued funcfunc= pew' tions in the annular hence annular region, region, and hence
£,logpdlft~ fr,togpdtfr.
. .. 000(81°2)
(c) If curves7' andy" andj' are (c) Ifthe curvesy' are such such that that the relationz"= rela.tionz" =!,transforms transforms z
y' y", it is easy to see see that y' into y",
f logpdljl+llogpdtfr=O. J.,.. y
•ooooo(81"3) (8r3)
From (81 2) and and (81"3) (81 3) itit is seen From (81"2) seen that in this this special special case
f log p d.p ~ Oo
}.,·
... 00o(81°4)
where 4q, (z) is F(z) = zq, (z), (z), where is regular regularand anddiffers differsfrom from zero zero (d) Let .F(z) throughout the theclosed closed circle circle IIz I ~ 1. We write throughout (0) ff/tt11l. F(e") == p (6) . .. ooo(81"5) (815) The hypothesis 4 (z) (z) isis regular regular for for Izz I ~ 1, hypothesis shows shows that the the function function log log q, 1, and, if 1z i =1, then log q, (e111) =log (F(el') r 111) =log p + i (tfr 6). By and,
(8r1) (81"1)
£Iogpd(tfr6)~0, jlogpd&i_9)>0,
where x" is the circle IIzI= z I= 1. 1.
(816) .... 0.(81"6)
______
§ 82] 82)
KOEBE'S CONSTANT CONSTANT
49 4:9
On by the the Mean Mean Value Value Theorem, Theorem, On the other hand, by
r2· logpd8,
1 logicf»(O)I=I0ogcf»(0)= 27T}o
(811) ...... (81"7)
(0) = F'(O). F' (0 ). Comparison Comparison of(8P6) of ( 81 ·s)and and (811) (81"7)shows shows that and, and, further, further, cf»4(0)=
F'(O) log II F"(O)
I~ 2:.ilogpd.r. I
. ..... (8r8) (81"8)
82. Let 1(z) Let/ (z)be bethe thefunction function considered considered at the end of of§§ 80. 80. If If rr << 11 we we write
1 g(z)=—f(rz) g (z) ==/(rz) 1
...... (82"1) (821)
a
and
F(z)= 1  JIg(z) 1 + J1g(z)
_
_fliz_~~ ....... (82"2) (822)
{1 + .11 .J1g(z)l2
By (66 3),thefunctionF(z) the function F(z) transforms transforms lzl z ~ 1 conformallyintoasimple conformally into a simple By(66'3), domain T2* T.• which whichlies lieswithin within the the domain T9 of§ of § 66. Thus F(z) does not
z
vanish in the the unitcircle. unitcircle. If therefore
F F (eill) === pe'"', the inequality inequality(81"8) (8r8) holds. holds. By By (822) (82"2) and (82.1) (82·1)
ff'(o). 4a
F' (0) = g' (O) = .!___I' (0). 4
Thus . log
Irj~~O) I~ ;11" flog
flog p p. i/i' ilog p. "'' (0) (8) dO dO= i,log = 2~ —
a.r. ...... (82"3)
Reference now shows showsthat that y' y' has has all all the the properties properties Reference to the last last part part of of§§ 66 now (b) and and (c) (c) in in§ so that that (81·4) (81 4)holds. holds. From assumed for for this curve in (b) § 81, so and (82'3) (823) it follows that If If' (0) 4a/r, and, since this and follows that since this holds holds for (0)1I ~ 4afr, all rr< <1, 1, . ..... (82"4) (82'4) I/'
4z w=f(z)=aw =1(z) = a + z)2 z)• (1 + transforms the thecircle circle II z I <<11 into a simple simple domain domain whose whosefrontier frontierisisat at aa transforms distance a from from w ==0. 0.
50
iv (CHAP. IV
SCHWARZ'S SCHW .ARZ'S LEMMA
Representation upon anotherof containing 83. Representation upon one another of domains containlng circular cfrcular areas. Let R,,, R., and simplyconnected domains domains which lie entirely entirely and R,. be two simplyconnected
,._1
In the the first firstplace, place,by bySchwarz's Schwarz's Lemmas Lemma,
l'w(z)l~! z . It
zl
at all all points points of of B,,. R,.. z
i; 1\
...... (83'3)
A
IIw—zI=I—1 w  z I = Iz i zl ~ kr
I; 1,. w z
.
.. .... (83'4) (834)
bound for for this this expression expression we we now now write To obtain an upper bound It
z = At, Z== t,
w (z) = = FF (t), F(t) = 4, (t), (t) (t), }/(0)==.,. z '
...... (83·a)
so that, by by (83'3), 1k
IF(0}11 ~ r• ...... (83'6)
§§ 8385]
51
RIGIDITY OF RIGIDITY OF TRE THETRANSFORMATION TRANSFORMATION
the last last inequality inequality holding holding since since F(O) F(O) = w' (0) was supposed supposed real and (0) was positive. positive. With With this this notation notation
I~ 11==·1F(t)11 ~ IF(t)F(O)I + IF(O)II' and hence, by (83·6), l
1k z I I~ 14> (t) II kI + T·
w
1
...... (83"7)
84. If a is an arbitrary arbitrarycomplex complex numnumber, represented (say) P, ber, represented (say)by by the the point point F, equal to to the the number aI a—  1I I, being being equal length of the segment segment UP, UP, is is not greater than the length length UHF, UMP, where where MP is is an arc of aa circle circle with withcentre centre0. 0. We We theretherefore have fore have the inequality 4 Iall IIa—i a  J I ~ JI IIaI a I  II + a II IEflog log aa.i . ...... (84i) (84"I)
Fig. 22 22
I
86. By (841), (84"1), 85. By —1 14> (t) 11 ~ 114> (t) III+ 14> (t) IIjf lE Iog4> (t) I· 141(t)—il
...... (851) (85"1) the branch %fr (t) =log = log 4> 41(t), But since, by (83"5}, 4(0) 4> (0} = 1, the branchofofthe thefunction function 1/1 (t) (t), satisfies, in in consequence consequenceofof(83"6), (836),the theinequality inequality which vanishes fort== for t = 0, satisfies, lllil/l (t) I~ 2log k; 76, if if It I ~ r, and therefore, therefore, by §§ 76,
<_4logh10gi_+r jflog41(t)j lllog4> (t)l ~  4 ~ogklog ~ ~~:
Also, by by (83"6}, (836), Also, 1 141(t)I 14> (t) I ~ Ji2'
114> II (t) 1 II~
...... (852) (85"2)
i—hk22 I
Ji2 .
. ..... (853) (85·3)
It from(85"I), (85i), (85"2) and and (85"3) (853) that It now now follows follows from l4> (t) _ II ~ I1—h2  k 2 _ 4logh 4 log It log 1+r I +r k2 7rk2 I r' (834) and and on using (83"4) and (831) (83"7)we we obtain obtain finally finally jw — z I ~m(k, m (ii, r), r), lwzl
(1—h3)'r 4r h3 ) r _ 4rlogh1 4rlop: k 1og .i!:_ (k ) =(Im 'h ', r) r — k" "Irk• og 1  r · —
l
•
0
••••
(85"4)
From this this e::pression expressionitit isis seen seen that that not From not only only isis urn lim m m (h, (k, r) ==0 0 for for h.1 h—i
every fixed fixed r, as was was to be be proved, proved, but but also also urn lim m m (h, (k, h) h)==0. 0. h1
BCHWARZ'S iwMMA SCHWARZ'S LEMMA
52
Iv [CHAP. IV
Bylonger longer calculations calculations than than the theabove above itit can can be be shown shown that Remark. By that function m m (h, (85·4) can can be be replaced replaced by aa much much smaller smaller the function (h,r)r) of of (854) function.
86. Problem. Problem. 86. wake use use of the results of The reader may now now make of§§ 82—85 8285 to establish the following following theorem. theorem. · the the tiplane Uplane which two simplyconnected domains of the Let Rv. and S,. be two in their their interiors interiors and and which which are are represented represented on on contain the point u ==00 in the interior interior of of the the unitcircle unitcircle IIz I <1 < 1 by by the functions uu=f(z), (f(O)0, the =1(z), (1(0) —0, (0)>> 0), respectively. respectively. It f' (0) (0) > 0), and u = = g (z), (g (0) == 0, g' (0) Itis is supposed 1' =1(z) and u = that the the functions functions u =f(z) = g (z) represent the circle circle Iz I <
less than !log log less
commonto to both both the the domains. domains. Then it is to be !1+h ~! are common are
is
hr < ii, there is (h, r), proved if Ilzl is aa relation relation 11(z) 1/(z)g(z)l proved that, if z ~hr
relation? relation7
87. Extensions 87. Extensions of of Schwarz's Lemma. again denote denote aa function function which whichisis regular regularand and such such that that Let w ==1(z) /(z) again all points of the circle z <1. Together with any nonIf (z) I < 1 at all points of the circle I z I < I. Together with any non11(z) I <1 Euclidean circle O(z) C(z) with with nonEuclidean nonEuclidean centre centre zz and andnonEuclidean nonEuclidean Euclidean circle we consider consider the the circle r (z) circle I' (z) in in the thewplane wplane whose whose nonnonradius radius pp (z) we If A.., denotes w =1(z) =f(z) and and pp (z). If Euclidean cent.e and radius radius are are w denotes the points belonging belonging to to an an arbitrary arbitrary set of such set of all points such circles circles C(z) 0 (z) (and their interiors), and Aw that that covered covered by by the the set setofofcorresponding corresponding circles circles P (z) Theorem 3 of§ of 70 that any T (z) in in the thewplane, wplane, ititfollows follows from Theorem any interior point of A2 by the function =1(z) iLto point A. is transfbrmed transformed by function w w =f(z) it: to an interior interior point of Aw. If the set (z) isis enumerable enumerablewe wecan can also also consider consider the the set If set of of circles circles C 0 (z) set of points B3 consisting of all all points which B. consisting which lie in all but but aa finite finite number number of the circles (s), and compare 0 (z), compare it with the set set of of points points B.., obtained circles C in the the same same way way from from the the corresponding corresponding circles circles rI'(z) (z) in in the thewplarte. wplane. It fromthe the theorem theoremquoted quoted that that w =f(z) =1(z) transforms It follows follows from transforms any point of B. B3 into into a point of B'ID. Various applications of of the above can be made, Various applications above considerations considerations can made, and and those applications are particularly particularly interesting which which throw throw light light on on the the applications are behaviour of f(z) in the = I.1. behaviour of/(::) the neighbourhood neighbourhood of the frontier Iz z I=
§§ 8689] 86—89)
JULIA'S THEOREM THEOREM
53
88. Suppose first first that the the centre centre zz of of the thecircle circle C(z) 0 (z) describes describes that 88. Suppose diameter of the circle diameter circle 1IzzI <1 < 1 which which lies lies real axis, and that that all all the thecircles circles along the real M (z) have have the same nonEuclidean radius. 0C (z) In this this case case the the domain domain A., consisting of all points interior to any any of of these these circles circles AA B O(z), C(z), is the area area bounded bounded by by two two circular and is is symmetrical with respect to the arcs, and the real axis (see (see §48). § i8). If then then f(z) is a function function which which is is. Fig. 23 regular < 1 if if II z I < I, Fig. 23 regular and andsuch suchthat thatlf(z) f(z)(I <1 1, and if if also also 1(z) f(z) isis real realwhenever wltenever zz is ts real, real, the the circles circles rr(z) (z) in and the wplane to these circles wplane corresponding corresponding to circles C 0 (z) will will have their their nonnonEuclidean centres axis, and, and, since since their theirnonEuclidean nonEuclidean Euclidean centres also also on on the the real axis, radii are the same radii of the circles C(z), same as the nonEuclidean nonEuclidean radii circles C (z), must lie inside precisely the same same area A., but in in the wplane. u·plane. We We therefore therefore theorem:: have the following following theorem THEOREM. Let 1(z) be is regular Let f(z) be aa function function which whick is regular and such sud that also that that f1(z) when zz is real. DeIf (z) I
<1I are :r, (:r <
(1—x)(l4li)(1—y)(1+k) (1:r)(1 +h) (1y) (1 +k). (1+:r)(1k) (1÷x)(1—/t) (1+y)(lk)' (1+y)(1—k)'
. ..... (89"1) 891
or, writing writing 11—h=u, It= u, 1—k=v, 1  k = v, if
1—yv(2—u)l—x 1y v(2u)1:r I1+y +y = u(2v) I +:r" u(2—v)14x
...... (89"2)
Consider now two sequences of positive real u,, real numbers numbersU1, ul> u u3, ... and 2, u 3 , ••• v1, v2, vh t'2o v3 t•3 , ••.. satisfying the following following conditions: . v,. vn IliD hm—=cc, =CX, lim u,.=lim v,.=O, ...... (89"3) n~oo
U,.
where ex is Denote by by K,1' K,.' the  u,. as is finite. finite. Denote the circle circle having having the the point It,.= = 11 —
54
SCHWABZ'S SCHW .ARZ'S LEMMA
[CRAP. IV IV (CHAP.
nonEuclidean centre and the and by rF,' the point point zz on on its itscircumference, circumference, and ,.' a 1 — v,. circle circle of the the same same nonEuclidean nonEuclidean radius radius and and having having the the pointk,. point k, = 1V,1
as nonEuclidean centre. centre. circle rF,' ,.' cuts the the real real axis axis of of the thewplane pointy,. which is The circle wplane at a point y, which obtained obtained by replacing replacing u, v, y in (89"2) by u,., circles K,,' K,.' u,,, v,., v,, y,.. The circles
Fig. 24
converge,in inconsequence consequenceofof(89"3), (893), to toan an oricycle oricycleKK of the nonEuclidean converge,
plane plane Izz <1, I< 1,passing passingthrough throughthe thepoints points zz == 11 and zz = z, and the circles circles rF,,' ,.' to an oricycle rF of of the the nonEuclidean plane II w 1 <<1, 1, passing passing through y,,, is given by the equation = 11 and ww == y; here y, being the limit of of y,., w= (see [see (89"2) and (89"3)] 1y 1z =ot. ...... (894) (89"4) l+y 1+z 1+a 1+y If If rr and and pp denote denote the the Euclidean Euclidean radii radii of of K K and andF,r,then thenz z=1— = 12r, 2r, y= by (89 (89"4), 1 — 2p, so so that, by y = 12p, p=
otr 1r(1ot)"
,;c;
...... (89"5)
90. Now (z) bebeanananalytic Nowlet! let/(z) analyticfunction functionwhich whichisisregular regularand and of of modulus less than than unity in the less the circle circle Iz I<1, < 1,and andsuppose suppose further further that there is ... such that a sequence sequence of ofnumbers numbersz1, Zt,Z2, z 2 , z3 , ••• (901) limJ(z,)=i, lim z,. = 1, lim/(z,.) = 1, ...... (90"1) fi ....OO
ft.~CIO
and
1 1l/(z,.) I lim 1m I I ==ot,
I.
I
,._.., 1
Zn
...... (90"2)
where ot a is finite. finite. It It follows follows from from (72"3) that
1l/(o) I
ot~ 1 + 1/(0)1 >O.
...... (903) (90"3)
Write now now u,. = 11 I v,. = 11 — If (z,.) I, and construct construct the circles circles Ks', K,.', — z,. I,, v,, Kand rF,,' ,.' of of §§ 89. Still using K and I'r to todenote denotethe theoricycles oricycles to to which which K,,' K,.' and FR converge, converge,let let Kn K,, and and r,. F,, denote denote circles circles with with the same nonEuclidean 1',.'
but having centres at z,, radii as K,,' Kn' and F,,', r,;, but having their their nonEuclidean nonEuclidean centres z,. and z,, and f(z,,) respectively, instead instead of of at at lznl I. Thus K,. f(z,.) respectively, 1/(z,.)l. obtained K,, is obtained
§§ 9092]
THEOREM JULIA'S THEOREM
55
from but since, since, by (90'1), from K,.' K,.' by by aa rotation rotation about about the the origin; origin; but (901), the angle infinity, the circles circles K,. K,. tend to of rotation rotation tends tends to to zero zeroas as nn tends tends to infinity, the same same limiting circles K,.'. Similarly the the circles circles F,. 1',. limitingfigure figureKK as as the circles tend to F. 1'. =1(z) cannot Thus, by by §§87, point zz lies lies inside inside K, the the point point w w=f(z) cannot 87, if the point lie outside 1'; we can of lie outside F; we can in in fact fact show, show,by byan anargument argumentsimilar similarto to that that of §§ 67, thatf(z) 1'. that 1(z) must lie inside I'. The above result constitutes Julia's Julia'sTheorem Theorem• * (16).
91. The work work of §§ 90 may may be be completed completedby byadding addingthat that ifif there there is aa point z1 of the frontier which is transformed of point z1 frontier of of K K which transformed into into aa peiatf(z point f(z,) 1) of 48, the frontier frontier of of F, 1', then then 1(z) f(z)must mustbe beaabilinear bilinearfunction. function. For, by §§48, touch the circle any two two oricycles oricycles which which touch circle Iz I== 11 at zz = 11 are parallel parallel curves thesense sense of ofnonEuclidean nonEuclidean geometry; geometry; also also equation equation (89'4), curves in the which nonEuclidean motion, motion, transforms the two two oricycles oricycles which represents a nonEuclidean into the the two oricycles oricycles through '!h through the the points pointsz1 .x1and and x2 .:v2 (v1 (.r1 <
92. Julia's Julia'sTheorem Theoremmay maybebeinterpreted interpretedgeometrically geometricallyas asfollows: follows: in m Fig. 25 we see see that
AFAr AP A.x 1z 1.:v
and similarly that
PB.xE .xE= 11 +.x' FB +z' A1P1 ~ 1y P1J31 1l+y PlBI +y"
Thus, 1(z) respectively, Thus, if if F, P,P1 P 1denote denotethe thepoints pointss,z,f(z) respectively, Julia's Julia's Theorem 'rheorem gives, gives, on using (89'4), the relation A,P1 A1P1~ AP AP ...... (92'1) (921) PJJ1 "'(J.PB· his own proof of of this this theorem theorem Julia. Julia requires requirestha.t that f(z) * In ,his own proof f(z) should be regular regula.r at z= = 1. 1. It is is remarkable rema.rka.ble that tha.t ititshould should be bepossible possible to to prove prove the the theorem theorem without without = 1. regula.ri ty at a.t zz = assuming regularity E
06 56
SCHWARZ'S LEMMA
(CHAP. Iv IV
But by elementary geometry
AP PB
=
f1_z122 AP2 AP AP l1zl AP.PJJ= CP.PD=llzl 2 '
...... (922) (92"2)
so that that Julia's so Julia's Theorem Theorem is is equivalent equivalent to to the thefollowing following inequality: ...... (92"3) ( 923 )
E
E1
Fig. 25 25
93. AAnumber numberofofinteresting interestingconsequences consequencesfollow follow from from Julia's Theorem. Julia's Theorem. continue to to denote, denote, as as in in (89"5), (895), the Euclidean radii of If we we continue of the circles circles by r,. and and p, p, then, by the theorem, theorem, the real real point point K and Fr by
x=1—2r 12'1"
...... (93i) (93"1)
:X=
corresponds pointf(z) lying corresponds totoaa pointf(:x) lyinginside insideor oron onthe thefrontier frontierofofF,I',i.e. i.e. such such that 1lf(:x) I ~ 11f(:x) I ~ 2p. Thus, by by (89"5) (895) and and (93·1), (931),
1—If(x)t
1
.X
z1
1
.X
...... (93"2) (
· · · ... 93"3
)
may be be any any number number for for which which an an equation equation such such as as (90"2) (902) holds; But a: may will be made made by by choosing choosing the thus the the most most favourable favourable choice choice of of a: will sequence z,. of of {90"1) (901) and and (90·2) (902) so so that that a:=lim x=lirn !=J/(.x)l_ x... l 1.x
§93] § 93]
57 57
THEOR.Elrl JULIA'S THEOREM
When is done, done, inequalities inequalities (932) (9_?·2) and (933) (93"3) at atonce once establish establish the the When this is existence of of the the limits existence = Ii1 /(x)l =Ot. lim =lim . ..... (93"4) urn1l1/(x)l = urn 1 z1 x—n.1
11—x X
z1
11—x X
Since, last equation equation shows shows that Since, by by (90"3), (903), at> 0, the last .
I1/(x}l
!~11/(x)l
But, if if we we write
...... (93"5) (935)
I.
1f(x) = AeUI,
...... (936) (93"6)
where we have where A> A> 00 and and 06 is real, we 20080—A I1/(x)l_ 2cos6A . 1—If(x)I 11=/(X}T 1 + J(1 2A. cos 8 + A.2) ,
. ..... ((93"7)
also, when xx tends tends to unity, by (93"5) (935) also, when unity, A A tends to to zero, zero, and therefore, therefore, by must tend tend to to unity. unity. From this it and (931), (93"7), cos cos 06 (and (and hence hence also eUI) must From this follows, onusing using(93"5) (935) a.ud and (93"6), (936), that follows, on 1 —f(z) —f(x)I 1 f(x)leUI lim =at. . ..... (938) (93"8) = urn 1Ii urn 1 f(x)=lim 1—z 1—x 1x • 1x We may what has has been proved in in the form may collect collect what been proved form of of the thefollowing following theorem: .,_1
THEOREM. regularand and suck suck tkat that 11(z) I <1 THEOREM. If 1/1(z) f(z) isisregular lf(z) I< 1 in the tke circle circk exists inside inside tke the unitcirck unitcircle aa sequence of points zz1, IzzI <<1, 1, and sequence qf and if jf there exists 1, = 1, and urn z,.== I, 1, lim/(z,.)=1, ~ • ... for /0'1"which wkicklim . 1l/(z,.)l
lrm ,._... 1 I z,. I
( ...... (939) 93·9)
itt finite, tken tke is finite, then the following following limitequations limitequationskold, hold,wherein whereinxz tends tende to to its limit through arbitrary real tkrougk arbitrary realvalues values and and at is a real realpoeitive positive constant = 11.m II/(x)l ==Jim l•"m 1/(x) lim 1l.f(x)l =lim .,_. 1x .,_1 1x .,_. 1x 1 —x 1 —x 1 —x
Ot. . ..... ((9310) 93"10)
CHAPTER V V CHAPTEB
THE FUNDAMENTAL THEOREMS THEOREMS OF OF CONFORMAL CONFORMAL THE FUNDAMENTAL REPRESENTATION
94. Continuous convergence. convergence. 94. Continuous If an arbitrary arbitrary set setof ofpoints pointsininthe thecomplex complex plane, plane, we we consider consider If A isis an complex functions}; functionsf1 (z), (z), ff2(z), (z), ... which are defined defined and aa sequence of complex 2 and we we introduce introduce the the following fol1owing definition: definition : if finite at all points of A A;; and z0 limitingpoint of of AA the (z) will will be be said said to be z 0 isis aa limitingpoint the sequence sequence f,. (z) be concontinuously. convergent convergentat at zz0 if the the limit limit 0 if lim f,.(z,.)=f(z0)
••••••
(941) (94:1)
exists for of AA having z0 as its its limit. for every every sequence sequence of points Zn z,. of z0 as It proved that the limit! limitf(z0) (z0 ) is independent of the It is readily readilyproved that the the particular sequence z,., z,, and and that that every subseq uencef,.1 ,j"a, ... of the given sequence sequence also converges converges continuously continuouslytoto! f (z0) at the point Zq, i.e. i.e. that that point;, (942) Jim f,." (zk) =f(z.), (n1 (n1 <
The following theorem also also follows followsimmediately immediatelyfrom fromthe the definition definition of following theorem continuous convergence: ThEOREM.IjIf (i) (i) the sequence THEoREM. sequence 0/ of functions w =f,. (z) is defined defined in the the points A A and continuouslyatat the the point z0 set of points and converges converges continuously z0 to to the the finite W0 = f(z 0 ), (ii) A and andfor for all allvalues values of of nn the the point limit w0 =f(z0), (ii) for for all points z of A point W =f,. (z) (z) belongs belongsto toaa set setof ofpoints points BB oj of thewplane, the wplane, and w and(iii) (iii)the thesequence sequence of functions cp,. (w) defined in BB and andconverges convergescontinuously continuously at at.w0, w., functions 4,, (w) is defined then the the sequence sequence F,. (z) = cp,. {f,. (z)} converges continuously atat;. z0 • · converges continuously
95. 95. Limiting Limiting oscillation. oscillation. functionswhich Letft (z),j2(z), (z), .....  be arbitrarysequence sequence of of complex functions which Let!1 (z),f2 be an arbitrary are defined defined in a domain B R and and unifbrmly uniformly bounded in aa neighbourhood neighbourhood point of of R, B, let C01 , 0(2), of every every point Zo of B. R. Denoting Denoting by by z0 Zo aa point Cl 2l, ... ••• be a sequence of circles circles lying lying in in R, B, having their their centres at z0, and such such that that Z 0 , and their radii decrease decrease steadily to the limit Z')rO. Further, let o,.(k) denote the oscillation oscillation off,, of{,. (z) in the the circle circle Clkl, i.e., O,.fk)
= sup If. =sup
(z') —f,,(z") (z')f,. (z") 1.,
. ..... (95"1)
§§ 9496]
coNTiNuous CON CONT~OUSOONVERGENCE
59
where z', z" are any two points of the where z, the domain domain 011:1. If If we now write .., ..•..• (952) (95"2)
__
limit we have, since 011:+ 11C ()1.1:1, wli:+JI ~ wiA:I; hence the limit w (z = 0) = (Zo)
lim urn
U~(l:)
koo
exists. We see also that that this thislimit limitdepends dependsonly onlyon onthe thesequence sequence.1,. /,.(z) (z) and on on the thepoint pointz0, z 0 , but butnot noton onthe thechoice clwice of ofthe the circles circles 01"1•
The number number ww (z limiting oscillation oscillation of the the sequence sequence (z0) is called the limiting 0) is /,. (z) pointz0. z0 • f,1 (z) at at the point
96. We Wenow nowprove provethe thefollowing following theorems: theorems: ThEOREM (z0) denote respectiuly respectively the the limiting limiting oscillaTHEOREM 1. 1. Jf w (z0) (z0) and ou (z oscilla0 ) denote of a sequence sequence of functiuns f,. f,. (z) (z) and andofofaasubsequence subsequence f"r (z) at at the the tions of offunctions point point z0, z 0 , then then (961) CT (z0) ~ w (zo). ...... (96"1) For, For, by (95·2), for every circle circle ()1.1:1 ulk) ~ wlk).
z0zthe limiting (z0) THEOREM 2. 2. If .({atata point a point limitingoscillation oscillation w (z 0, the the 0 the 0 ) >>0, sequence f,,(z) (z) cannot cannot cunverge convergecontintWUSly continuouslyat at zz,. sequence/,. 0• hypothesis, for for every everyinteger integerlchan functions of For, by hypothesis, an infinite number of functions — 4u (z0), thesequence sequence have, have, in the the the circle ()1~>1, an oscillation greater thanU~It) !w(z 0 ), (z0)) i.e. (since wltl ~ w w(z Hence there there are are two two sequences (z0). 0 ). Hence 0 )) greater than !w(z of points z,",..., points z1', z1', zz2'..., zt, ~", ... , and also an an increasing increasing sequence sequence of of 2' ••• , and z1", inkgersn1 ~
k....,oo
lim z~:'' == z0, z0 , kse
]& ..... oo
I/,." (h') (z~:') 
(Zt") II > !w (z0). j "" (zA:") (zo).
...... (96"2) (962)
Hence, of the two two sequences sequences
w,: (z~:'), wi" w~:" =/,.A: (zt"), (zA:"), ...... (963) (96"3) Wh' = f,." (zk'), either at least one is not convergent, or, if both converge, their least one convergent, or, converge, their limits are unequal; and and each each of ofthese thesepossibilities possibilities excludes excludes continuous continuous conconvergence, by by §§ 94. J,.(z) (z) cunverges converges cuntintWUSly continuously at at z0 THEoREM 3. The Tlte sequence sequence/,. Zo provided provided that contains aa point point ' such tlw.twU~ (z0) (z0 ) ==00 and andthat tlw.tevery everyneighbourhood neiglthourlwod of of z0 Zo cuntains BUCk that tlw.t the tAe seqtteneej,.(') is contJergent. (z0) ==00 Titus, in particular, tAe sequence converges continuously if w (zn) and urn and is lim J. (Zo) exists and is finite. finit8.
·...
60
THEOREMS OF CONFORMAL REPRESENTATION
[CHAP. [crt.&p. V v
z,,... denote Let Zz1, denote an an arbitrary arbitrarysequence sequence of poiiits poi11ts of B R having having z0 Zo 1 , Zto··· as its limit. limit. We Wehave havetotoshow show that, that,when whenthe thehypotheses hypothesesof ofTheorem Theorem 33 are satisfied, satisfied, the the sequence sequence w,.=f,.(z,.) ...... (96'4) (964) is convergent. Denoting Denotingby byE an arbitrary arbitrarypositive positive number, number, we we can, since c an 0, define lim (1)11:1 ==0, urn defineaacircle circle()1.1:1 for which k._CII
(A)(k) <}E. .. .... (96'5) In ()l.l:l there is, is, by hypothesis, hypothesis, aa point ( at which which the the sequence sequence f,. (z) converges; N,1 such converges; and and hence hence there there exists exists aa number number N such that, that, ifif n> n >N1 N 1 and and m m>N1, (966) If,. fm I > N N,2 the oscillation that for oscillation 0,.11:1 < tE; finally, suppose suppose that for n> n >.Y3 N 3 the the point pointZn z,. denotes the greatest of the numbers numbers N1, N1, N2, N,, Na, N3, lies in ()1.1:1. Thus if N denotes we have, >Nand m > N, in addition addition to (966), (96'6), we have,provided providedonly onlythat thatnn> N and m> inequalities the inequalities lf,.(z,.)f I< !E, lf,.(z,..)f ... I< 1€....... (967) (96'7) if,, (z,,) .. Ifm (Zm) fm In view (964) to (96'6), view of (96'4) .. .... (96'8) I w,.w,.l <E (n,m>N). Hence, by by Cauchy's Cauchy's criterion, criterion, the sequence sequence w,, w,. converges, converges, so Hence, so that Theorem 3 is proved. proved. THEOREM 4. Q B denotes denotes tke whick the tke sequence sequence the set set of ofpoints pointszzof ofAA at which THEOREM (z) converges convergestotoa afunction f(z), then thenevery everylimiting limitingpoint pointZ;0 of f,. (z) function f(z), of B such suck that tkat w(A)(z = 0 belongs belongs to tkefunctionf(z), whick is is defined defined in to B, B, and the functionf(z), which in B, (z,) 0) =0 is continuous at z,. continuous at z0 • From 'l'heorem the functions functions In f,. (z) (z) From Theorem 33 we wehave haveat at once oncenot not only only t.hat that the converge at theconvergence convergence converge at zz0, sothat that zz0 isaa point point of of B, B, but but also also that that the 0 , so 0 is at iscontinuous. continuous. Now consider an arbitrary z,, ... at z0 z0 is arbitrarysequence sequence of of points points z1, z~> Zt, of B, having as limit; to 1(z) of having z0 z 0 as to establish establish the thecontinuity continuity of off (z) at at z,zowe we have only only to to prove prove that ...... {96'9) (961) limf(z,.) =f(zo)·
m m
I
m
m I
I
if
nco
Since the sequence sequence/,. each of of the the points pointsZk, z~;, we can find I,, (z) converges converges at each .an increasing increasing sequence ... , such that for for every k sequenceofofintegers~. integers n1,n,,n,,..., 1 lf,.t (z~:)f(z~:) I < k. I
(9610) .. .... (96'10)
Equation {96'9) now follows follows on fact that, that, Equation (961) now on combining combining(96'10) (9610) with with the the fhct on account of the continuous convergence of the the sequence sequence at at z,, convergence of zo. (9611) lim f,.t(z~;;)=f(zo)· ...... (96'11) urn fnk(zk)=f(zo). kco
§§97, 97, 98]
61
NORMAL NORMAL FAMILIES
97. Normalfkmilies ftunilleaof ofbounded functions(17) (17) 97. Normal bounded fUnctioni
(18).
We consider definite class {/ (z )} offunctions consider a definite which are defined of functions f(z), 1(z), which
in a domain R and and are areuniformly uniformly bounded bounded in incertain certainneighbourhoods neighbourhoods of every point such aa family family of offunctions functions we we can can define define for point zz0 ofR. B. For such 0 of each point we shall shall call call the the limiting limiting oscillapointz0 z0 of RB a number ow (z0), (z0), which we tion of the family. We Weagain againconsider considerthe thesequence sequenceof ofcircles circles 0<" 1 defined we denote denote by by ww tkl the upper upper limit limitof of the theoscillations oscillations in ()t"l in §§ 95, 95, and we all functions functions /(z) given family; family; thus w<"1 has the following following of all f(z) of the given two number of of distinct distinct functions functions two properties: ifif p >> w<"l at most most a finite number belonging to to {f(z)} have oscillations are greater greater than p, belonging oscillations in 0"1 which which are whereas if q < w 1"' an infinite number of distinct functions whereas functions of of the the family family greater than q. define the limiting have in 01"1 oscillations oscillations greater q. Finally, we define oscillation by by the equation oscillation w (z (z0) 0)
= lim w<J:I. = urn k._ao
. ..... (971) (97•1)
The following followingtheorem theoremisisobvious obvious: The : THEOREM 1. 1. Tke oscillation atatZo; of any of functions THEOREM The limiting oscillation any sequence sequence of functions which belong };. (z), /,. (z), (z), ... wkich belong toto{f(z)} f1(z), {f(z)}cannot cannotbe begreater greater than than the the limiting limiting oscillation (z0) of the family {f(z)} at z0. oscillation ww (z tkefamily {f(z)} at Zo. ) 0
definition: a family {f(z)} We now now introduce introduce the the following following definition: {/(z)} of of 98. We defined in anduniformli, uniform(y bounded bounded in detail (im kleinen) kleinen) is functions defined in R B and (z0) said to to be be normal on R tke limiting limiting oscillation oscillation ww (z at every every said B if the vanishes at 0 ) vaniskes point B. point z0 Z 0 of R.
We We have functions all (z), /, 1! (z), (z), ... is a sequence THEoREM 2. sequence of /unctions all belonging belanging 2. If if f 1 (z), normal on B, contains which is normal R, then tkenthe tke given given sequence sequence contains to a family {f(z)} which a subsequencef,,1 (z),f,,2 (z), (z), ... which at every subsequencef.,.1 (z)J~~s whick is continuously convergent convergent at point of qfBR (19). (19). poznt
Consider aa countable Consider countable sequence sequence of points points z1, z,, z2, z,, ... of B R which which lie lie everywhere denseon onR. B. By the diagonal everywhere dense diagonal process process ((Cantor Oantor and Hilbert), . can be picked pickedout out from from the given (z), f,,2 (z), .... subsequence /"'(z), f.,..(z), can be given a subsequence sequence sequence in in such such aa way way that, that, at each poie :f the countably infinite set, the limit limit ......(981) (98•1) limf,,k(zP)
exists. The 96, are are satisfied by this subt:xists. The conditions conditions of of Theorem Theorem 3, §§ 96, satisfied by sub· sequence at every B; this sequence at every point of R; this proves proves the the present present theorem. theorem.
62 62
THEOREMS OF OF CONFORMAL CONFORMAL REPRESENTATION TREOREMS
[CRAP. V V
99. ItItisiseasy easytotoshow showthat thatthethe two conceptions,(a)(a)continuous continuousconverconvertwo conceptions, set y, g61/Ce of gence of a.a sequence of continuous continuousfunctions functions at at all all points of a closed set and (b) (b) uniform uniformconvergence converg61/C6 of this this sequence sequence on on y, "f, are arecompletely completely equivalent equivalent(18). (18). theorem isis aa consequence of this this equivalence, The following following theorem consequence of equivalence, but we we direct proof. proof. shall give a direct qi functions converges THEOREM 3. 3. IfIf aa sequence sequene~~ J;. (z), .h (z), ... ... qf converges conconTREOREM f1(z), f2(z), tinuousli, to thefunctionf(z) thefunctionf(z) atatallallpoints set y, y, and and tinuousZ11 to pointsqfqfa abounded boundedclosed closed set if1(z) f(z) =F 0 on y, then there there is is apositive a positivenumber numbersri m and and aapositive positive integer integer N, such such that, tkat, atatevery every point point of of yy and and for for each eack nn > N,
lf,.(z) I >m. Ifn(z)I>m.
. ..... (99'1)
If this were not so, it it would would be be possible to to find an increasing sequence sequence y, such n1 < ~ <..., < ... , and a sequence sequence of points points z1, ~. z2, z~, ... of "f, of integers, n1 that ...... (99'2) (992) (k=1,2, ...). Ifftk(zk)l
By omitting some of the points Zk afresh those that some of z., and ordering ordering afresh that rerecan ensure ensure that that we can main we (993) lim z., = z0 • .. .... (99·3) limzh=zo. I:_..
and therefore Since y is closed, z0 z0 belongs to y and therefore the the sequence sequence continuously z0. Hence • /11(z), (z), f~ (z), ... converges at z Hence /2(z), 0 1 urn f,... (z.,) ==0, f(z,) f(zo) ==lim 0,
(99 •4) ...... (99'4)
k
and aa hypothesis hypothesis is is contradicted. contradicted. This proves the theorem. theorem.
100. Eldstence Existence of of the the solution in certain 100. certain problems problems of the calculus of of variations. Weierstrass' Theorem that thataacontinuous continuousfunction functiondefined defined on a bounded closed space attains attains its maximum maximum at at some some point point of of closed set in ndimensional space the set setcan canbebeextended, extended,inincertain certaincircumstances, circumstances,totofunctionspace, functionspace, that that is, is, to to space space in in which which any any element element is is defined defined not by a system system of n definite fun<;~~vn. For instance, instance, suppose that thataa family family numbers but by a definite {/(z)} of functions is is given, {f(z)} of complex complex functions given, all the functions being being defined in a domain domain R. The Thefamily family isissaid saidtotobebecompact compactififevery everysequence sequence f1(z), functions of the / 1 (z), /2(z) / 2 (z) ..., ... , of functions the family, family, contains containsaasubsequence subsequence (z), 4(z), convergesininRfl to a functionf0 (z), where.fo wheref, (z) f"l (z),fs (z), .... which converges function/0 (z), also belongs oolongs to to the thefamily family(20). (a)).
99—101] §§ 99101]
OF THE THEMAXThItJM MAXIMUM EXISTENCE OF
63
Secondly, we (f), by by means means of which which aa finite finite Secondly, we consider consideraafunctional functionalJ.1(f), f (z) (z)}. The functional functional number number is is associated associatedwith witheach eachmember member! (z) of of{f {f(z)}. .1(1) J (f) isissaid saidto t<>be becontinuous continuous if the the convergence convergence in B R of of the thesequence sequence ft (z),f the functionf0(z) function/0 (z)of ~J (z)} —J (fo). (fe). of {f(z)}alwaysimpliesJ(f,.) 2 (z), ... to the /2(5), fi (z), ·We We now now have: have: THEOREM. If J (f)isisaacontinuousfunctional continuqusfunctionaldefined defined in theeompactfamily tlze.compactfamily TilEoRmi. 1/1(f) {f(z)}, then tken the the problem problem {f(z)}, I J(f)l ...... (1001) (100'1) J(!)I ==Maximum Maximum kas a solution solution within the thefamily has family {f(z)}. I
That is is to to say, say, there is is at least least one one member!0 member fo (z) (z) of the family family {f(z)} such such that that all membersf(z) of {f(z)} satisfy I J(f)l ~I J
I
For if if IXa denotes denotes the the finite finite(or (orinfinite) infinite)upper upperbound bound ofof ! J.1(f) (f) 1. asf(z) varies in the the family family {f(z)f, then there there are aresequences!2 sequences .It (z),f2 (z),,/; (z), (z), ... of members of members of{f(z)}, {f(z)}, such such that that lim J (f,.) .. .... (100'3) urn II .1 (fe) II =IX. =
,._00
Now, since {f(z)} is compact, a subsequence/"~ (z),f,,2 (z), ... can be Now, since (z).f,.. (z), such a way selected from f, (z), selected from the sequence sequence .It (z), /f2(z), way that that the the subsub2 (z), ... in such The continuity continuityofofJ .1(f) (z) of {f(z)}. The (f) sequence converges converges to to aa memberfo memberf; (z) shows that that shows = a. I J(.fo)l =lim lim II J(f,..t)l =IX. I
k.o
I
This not that aIX isis finite. not only only proves the theorem but also shows shows that finite.
101. Normal families regular analytic flinctions. 101. fil.mUies of regular functions. families {f (z)} (z)} whose whosemembers/ members! (z) are all We now consider considerfamilies all defined, defined, analytic and regular, regular, in in the interior interior of of aa domain domain R. B. For For these these families families analytic the following following theorem holds: TRnOREM familyqf / analytic are all all THEOREM 1. 1. lfif{f{f(z)} {z)} isisa afamily analytic functions, functions, which whick are regula!r in in the the interior interior <j qfaadomain domain B, R,and andwhich whickare areuniformly uniformlybounded bounded regular on the circumference qf any circle circle lying then {f(z)} lying strictly witMn within R, B, then {f(z)} is normal in B, normal in R, and and the the same same is true true of of the the family family {f'(z)} {!' (z)} of ofderived deriwd functions, obtained from the thefunctionsf(z). obtained from functions f (z). Let z0 be a.a point of — z0 z0 be of B R and andzIzz0 I ~ p a closed closed circle circle lying strictly within B. R. There Thereisisaa.positive positive number number M such that thatall allmembers members of of {f(z)} satisfy the the relation relation If If(z)I satisfy (z) I < Mat all points of Iz  z0 I =p = p andhencealso and hence also at all at all points pointsof ofIz—z0I Iz Z 0 I
64
THEOREMS OF REPRESENTATION OF CONFORMAL CONFORMAL REPRESENTATION
[CHAP. [cuip. V v
being taken of Theorem Theorem 5, 5, §§ 71, it is is seen seen that, if r is is any any Account being number between between 00 and and 1, 1, the the relation
2Mr ~ MIz''z"I Mlz'z"l < 1r 2Mr If( 1/(z) —/(z., ) "'p(1r) z')f(~")l p (1 — r2) < I
(ion) ...... (101"1)
is satisfied by (z', z") z") of of points points of ofthe thecircle circlezIzis satisfied by every pair (z', — zz00 I < pr and by (z)}. From this ititfollows immediately that that by every every member member of {f (z)}. follows immediately limiting oscillation oscillation of {f(z)} vanishes vanishes at the point point z0. z 0 • By By§ the limiting at the § 98 the family {f(z)} is is normal. normal. family To prove the family family {f' {f'(z)}, (z)}, consisting consisting of the the derived derived functions functions prove that the
is also also normal, normal, we we need need only only r3mark r3lnark that that by of the given given functions, functions, is Theorem 4, functions!' (z) Theorem 4, §§ 71, the derived functionsf' (z)are areuniformly uniformly bounded in the circle lz—zoI lzzol <;,pr. 102. We shall now now show show that that the the boundary boundary function!0 function.fo (z) (z) of ofaa converconverfunctions of of the the family family isis analytic. analytic. For gent gent sequence!1 sequence/1 (z),12 (z),f2 (z), ... of offunctions as isis well well known, known, itit isis sufficient sufficient to to show show that!0 that fo(z) (z)isisdifferentiable differentiable this, as at every at every point of B. R. lying in in R, B, and closed circle circle Iz —  z0 I <;, p, p, lying and in in it it consider consider the Take a closed sequence analytic functions functions 4,,, cp,. (z), defined by by the the sequence of regular analytic (z), which which is defined equations ..~,. ( ) _f,,(z) _f,. (z)—fe f,.(zn) (zo) ...... (102"1) ..,..,. z' (z *' Zo), zz z — z00
...... (102"2) (1022)
=f,,'(zo).
The sequence (z) is uniformly bounded bounded on on the circumsequence of functions functions 4,,, cp,. (z) circumz0 I = p of the circle in question, question, and hence, hence, by by the result of ference IIz — Zo the preceding paragraph, it the it is is normal. The Thesequence sequenceobviously obviously converges converges other than than;,zO>and at all points of the circle circle other andtherefore thereforeititconverges converges concontinuously the point point z0 z0 itself itself (Theorem (Theorem 3, 3, § 96). Let the the boundary boundary tinuously at the function be be denoted denoted by by cp4,(z); (z);then then (102"1) (1021) and (1022) function (102"2) yield vield I
cp(z)=fo(z)fo(Zo), zZZ — 0
(z*'zo),
...... (102"3) (1023)
....•. (102"4) cp (z0) ==lim lim/,,' f.,.'(Zo). (zo)· at the the point z0. Further, 4, §§ 96, 4, cp (z) is continuous continuous at z0 • The The Further, by Theorem 4, 4,
above equations therefore therefore assert assert that above equations fo' (z0 ) = 4, cp (z0) ==,.._.., lim f,.' (Zo). (zo). . ..... (102"5) gives This gives THEOREM 2. IfIf sequence of Q{analytic analyticfunctions is is uniformly bounded in a asequence converges,tken thentke theboundary boundary/unction a domain domain in inwhich wkick f/se tke sequence sequence converges, function is
§§ 102, 103] 103] fi 102,
REGULAR CONVERGENCE CONVERGENCE
Go
65
analytic and and its its derived derived function is the tke limit of qf the tke derived derived functions of of the functions of tke functions qf the tke approximating approximatingsequence. sequence. Keeping this of brevity, brevity, deKeeping this theorem theorem in in mind, mind, we weshall, shall, for for the sake of scribe a convergent convergent sequence sequence of regular regular functions functions which which is is uniformly uniformly scribe bounded in detail detail as asregularly regularlyconvergent. con'INJrgent. Suppose that that a given Suppose given sequence sequence of of analytic analytic functions, functions, defined defined in aa. domain R, is uniformly uniformly bounded the sequence sequence is regularly regularly domain B, is bounded in in detail detail;; the convergent inin R provided only points at at which which ititconverges converges convergent B provided onlythat that the points possess aa limiting possess limitingpoint point in in the the interior of R. B. For, For,ififthe thegiven givensequence sequence did not converge converge everywhere, everywhere, it it would would be be possible possible to to select select from from it it two subsequences converging (z) and andgg(z). (z). convergingto totwo twodistinct distinct analytic analytic functions/ functionsf (z) ateachpoint — gg(z)) But, at each point atwhichthegivensequenceconverges, at which thegivensequenceconverges, (f(z) (f(z)(z)) == 0. These zeros zeros cannot cannot possess possessaalimiting limiting point point within within R. B.
103. The following regularly convergent sequences sequences followingtheorem theorem relating relating to regularly analytic functions functions is is a special theorem due to of analytic special case of a wellknown wellknown theorem Hurwitz: THEOREM 122 (z), ... THEOREM 3. 1./, in aa domain domainB,R,the thesequence sequenceofoffunctions functionsf1J;(z), (z),/ converges regularly none of oftkefunctionsf,.(z) vanconverges regularlytototkefunctionf(z), thefunctionf (z), none thefunctionsf,, (z) vanishing at at any point of R, B, then eitherff(z) (z) 0 or f(z) 1(z) does not vanisk vanish at ishing tken either does not any point of of B. R. For suppose supposethat/ that 1 (z) (z) does doesnot not vanish vanish identically identicallyin in R. B. CorrespondCorresponding to each each point point z0 z0 of B R there there is is at at least least one one circle circle Izz— z0 I ~ p lying on its circumference circumference f f(s) (z) =1= 0. within R within B and such that on By Theorem 99,there theremust mustbe beaanumber numbermm> suchthat that Theorem 3 of §§99, > 00such
=
(1031) If. (z) I> m ...... (103"1) If,,(z)I>m for all points points on on the circle circle II zz — z0 I = p, provided that nn isissufficiently sufficiently circle/,. (z) =1= if n is is sufficiently sufficiently large, large. large. But But in in this this circle!,, large. * 0, and therefore, if (1032) If,. (Z0 ) I~ m. . ..... (103"2) Thus lf(zo)l ==lim lim If,,(zo)I lf.(zo)l ~m, ..._
..
that and it it follows follows t~tf(z * 0. 0)=1=0. this theorem theorem is is:: An immediate corollary corollary of this THEOREM dinnain B functions!1 (z),f2 (z), (z),... THEOREM 4. 4. IfIfininaa domain R aa sequence sequence of functions .f. (z),f, ... convergesregularly regularlytotoafunctionf(z) afunctionf(z) which is not not a constant, then any any con·veryes wkick is constant, then of R B contains points z,, such tkat that neighbourhood neig/WourltOOd N zo of qf aa point point z,, z 0 of z,. suck f,. (z,,) (z,.) = f (zo), f,, if n isis sufficiently sufficiently large. If If this this were were not so, so, the given given sequence sequence would would yield an an infinite infinite subsubthat, in in the the neighbourhood N,,,ofof;, sequence, sequence,/,.1 (z),f"' (z), .~., such such that, neighbourhood N,.. z0 , (z), (z),..., (z) .f(z,)) zero. Since the the functions functions (f.~ (z)f (z.)) all differ differ from from zero. the boundary boundary
66
REPRESENTATION THEOREMS OF CONFORMAL CONFORMAL REPRESENTATION
[CHAP. v V
function (1(z) (f(z)f(z function —1(z0)) ofthis thislast last sequence sequencevanishes vanishesatat the the point zz,, 0 )) of 0, Theorem 33 shows showsthat that (f (f(z) —1(z0)) Theorem (z)f (z0)) vanishes identically, i.e. the the funcfunction 1(z) is tionf(z) is aa constant, constant, and and aa hypothesis hypothesis is is contradicted. contradicted.
104. Application to conformal representation. 104. followingtheorem theoremisisofoffundamental fundamentalimportance importanceininthe thetheory theory of of The following conformal representation representation:: conformal Iff,(z), functions which THEOREM. If / 1 (z), /f2(z), sequence of offunctions whick converges converges 2 (z), ... is a sequence regularly domain B, R,and andjfifthe thefu11ftions giveconformal conformal transfortram(orregularly in a domain functions give which are are uni... respectively, whick mations qf of B matiws R into into simple simple domains domains SlJ 8S2, 21 ••• formly bounded, then, either either the the boundary boundaryfunction functionf 1(z) bounded, then, (z) is aa constant, constant, or it gives gives a conformal cMiformal tran:iformatioo simple domain domain S. transformation of RB into a simple S. By hypothesis, f,. (z) f,, (z0) z0. hypothesis,/,. (z) *=Ff,. (z0) when z and andzz,0 are are points points of of B, R, zz =Fz •. The functions 4>,. (z) ==f,. (z0) donot notvanish vanishininthe the pricked pricked (punlctiert) (punktiert) (z) — f,. (z 0 ) do f,, (z)domain R—.z0. Thesefunctions functionsconverge convergecontinuously continuouslyinin this this domain, R z 0 • These domain, By Theorem Theorem 33 of § 103 103 either either towards the the function function41(z) 4> (z) =1(z) =f(z)—f(z0). f(z0 ). By 4> (z) isisidentically then aaconstant, constant,oror414> (z) (z) isis different different 41(z) identicallyzero, zero,andf andf (z) is then from zero zeroand andsof(z)=Ff(z.). sof(z) from 105. The main maintheorem theoremofofconformal conformalrepresentatIon representation (21). 105. The (21). Let B R be be an an arbitrary arbitrarybounded boundeddomain domain in in the the zplane, zplane, containing containing the point z = and therefore therefore also a circular area K defined by =00 and (105i) Iz I < f' ...... (105'1) interior. No No assumption assumption is made made as connectivity of in its interior. as to the connectivity of R. B. Consideraafamily family{f {f(z)} offunctions functions which which are are regular regular in in the circle Consider (z)} of circle (105'1). The The family familyisis assumed assumedtotobe bemade madeup upofofthe thefunctionf(z) functionf(z) 0 1(z) which satisfy the following conditions: and also all functions functionsf(z) following conditions:
=
(a) .f(0)=0, f(O) =0, analytic continuation continuation of possible along off(z) 1(z) is is possible along every every path path y'Y (b) analytic B and I (z) within R and the thefunction function/ (z) isis always always regular, regular, two paths paths joining joiningzz==0 (c) if y' 0 to the points z' and z" y' andy" and )/' are two (z") are are the the values values obtained obtainedat.at z'I respectively, and (z') and and '('F (z") respectively, and if if .,.F .,F (z') z" along these these paths, paths, then then ifif z' z' =Fz" and z" z" by by continuingf(z) continuingf(z) along .,.F(z')=F.,.. F(z") ..... (105"2) provided z#0), (in particular, .,F (z) =F 0 provided z =F 0}, (d) with with the the above above notation notation I.,F(z)l< 1. 1. ...... (1053) (105'3)
106. ItItwill willfirst first be be proved proved that that the thefamily family{f(z)} {f(z)}isiscompact compact(§ 100). 100). Condition (d) shows showsthat that any any sequence sequence of offunctions functions of ofthe thefamily family conta.ins contains
§§ 104107] 104—1071
67
THE MAIN MAIN THEOREM THEOREM
a subsequence subsequence 11 .it (z), (z), /122 (z), (z), ... which satisfies the relation lim ...... (1061) (106"1) urn /,, (z) =fo (z)
,._..,
K. We have to show show that that 10 fo (z) belongs to the the family family {f(.:)}. in x. so that either It is obvious obvious that/ (0) = 0, so either condition condition (a) (a) isis fulfilled fulfilled or It is that!00 (0) f(z)n f(z) 0. 0. must show that ifif y'y' isis aa path path within B To verify condition (b) (b) we must show that within R point z', z', and and ifif for for every every point point { of y', y', other z', other than z', joining z = =00 to aa point (z) gives a function .,.F0 which exists, the analytic continuation of fo (z) yFo which is regular and can be be obtained as the limit of the analytic continuations .,.F,. ('> of the (z), then all these conditions conditions are satisfied satisfied at the functions f,. (z), the point point z' z' itself itself and and in in aa certain certain neighbourhood neighbourhood of that point. point. It isis easy (z) are are regular regular in in a It easy to to show show this, this, for for the the functions functions 1F,. yFn (z) certain neighbourhood neighbourhood of of z' and bycondition by condition (d) (d) they theyform a normal family certain which converges ,,F0 (z) (z) at at all all points of a certain portion converges to iFo portion of of yy' (§ 102). To prove condition (c), (c), consider considertwo twopaths pathsy'y' andy" and y" with with distinct distinct endz", and and let Nz' and Nr· be nonoverlapping neighbourhoods points z' and z", neighbourhoods respectively. By Theorem Theorem 4, 103, there points z0' z,.' in of z' z' and and z" respectively. 4, §§103, there are points Nz' and z,." in Nz'' such such that the equations equations . (1), 1F,, 1F,. yFn (zn') (z,.') = = 1F0 yFo (z'), y"F,.(zn") (z,.") == 1F0 yFo (z") ...... (106"2) hold simultaneously, being suitably chosen. chosen. By simultaneously, nn being By hypothesis hypothesis the terms these equations are unequaL unequal. The 'fhe required required result, result, on the left in these yFo (z') * y"F', (z"), follows. Finally, obvious that/ (z) satisfies satisfies condition condition (d). (d). Finally, it is obvious that!00 (:)
=
m
Consider now now aa particular functionf(z) 1(z) of of the the family and its 107. Conf?ider particular function family and analytic continuations yF(z} in B. R. Suppose Suppose that thatthere thereisisaanumber numberw0, w0 , where 8 w0=he'°, w (O
hf,(z) ex (z) = i lt.f. (z)'
...... (107"4}
$(z)='Jcx(z), (/3(0)=÷.ih), {3(z)=Jcx(z), ({3(0}=+Jh),
...... (107"5)
JJI{3(z) g(z)= 1Jk{3(z)'
...... (107"6) (1076)
proceed to investigate their their properties. properties. and proceed
68
REPRESENTATION THEOREMS OF CONFORMAL CONFORMAL REPRESENTATION
(cHAP. V V [CRAP.
The function the family family {f(z) {f(z)}, }, and neither function ff1(z) belongs to the 1 (z) clearly belongs value k. The the function function nor analytic continuations continuations takes nor its analytic takes the the value h. The connected with withjj {z) is connected _A (z) (z) by by aa Möbius' Mobius' transformation transformation which which function alX (z) (b),(c), (c),(d) (d)of of§ § 105 transforms the unitcircle into into itself. itself. The Theconditions conditions(b), are all satisfied satisfied by bylX{z). Thesame same is is true trueof of/3(z). f3(z}. Neither NeitherlX{z)nor a (z). The a (z) nor its continuations continuations take the the value value zero, zero, so that that/3(z) f3(z) isis regular regular on on every every path y. To condition (c) we notice that if analytic continuation notice that if analytic continuation verify condition (C) path y. P(z) (z} along leads to to coincident coincident values z", of /3 along paths paths y'y' and and y" j' leads values at at z' z' and z", of same must hold for conclusion that the same for lXa (z), (z), which whichwould wouldlead leadto to the conclusion that
zz'=z". =z. we see, see, from g (z)satisfies satisfiesthe theconditions conditions(b), (b), (c), (c), Finally we from {107"6), (1076), that that g(z) and (d), and and further, further, since since /3(0) f3 (0) = = Jk, the function g (z) vanishes at = 0, at zz = so that that ititisisaamember member of of the the family family {f(z)}. {f(z)}.
a(z),jj 108. IfIfthe theequations equations (1073) (107"3) to to(1076) (107"6)are aresolved, solved, /3(z), f3 (z), lX {z), .h (z) being successively determined determined as as functions functions of bein.g successively g(z)=t, is found found that it is
...... (108"1) (1081)
/(z)=ef8t2 .Jk(I +k~t. . ..... (I08"2) (1082) (I +k)2 Jkt The expression on the the righthand side of expression on of this this equation equationcoincides coincides with one of the functions § 56. in§ 56. It It satisfies satisfies all the ihe conditions conditions functions investigated in of Schwarz's Lemma (as (as can can also also be be verified verified by by direct calculation) calculation) and and Schwarz's Lemma therefore, for for every every value value of t in the the domain domain 0 << II tt II <<1, I, therefore, 2 Jit(I +;> tl
...... (108·3) (1083)
(108·3) ititfollows follows that From (108"I) (1081) to (1083) •..... (I08"4) (1084) lf(z)l
109. 109. Let !Jet z,z1be beaafixed fixedpoint pointofofK,K, zz11 =F 0, and and define define aa continuous continuous #0, functional J(f} (see§ equation functional J(f) (see § 100), by means of the equation J(f)= = 1/(zi) J(J) If(zi) I.J.
••.... (109"1) (1091)
§§ 108111]
69
TEE MAIN THE MAIN THEOREM THEOREM
Since the family Lf(z)} Since if(z)} is compact, compact,§§ 100 shows shows that least one one that it has at least member .fo (z) · member!0 (z) such such that that (1092) lfo(zi)I ~ lf(zi)I ...... (109·2) f(z) varies varies in in the the family. family. The The domain domain B R being being bounded, bounded, the the as 1(z) Az belongs functionf(z) functionf(z) == .\z family, if A belongs to to the family, A isis aa positive positive number taken sufficiently small, and and thus, thus, from from (109'2), (1092), sufficiently small, l.fo (~)! ;?: A Iz1l > 0, which shows showsthat.fo(z) that!0 (z) isis not not a constant. Using of§§ 108 Using the theorem of 108 we weobtain obtain at at once once:: T*IEOREM The family{f(z)} {f(z)}contains containsat at least onefunctionfo onefunctionf0 (z) THEOREM 2. 2.The family (z) which, wkick, with takes every everyvalue valueinin tke the circular circular area witk its analytic analytic continuations, continuations, takes
)
lwl <1. IwI<1. 110. 110. We now now consider functionsf(z) functions 1(z) which which satisfy satisfy conditions conditions (b), (b), (c) (c) and (d) (d) but but not notnecessarily necessarily condition condition (a) (a) of of §§ 105, and which which have in in property: addition the the following following property: (e) To of the w <<11 there (e) To every every point point w0 Woof the unitcircle unitcircle JIwl there corresponds corresponds a path y 0 , joining z == 0 and some some point zz0, notation such that, that, with the notation 0 , such used above, 'I' F(zo)=wo. .. .... (110'1) 0 The last theorem proves the existence existence of of such such functions. functions. Further, by in (110'1) (1101) is a singlevalued function of the property (c), (c), the number z0 z0 in singlevalued function w0, defined throughout throughout the interior, wII <<1, w interior, Iw 1, of the unitcircle. unitcircle. This This 0 , defined function is the the inverse inversefunction functionofofY0;F(z) F (z) = w in some neighbourhood of of function is each point point ww0, anditit isis therefore thereforeregular. regular.Thus Thusitit is is seen seen that that the each 0 , and which we denote by 4) rp (w), function, which (w),isissinglevalued, singlevalued,regular regular and and analytic analytic throughoutthe thecircle circle1w IwII<<11 (~ 63). The equation throughout zz=4)(w) = rp (w) ...... (110'2) (1102) therefore gives givesaa conformal conformaltransformation transformationofofthe thecircular circulararea area IjwwI <1 therefore <1 into a Riemaun Riemann surface surface which which covers the whole whole domain B. R. <
(d) and and (e), (e), yields yields an an 111. AAsecond secondfunction function gg(z), (z), satisfying satisfying (b), (b), (c), (c), (d) inverse function z=ifr(w) ...... (111'1) Z=t/t(w) which has the the same same properties properties as as the thefunction function4)rp(w) (w) of of (1 (110'2). By which means of of (1 (110'2) a correspondence is set up between the the points pointsw0 w0 within the unitcircle z 0 , with with associated paths paths y 0 • The paths may ma.y unitcircle and and points points z0, be obtained, for instance, instance,as as the the paths in the zplane obtained, for zplane resulting resulting from the the transformation of straight If we straight lines lines joining joining00totow0. w 0 • If we write w0 = =Yo G (z0),
70
THEOREMS OF CONFORMAL REPRESENTATION CONFORMAL REPRESENTATION
[CHAP. V [cK&P. V
then w0 then wo is a singlevalued regular function In what what has has just just been been function of of ww0. 0 • In said ww and w interchangeable, and ititfollows follows from from §§ 68 that the the so are are interchangeable, equation 4,cp(so) = çb (w) gives a nonEuclidean equation (w) =.; nonEuclidean motion the unitcircle unitcircle motion of the on itself.
shows that that if the 112. The Themonodromy monodromy theorem theorem (§ 63) shows the given given domain domain B isis simplyconnected the inverse R simplyconnected the inverse of the function function (110"2) must be a regular singlevaltMd sin glevaluedfunction functionofofz zininR. 1?. The The equation z == 4, cp (so) (w) then gives a oneone conformal conformal transformation transformation of the the interior of the unitcircle domain B. R. on the domain It has It has already alreadybeen been shown shown(§ 64) that that any anysimplyconnected simplyconnected domain domain with more than one one frontierpoint frontierpoint can can be be transformed conformally into a bounded domain. Combining this with the results just Combining this justobtained, obtained, we we have THEOREM 3. The THEOREM The interior of of any anysimplyconnected simplyconnected domain d<Jmain I? R with withmore m(ffe than frontierpoint can onthe theinteri(ff interior of the the unitcircle unitcircle than one frontierpoint can be be represented on oneone conformal The representation by means means of of aa oneone conformaltran._yormation. transformation. The representation is not unique. AAgiven git·endirected directedlineelement lineelementin in11 R can can be be made made to to correspond C(ffrespond to an an arbitrary arbitrary lineelement such aa correspondence to lineelement ininIwj w I <1. < 1. JVhen JVhen such C(ffresporuknce is assigned the transformation transf(ffmation isisdetermined determineduniquely. uniquely.
113. On 113. On the the other otherhand, hand, ififthe theinverse inverse of ofthe thefunction function (110·2) is a singlevalued function of of zz in B, R, then thenby by§§ 63 the domain domain B R isis simplysimp\ysinglevalued function connected. Thus Thus ififBR isismultiplyconnected multiplyconnected there must be be at at least least two two connected. 4, ( soI<1 poilits (w"). points w' and sv" 1v" in Iw < 1 such such that that4,cp(w') (w') == cp w"). Hence Hence there is is at at least least one one nonEuclidean nonEuclidean motion motion of the circle circle Iw I <1 < 1for for which which4,cp(so) (w) The aggregate is invariant invariant (§ 111). lli). The aggregate of of such such nonEuclidean nonEuclidean motions motions forms function. forms a group, group, and 4,(w) cp (w) is is an an automorphic automorphic/unction. The points w 4, (so0) are called called w of of the the circle circle II wl <1
THEOREM 4. 4. IfIfthe multiplyconnected, then THEOREM thedomain domain RB is multiplyconnected, then the function function (1 (110·2) is automorphic, invariant for forcertain certaintransformations transf(ffmations automorphic,and anditit is invariant
112—115] §§ 112115]
71
AUTOMORPRIC AUTOMORPHIC FUNCTIONS FUNCl'IONS
wlwse infinite group of of nonEuclidean nooEuclidean translations translations wlàose aggregate forms an infinite and limitrotations. ancllimitrotatimzs. thecase casewhere whereBRisisdoublyconnected doublyconnected it is is easy easy to to form form the the 114. InInthe of transformations. group of Supposefirst first that that B which the the Suppose R isis aa simplyconnected simplyconnected domain domain from from which point z0 z0 has Thenaaoneone oneone conformal conformal transformatransformasingle point has been been removed. removed. Then transform R the pricked pricked (punktiert) (punktiert) circular tion can can be be found, found, to transform B into the domain ...... (114'1) (1141) O
*
115. Next Nextlet letthe thefrontier frontierofofBRconsist consistof oftwo two distinct distinct continua continua C, 0 1 and C,, each containingmore morethan thanone onepoint. point. Regard Regard 0C, as the the frontier frontier of C,, each containing 2 as a simplyconnected simplyconnected domain containing containing 01. 0 1 • This domain domain can be be conconformally transformed into the the interior interiorof ofaa circle. circle. In this transformation B whose frontier frontier consists of a conR becomes becomes a doublyconnected doublyconnected domain whose F
72
THEOREMS OF CONFORMAL REPRESENTATION THEOREHS
(CHAP. V V [CHAP.
tinuum tinuumC1' 0 1' and a circle (J,'. 0,'. The simplyconnected simplyconnected domain to domain exterior to andcontaining containing C,' 0C1' 0 1' in its interior is now transformed transformed into into the interior 1' and
transformed into into an unitcircle. In of the unitcircle. In this thistransformation transformation Cl' 0 2' is transformed analytic curve. curve. It can therefore therefore be be assumed, assumed, without without loss loss of generality, It can that as its frontier that the thedoublyconnected doublyconnected domain domain 1? R has as frontier the unitcircle unitcircle 0C11 and an analytic analytic curve curve C2, 0,. without without doublepoints, doublepoints, surrounding surrounding the point zz=O. = 0. By periodic curvilinear By means means of of the transformation zz = e", s.a certain periodic strip Ru 1n In the uplane transformedinto into the domain B. uplane is conformally conformally transformed R. corresponding to a given given point point z0 z0 is of the form form The set of points corresponding U~;=U0 +2ikr (k=0, (k=O, ± ±1, ... ). 1, ±2, ±2,...). uk=uO+2jklr are generated by a cyclic cyclicgroup. group. Since Thus the points equivalent to to u0 u 0 are 11,, simplyconnected,aa oneone oneone conformal conformal transformation transformation uu ==4cf> (w) R,. isis simplyconnected, ( w) can wI<1 can be found found which which represents represents it on on Iw I < 1. The points of Iw < 1 which are equivalent to w0 areobtained obtained as as the the images images of of the points which are w 0 are points u~;. 'fhe corresponding corresponding group The group isis certainly certainly cyclic cyclic and and is generated by repetition of a certain tion certain nonEuclidean nonEuclidean translation, translation, the possibility possibility of a limitlimitthe result resultofof§ 114. rotation being excluded by the § 114. I
116. 116. A l\ nonEuclidean transformation with two two fixed fixed nonEuclidean translation translation is a transformation points, A B, on on the thecircumference circumference of of the the circle. circle. Let points, A and B, w = (t) w=lf!(t)
117. Let LetRRbebea asimplyconnected simplyconnecteddomain domainon onwhich which the the circular circular area area Iw I <1 < 1 is is conformally conformally represented. represented. Two Two fixed fixed points, points, z1 Z1 and z2 z 2 of B, R, I
§§ 116119] 116119]
73
DOUBLYCONNECTED DOMAINS DOMAINS
correspond correspond to two two points points w1 w 1 and w2 w1 which which are uniquely determined for for any p088ible possibleconformal conformalrepresentation representationofofthe the circle circle on on R, B, and whose nonEuclidean distance number nonEuclidean distance D(wl> D (w1,ww2) alwaysthe the same. same. This number 2 ) isisalways distance of of the points may therefore be be regarded as the l.lonEuclidean distance Zt and for B. R. z1 and ~. z2,and andin in this this way way aa nonEuclidean nonEuclidean metric metric is is set up for It metric for for aa multiplyIt is also also possible possible to set up a nonEuclidean nonEuclidean metric for here connected domain R, B, but the process connected domain process is somewhat somewhat different, different, for an infinite infinite number number of points points w correspond correspond to single point z. Let LetYz y. to a single be any rectifiabld It is is transformed transformed into into an an infinite infinite number rectifiablecurve curveininR. B. It of wplane. All ofcurves curvesYIL'l, y"'l, y"", ... in the thewplane. All these these curves curves can can be be transformed transformed motions, and and thus thus they all have the the into one another by nonEuclidean nonEuclidean motions, nonEuclideanlength. length. This length associate with with the curve y •. same nonEuclidean length we we associate
In In the multiplyconnected multiplyconnected domain dmnain1? R curves curves have length, but but the the coneeption rxm.ception of a distance distancebetween between two points has has no no meaning. meaning. of
118. Consider an annular region region ...... (118'1) r1 < lwl
119. Normal of functions ftznction. which which transtrans. 119. Nonnal tkmilies families composed composed of form form simple simple domains domains into into circles. wplane, Let {TT}} bea be a familyof family of simple simplyconnected domains in the thewplane, w = 0 while contains domain containing containing the point point w each domain while none none of them contains } is formed formed of analytic functions the point point w w = oo. A family {f(z) {f (z)} functions by by means of of which which the the domains domains Tare T are transformed means transformed into the the circle circle Iz I
74: 74
OP CONFORMAL REPRESENTATION THEOREMS OF
[CHAP. [CRAP. V
=
The theorem of 104 shows showsthat that the the addition addition of the functionf(z) function 1(z) 0 of§§ 104 makes makes these these families familiescompact. compact. In In case case (/3) (/3)there thereisisno noneed needtoto add this the condition If I!' (0)1 stricter condition function if the (0) I ~ Mis Mis replaced replaced by the stricter 0< m ~If' (O)I ~.M.
120. kernel of of a sequence sequence of of domains. domains. 120. The kernel Take a family family {TT} of domains, domains, and, ideas, suppose it } of and, to fix the ideas, it satisfies sequence T1, T1 , T2 , ••• fies condition (at) (se)of§ of §!l9. 1 j9.In In it it choose a sequence ... of domains such that thecorresponding correspondingfunctionsf (z),f that the functions!11 (z), f2(z), 2 (z), ... converge regularly (z) in IzI to a function!0 functionf.(z) lzl
"k
(z0) = lim f.,:lr (zn',,). fo (z (z,.· 11 ). 0) = koc
belongs to to the the closed set A.,. A,,. Reference Also the the point pointurn limW,,'k w,.· 11 = w,, w. belongs closed set Reference koo
shows that that this to the construction construction of the circle circle IzzI = = nri shows this isis impossible. impossible.
121. We We now now assume assume that that T* T*isisaadomain domain which which contains contains the point point w = 0 and and has the property property which which has just just been been proved proved to hold hold for for T0, namely, any any continuum continuumAA,,, whichlies liesinin T"' T* and contains T contains the the 0 , namely, 10 which point w ==0 lies in in every domain T,. T,, from from some some nn onwards. 0 lies every domain onwards. We We shall shall
120—122] §§ 120122]
75
THE KERNEL KERNEL
show, first .f. (z) then show, firstthat thatf0 (z) isis not not aa constant, constant, and, and, secondly, secondly,that thatT* T* is contained contained in in TT,. 0• Let to0 w 0 be Consideraa domain domain A., containing Let bean anarbitrary arbitrarypoint point ofT*. of T*. Consider points w0 w0 and w w = 0 and such that both both A,1, A. and frontier are the points and such and its frontier whenever nn is contained in Byhypothesis, hypothesis, A,,, A., is covered covered by T,. whenever contained in T*. By the inverse inverse functions functions .p,. (w) (w) of sufficiently large; sufficiently large; for these values of n, the defined in in AA,,,. r in absolute f,. (z) are all defined •. These functions are all less than rinabsolute value and so so form form a normal normal family. family. functions The sequence sequence {q,,. (w) (w)}} yields a subsequence subsequence {{q,,..t (w) (w)}} of functions A. and converge converge continuously continuously to function which are all all defined defined in in A,1, which to a function § 104, the function .p (w) is either either aa constant, constant, and, and, in that that case, Y, ((to). w). By By§ (w) is case, i,1'(w) (w) gives vanishes everywhere, everywhere, or equation zz = Y, since Y, (0) == 0, it vanishes or the equation conformal transformation intoaadomain domainlying lyingwithin within the the domain domain a conformal transformation of of Aw A into z
••••••
(12r1) (121"1)
it that it follows follows thatzkz sincez pointwithin within < r,f,.,.(zk)fo(zo) ;is0 isa apoint j zIzl
from which which it follows follows that that to0 =10 (z,,). Wo=fo (zo).
Now Wo w0was wasananarbitrary arbitrarypoint pointofT*; of T; thus thus /0 fo (z) is not constant, constant, and, further, w =fo (z) gives aa conformal conformaltransformation transformationof ofIz I< r further, the the equation equation to =f0(z) which must must have have T* as into a domain domain T,, To which as aa subdomain. subdomain. The domain T 0 , obtained from means of of the traniformation from IIz I <,.by
76
TREOREMS THEOREMS OF CONFORMAL CONFORMAL REPRESENTATION REPRESENTATiON
v [cHAP. V
the single single point point w == 0. In of the In all allother othereases cases the the kerrwi K of ofthe thesequence sequence is the largest domain domain having having the property that that every every continuum, continuum, containing w w = 0 and and contained contained in inK, lieswithin witkinevery every domain T,.for sufficiently K, lies (2'l). large values ·valuesofofnn(22). large The kernel of a sequence of domains domainsisisuniquely uniquelydeter,uined. determined. Suppose that thatw0, w0 , aa point pointof ofthe thewplane, wplane,has basaaneighbourhood neighbourhoodwhich which is is covered covered by all the onwards. Let Kw0 be the largest circle with the domains domains T,. T,. from some nn onwards. centre w, W 0 and such such that thatall allsmaller smallerconcentric concentric circles circles are are covered covered by by T,. for Now let leti.e0 w 0 vary within the thewplane. wplane. for sufficiently sufficiently large large values values of n. Now interiors of of all all the the circles circles Kw0 is either the nullset or an The sum of the interiors open ofnonoverlapping nonoverlapping domains. domains. open set which which may may be be regarded regarded as as a sum of If one of of these these domains domainscontains containsww==0, then that that domain domain is the the kernel kernel K. K. If 0, then Otherwise single point point w w = 0. Otherwise the the kernel kernel is the single 0domains is said to its its kernel kernel K KIJijevery 123. A sequence {T,.} ofdomains said to to converge converge to every subsequence T"'t., T,..., ... has thesamekernelas T,,}. theoriginalsequence the same kernel as the original sequence {{T,.}. domains 7',, T,. are assumed to be be simplyconnected, simplyconnected, there are, Since the domains by transformations by§ 112, functions.ft (z), /122 (z), (z), ... which give conformal transformations of 1;, T2, T2, ... on the circle circle Ijzj z I <1 < 1 and and satisfy satisfy ofT,, (1231) ff,,(O)=o, .. {0) = 0, f ..' {0) > 0. . ..... {123'1)
converge regularly regularly to to a function Suppose that the the functions functions f,, f,. (z) converge function Suppose fo (z) (z) in in this this circle. circle. Then, if/,. iff,,11 (z)J,.. (z),f,, (z), (z), ... is any subsequence, subsequence, limf,.t (z) =fo limfflk =f,, (z).
.. .... (123'2)
k
The The function function.fo .fo (z) transforms < 1into into the the kernel kernel.K' of the transformsthe thecircle circleIzzI<1 K of sequence {T,.k}. From is, the the sequence sequence follows that From this this it follows that lC K == K, that is, {T,.} {T,,}converges convergesto to its its kernel. The converse converse of this result result also holds: the the convergence convergence of a sequence sequence domains to its its kernel kernelimplies impliesthe theregular regularconvergence convergenceof ofthe thesequence sequence of domains the circle circleIIz I <<1. 1. For Forthe thefunctionsf,, functions/,.(z) (z)form form aa normal normal family. family. {f,. if,, (z)} in the If the If the sequence sequence {f,, {f,. (z)} were not convergent ititwould would be be possible possible to to find find two subsequences, ...... (123•3) ..,' (1233) f,.l (z), f ... (z), ...
...,, fm 1 (z), f,... (z), ...
...... (123•4) (1234)
converging regularly in converging regularly in Izz I<1 < 1tototwo twodistinct distinctfunctions functions 4cp (z) (z) and and 1{1 (z) respectively. But this this gives gives a contradiction, contradiction, for for both functions 4cp (z) (z) respectively. But 'p (z) (z)transform transformthe the kernel kernelof ofthe thesequence sequence{T,.} {T,,} into circle and 1{1 into thethe circle Iz zI <<1, 1, (1231), and also, also, from from (123'1), 'p(O)=O, cp (0) = 0, cp' (0) > 0; "'{0) = 0, 'p'(O)>O 1{1' (0) > 0 ....... (123•5)
§§ 123—125] 123125]
77
THE KERNEL
The uniqueness uniqueness theorem theorem for for conformal conformal representation representation (§ 68) shows
that q, (z)
=1/J (z).
124. 124. Examples. Letthe thewplanc wplaMbe becut cutalong alongthe thenegative negative real real axis axis from from the the (a) Let to the point point—  ! . Denote by In (z) (z) the function which which gives gives byf,, n transformationofofthe thecircular circulararea areaIz I <<11 into this a conformal conformal transformation this domain, domain, In conditions (123·1). (123i). Then Then 1.. (z) converges converges continuI,, (z) satisfying the conditions ously to zero, zero, for for the thekernel kernelof ofthe thesequence sequenceof ofdomains domains is is aa single single point. point. Actually, we find point — point 
a:;
4z f,.(z)= n(lz)~·
...... (12U)
(b) Let the wplane wplane be be cut cutalong along those those parts partsof ofthe theimaginary imaginary axis axis which lie above above the the point point i/n i/n respectively. which lie ifn and below below the point point—  ifn (z) — Consider Consider the function I,, fn (z) which which transforms transforms the halfplane halfplane .li (z) > 0
into this domain domain and satisfies the conditions conditions
.f,,(I)=1, 1,,'(l)>O. f,. ( I ) = 1, I,.' (I ) > 0.
. ..... (124•2) (1242)
The general theory shows (.z) converges converges continuously shows thatf,. that!,, (z) continuously to to zz in in the halfplane. This can that halfplane. can be be verified verified at once; once; for for calculation shows shows that f,. (z) =
1) + ..jf+n2( 1). 21(z + z ~ z:z
(1243) . ..... {124·3)
(c) Let (c) Let the the wplane wplane be be cut cutalong along an an arc arc of of the the unitcircle uni~circle joining the points e"•tn and ei"1", and of — 1/n). of length length 2n271" (1 (1A simplyconnected simplyconnected domain is thus formed, domain is formed, with the point point w w = a:; as as an an interior interior point. point. Let the interior interior of of the the unitcircle, unitcircle, Iz I <<1, 1, be transformed into domain be transformed into this domain function/,. (z) which satisfies the conditions (123'1). The by means of a function!,, problem is analogous analogoustoto (b), (b),and and itit is is found found that that problem is
lzcos,r/2n
_ 1  z cos 1rj2n /,n (z) =z • I,, ( Z) Z
cos 1rj"l.•z z
...... (124'4) (1244)
As we might anticipate, has a pole As we might anticipate, each each of these these functions functions has pole within within
Iz I<1. < 1.However, However,they theyform form aanormal normalfamily. family. This 'fhisisismost mostreadily readilyseen seen
if we that our domains can all be into those we observe observe that domains can be transformed transformed into those of example (b) by by the use of aafi.red fixed Möbius example (b) Mobius
h<~.nsformation.
125. Simultaneous Simultaneous conformal conformal transformation transformation of domains 125. domains lying each within another. problem is is important for The solution of the following following problem for the the theorems theorems by which which arbitrary analytic functions functions are are represented represented by by singlevalued singlevalued
78
THEOREMS CONFORMAL REPRESENTATION THEOREMS OF CONFORMAL REPRESENTATION
[CHAP. V V
Oonsider an an infinite sequence functions Chap. viii). vm). Consider sequence of of complex complex functions (cf. (cf. Chap. plane.'!, z,, z2, .::2 , ..., ••• , and let each each plane plane contain contain two two simplyconnected simplyconnected planes, z1, domains, origin. domains, one onewithin withinthe the other other and both containing the origin.
.r,.plane
Fig. 26
z,.+rplane
z,.plane are denoted by by 0~") and o~nl) respecThe two domains domains in the znplane 21, •.• tively. Takeaawplane wplaneand andattempt attempttotofind findininit ita asequence sequenceOw, 0 11,0 tively. Take Ow, of domains all containing containingthe thepoint pointw w == u, o, and domains each each lying lying within within the next, all function cp<"l (z,.), such such that the function (zn), with cp(") (0) = 0, (0} > 0, 0, cp'(ft) (0) which which transforms o<:> into 0"1, also transforms o~l) into O!Rl). The be such such that thatthis thisholds holds for for all all values values of of n. n. sequence {01"'} is to be into a domain Suppose that that the the domain 0~"> is transformed transformed conforinally conformally into r~"> in the the same same plane, plane, by by means means of a function function if! (z,.) with ifr if! ((0) 0) ==00 and 1/r' (0) (O) >>0, 0, and that that the thesame samefunction function transforms transforms 0~" 1> into the the domain domain 1 r~ >. Then Then the thetwo twooriginal originaldomains domains may maybe be replaced replaced by by their their images images ri:—'). affectingthe the problem problemjust juststated. stated. From without materially affecting From this itit follows follows that there in the assumption that the there is is no no loss loss of generality involved involved in assumption that domains o
lz,.l
.f.:t
.r<:!
where it is assumed assumed that where it 0. If r,, ,.,. is is fixed, fixed, = 0 and/~"~ 1 (0) >>0. 1 (0) = and can be determined so as t.o to then /'~"! 1 (0) is proportional to (0} tor,.+., andr.. +, determined so as ensure ensure that
f("l (0) = 0 n+l '
f,(n) (0) = n+l
1.
...... (12.5'3)
We take and require (1253) shall shall hold for all take r1 r 1 = 1, 1, and require that conditions ~onditions (125'3) hold for values of of n.n. Then values '11hen the the numbers numbers r,. are all uniquely uniquely determined. determined.
§§ 126128]
79
SIMULTANEOUS TRANSFORMATIONS SIMULTANEOUS TRANSFORMATIONS
126. The system of equations (n) (z) =f(n+l) (f(n) (z )) f n+2 n n+2 n+l " ' (n) (z,.) J
...... (126"1) (1261) ...... (126"2)
(zn) which can all be calculated calculated in in succession succession defines new functions f !:'~k (z,.)
are known known for for all all values values of of n. n. From if the functionsf!:'! 1 (zn) (z,.) are From (1 (125"3) it is seen seen that (1263) ...... (126"3) fhlZ+k(0)_0, ~n!k (0) = 0, f'~;!k {0) = 1. Now Now observe observe that that the relation w =f~~k (z,.)
...... (126"4)
into aa simple simple domain domain of of the the circle circle Iz,. I < r,. conformally conformally into transforms the n=1, 2,...), 2, ... ...,, n==1, 2, ... ), wplane. By(126"3), the functionsf~".!.k(z,.),(k==1, 2, so that they satisfy condition condition(/3) (/3) of of§ they form form a normalfamily. normal/amity. satisfy § 119, so given, aasequence sequence k~"1, k~n>, ... of natural numbers numbers can be be 127. IfIfnnisisgiven, kr, ...
found such that that the thefunctions functions (n)
fn+k(n) (zn) (z,.) p
(p=l, 2,...) (p=1, 2, ... )
...... (127•1)
form converges continuously continuously in in the circle form a sequence sequence which which converges circle Iz,.lj <
(n= 1,2,...) (n=1,2, ... ), ...... (127"2)
P+"'
within the z,,
(zn)
give conformal z,.l I <
was required. required. This is is exactly what what was hasjust 128. Theset setofofparticular particulardomains domains 01"1, which has justbeen been found, forms aa. 128. The figure a.bont about which some further further observations maybe may be made. made. From From (125"2)
80
THEOREMS OF REPRESENTATION OF CONFORMAL CONFORMAL REPRESENTATION
[CHAP. V [cEAP.
Schwarz's Lemma., Lemma, itit isisseen seenthat, that, if r,.. 1 ~ r,., then If'~"! 1 ((0)1 using Schwarz's 0) I <1. < I. that r,. < r,.+I and either Thus (1253) (125"3) shows shows that lirn r,. = lim
...... (12!:1"1) (1281)
OC),
or there is a finite number B such such that number R lim r,. = R.
. ..... (128"2) (1282)
These two two cases cases must be considered separately.
129. Denote by a,. the distance distance of of the the boundary boundary of of Cl"l from from the point 0. The Thefunction function ¢1"l (z,.) transforms the circle Iz,.l < r,. into aa simple simple w ==0. domain, and, and, by by (82"4), (824), domain, a,.> r,./4. . ..... {129·1) an>rn/4. infinity, so so that that Thus, if(1281) if (128·1)holds, holds, the thesequence sequence a1, ~. a2, a..z, ... tends to infinity, sequence of of domains Cl"l covers the whole whole wplane. wplane. the sequence j
130. Now Now assume assume that (1282) (128·2) holds. holds. The Theconstruction construction shows shows that B, so that, on (zn) I < r,..~: < R, If~!~~ (z,.) on proceeding proceeding to the limit, I¢1"1 (z,.) I < R 2 l, ... for all values of n. Thus 'l'hu!> the thedomains domain;; Ce), C(I), 0G12), ... lying each each within within the the next, are all contained in the circle circle Iw I <
I
>¥,.' (0)
~ ,.,. R .
. ..... (130·4)
shows that that Since ¢''"1(0) = 1, (130·3) shows >¥,.' (0) = "'' (0),
...... (130"5)
From (128"2) (1282) itit follows so .p' (0) (0) ~ Rfr,.. From follows that that 4" .p' (0) ~ 1, 1, and in in so that that 4" (1302) the the sign sign of ofequality equalityholds. holds. But But this this can can only only be be the the case case ifif D D (130"2) coincides < R, i.e. i.e. if if the the domains domains Cl"l tend coincideswith withthe thecircular circulararea areaIwwI
CHAPTER VI VI
TRANSFORMATION TRANSFORMATION OF OF THE FRONTIER FRONTIER
131. 131. An inequality inequality due due to to Lindelöf. Lindel8f. be an an arbitrary arbitrarydomain domain in inthe thezplane, zplane,containing containingthe thepoint pointz0. z0 • Let R be The domain R may be be multiplyconnected. Supposethat that there is an The domain R multiplyconnected. Suppose an arc of — z0 of the the circle circle IIzzz 0 I == r, subtending an angle 2,r 211"
— ot>n
at the the centre centre z0 z0 and domain R. Here nn isis aa positive positive at and lying lying outside outside the domain R. Here By means means of of rotations rotations about about z0 z0 integer. By integer. through angles angles 27r 4ir 2(n—1)ir
n'_tz_' ... •••''
the domain domain R R gives gives rise to to new new domains domains
R1, 14, ...,, R,.1respectively. The R2, ... Thecommon common ... R,._ 1 of all part RR1R2 RR1R2 ... all the thedomains domains the point is an an open open set set which which contains contains the point Pig. zo but which which has no point point of of the the circle circle Fig. 27 27 z0 has no z—z0I=r =r as asan an interior interior point point or or as as aa frontierpoint~ frontierpoint Among Izzo Among the the domains whose sum makes makes up this common whose sum common part there there is is one, one, 14, R 0 , which which contains the point z0. — zz00 I
1(z) bebea abounded 132. Let Let/(z) boundedanalytic analyticfunction functiondefined definedin inB, R,and andsuppose suppose
1/(z) ...... (1321) (132'1) If(z) I <M <M is an arbitrary B, lying B. Assume further that that if ~is Assume fuither arbitrary frontierpoint frontierpoint of R, in R. z — z0 within the the domain domain II zz0I <
(k=1,2, (n—i)) ....... (1323) ...,, (n1)) /~:(z)=/(zo +e.,. (zzo)), {k= I, 2, ... (132·3) .fk(z)=J(zo
82
TRANSFORMATION TRANSFORMATIONOF OFTHE TEE FRONTIER
[CHAP. VI [CHAP. VI
The function fk (z) (z) is analytic The function f~< analytic in in the thedomain domain11k. RJ<. It follows followsthat that the function F(z)=/(z)/I(z) ... fndz) ...... (132•4) F(z) =1(z) (z)
analytic in in R14. is analytic Thus, if {z,.} is a sequence sequence of of points points of of Ro 14 tending tending to 0 • Thus, a frontierpoint Cas limit, our hypotheses show show that that urn i
...... (132•5) (1325)
I
and hence, since (1325) (132·5) holds holds for for all all frontierpoints frontierpointsof ofB0, R0,
I F (zo) I ~ M" ; . By (132·3) and (132·4),
so that, finally, finally,
F(z F(z0) 0 ) = (/(z0 ))n,
1/(z.) I~
! j)f (';)".
. ..... (132•6)
133. The Thefollowing followingtheorem theoremfollows follows almost immediately immediately from Lindelof's inequality inequality (132·6): (1326):
Zf /(z) 1(z) is bounded and analytic analytic in in an arbitrary THEOREM. If bounded and arbitrary simplysimplycon necteddomain domain.!? whose frontiercontains contains mare merethan than one onepoint, point, and and if the1·e connected R wkose frontier there is a frontierpoint t with aa neighbourhood that If (z) neighbour/wad N~ such that (z) isis continuous continuou..~ at tko1<e points and frontierpoints of of Ii R which which lie lie within within N~, and takes takes the constant constant value value exa at tiwse frontierpoints, then then/(z) f(z) =ex. tkose/rontierpoints, suppose that point of of points points which which do do not not belong belong First suppose that Cis is a limiting point to B R or or to to its itsfrontier. frontier. Then, Then,ififz0z0isisany anypoint pointofof1? R sufficiently sufficiently near C, we z0 I ~ r lying entirely within N~ and having we can canconstruct constructaacircle circleIzz—; whichdo do not not belong belong to to R B or its points which its frontier frontier on on its its circumference. circumference. On On we obtain/ obtainf(z0) = a, applying applying Lindelof's Lindelöf's result result to the function function (f(z) (/ (z)— a) ex) we (z0) =ex, and the required required result resultfollows follows at once. once. If C does does not fulfil fulfil the above above condition, condition, we we consider consider the the function function F(t) — a.ex.Comparison in 65 F (t)=f =I+(Ct2) + f) Comparison with with aa similar similar substitution substitution used used in~ shows that this shows that this function function is is defined defined in in domains domains 14 Rt which which have have the the property limiting point The property that that t = 0 is is a limiting point of of points points exterior exterior to to Rt. 14. The argument used above above therefore serves serves for this this case. case.
134. 134. Lemma 1, on representation of of the frontier. Let Let aa bounded boundedsimplyconnected simplyconnecteddomain domainB,,, RtDbe begiven givenininthe thew—plane wplane and suppose suppose that the the function function w=f(z) .•.... (1341) (134:1) w—f(z)
§§ 133135] 133—135]
JORDAN JORDANDOMAJ.NS DOMAINS
83
represents itit confornially conformally on the domain domain R. in the zplane. zplane. We We assume assume that thefrontier frontier of ofB2 Rz is a closed closed Jordan curve the origins origins that the curve c,c, and and that the Ow and 0. lie within the respective respective domains domains Rw and B2, R., and are correcorresponding points in the the transformation. transformation. Let Yw be be aa cut into the interior By this this is is meant interior of of the the domain domain Rw. By a Jordan point ww of the Jordan curve curve joining joining an interior interior point point of of R"' to aa point frontier, and such that that any any infinite infinite sequence sequence of points points of of yfiJ either has frontier, at least one one limiting point within RfiJ or else else has uw as its only only limiting limiting point.
tran4formed into into a curve We shall prove prove that: that: The We shall The cut cut '"' is tran.iformed curve Vz y. which ofthe the densain dAJmain B2 R. bounded bounded by which isis aa cut into the interior of by the the Jordan Jordan cur·ve c. curve c. 135. we use use Jordan's Theorem, which states states that that 135. To prove this lemma lemma we Theorem, which c divides the the plane plane into intotwo two and andonly onlytwo two domains, and also also the the simplest
properties of the the domain domain R B2. point of the the Jordan •. These 'rhese are, are, that every point properties that every of points exterior toR., and that every point curve c is a limiting point of to interior. of ccan c can be bejoined joinedto toany anyinterior interior point point by by means means of ofaacut cut into into the interior. possible,that that sequences of points of Yz can be found, which Suppose, if possible, It will have M and H N of of cc as as limiting limiting points. points. It will be be have two two distinct points JI shown that this shown that this leads leads to to aa contradiction. contradiction. The two two points points.DI Atand andNH divide divide the Jordan Jordan curve curve cc into into two two arcs, arcs, The the positive positivesense sense and and choose, choose, cc1 andcc2. Imagine the the curve c described in the 2 • Imagine 1 and B, CCand and D, D, of of which which the thefirst firsttwo twolie lieon onc1 c1 and and in order, four points A, B, the others others on on c2. c2 • Let the the four four points points be N joined to· 0, 0, by by four four cuts cuts without without other other joined to intersections than that that at 0.. These 'l'hese cuts intersections than four subdouains. subdomains. By By hypohypo C divide divide R. B, into four thesis, thesis, we we can can find on Yw y,,,aasequence sequence of of points points which converges convergestotow,w,but but such such that the the which correspondingpoints pointsinin the the zplane corresponding zplane lie lie DA and in the alternately in the domain U=DA domain OZBU. An arc arc of y. O.BC. An y, joining joining two two domain successivepoints points of of the transformed successive transformed seseFig. 28 quence must of the the Fig. 28 quence must cross cross one one or or other other of domains 0ZA13and ando.cD. 0,CD. To fix the the ideas ideas suppose suppose that that there is domains O.AB To fix is an an infinite number of of these these arcs arcs crossing crossingO.AB, OZAB,and andlet letthe the parts parts of of the the infinite number arcs lie in in this and have O.A arcs which which lie this domain domain and have their their end end points points on on 02A (1) (2) (1) (2) • (1) (21 Ofl Yw are O,B be and O.B bfl ~, ,, a= , .... The corresponding correspondmg arcs aw, a,,,, ... on .
84: 84
TRANSFORMATION OF THE THE FRONTIER TRANSFORMATION OF FRONTIER
[ca&p. VI [CHAP. VI
!
all distinct and tend uniformly tow. uniformly to 111. The arcs ~1 1, ~1: 1, .•• ... tend uniformly to the the arc arc AB AB of of c.c. Consider Consider the domain O;AB. On AB AB choose choose a the domain OZAB. On point E whose pointE whose distance from from the the arcs arcs BO. and 02A O.A may be denoted by 217> distance from 2"7 > 0. In the the domain domain OZAB O.AB take take aa point pointzo z0 whose whose distance from E is is than "7, and with centre z0 drawaacircle circleKK of radius "7· The point less than z0 draw pointE E
Fig. 29
lies within K K and is a limiting point of points exterior to to 14, lies within limiting point points exterior R., so that there is an arc K1 of KK lying lying outside outside R14. K 1 of •. Let ex be the angle subtended z0 by choose aa positive 2rfn. at z0 by this this arc, arc, and and choose positive integer integer nn such such that ex> 2ir/n. Join 0. to z; 0by bymeans means of of an an arc arc yy lying within OZAB. O.AB. If Ifkk isissufficiently sufficiently k>k0, k 0 , ~~:>has say k> has no points in common common with with:y, y,so sothat that the the domain large, say O.F~ 1 :1 G (Fig. 29) is a subdomain of OZAB O.AB containing z 0 but not not containing any point of K1 K 1 either frontier. either in its interior or on its frontier. Let ~ be be an arbitrary arbitrary positive positive number. number. Denote Denote by 11 hi the the (finite) (finite) diameter of B,,, and choose choose kk greater greater than k0 and also also such such that the Rw and ko and the arc lies entirely within the circle ~':]lies circle Iw W1111
apply Lindelof's Lindelöf's inequality to the — w), connow apply thefunction function (1(z) (j(z}111), We now
G. The sidered sidered in the the domain domain O.F~ 1 :1 G. The points points of of the the frontier frontier of of this this domain which whichlie liewithin withinKK are are all all points points of ~ 1!1 , so that that we we may may write write domain m= m=Me". £ was arbitrary, that 11(z0) The relation (132 (132'6) shows shows that l/(z0)—  ww I<<Me. hk Since Since (was = c'i. /(zo)=w. gives the the desired contradiction; =1(z,) This gives contradiction ; for for the thepoint pointw0 Wo =I (z0) lies within Rw, whereas o,w is a frontierpoint of of this this domain. domain. B,,,, whereas
136. Lemma ~. 2. 136. Considertwo twocuts cutsrw' y,,,'and and',,,," intothe the interior interior of of the domain Consider rw" into domain B,,,. Rw. We suppose that that they We suppose they join join the point point Ow to the the points points w' w' and w" of the
§§ 136, 136, 137]
85
JORDAN DOMAINS DOMAINS
frontier and have have only onlythe thepoint point 0., 0,,,in in common. common. The The result result just just proved shows that thatVw' y.,' and ye" y.," are images of are the images of two twocuts cutsinto into the the interior interior of B2. R •. These These cuts cuts y,,' y,' and y." join the point point o. to the points points C rand(;' and C" of the Jordan Jordan curve curve c. c. We Wenow nowinvestigate investigatenecessary necessary and sufficient sufficU!nt conconditions ,, and r' should slwuld coincide. ditio,zs that C' curve (ye' (yz' + +yz") 14, into into two domains R 1 and R'!l. If(! The curve ye") divides R. If C' and C' r' coincide, then, since ccis curve,one oneof of the the domains, domains, say say R, and is a Jordan Jordan curve, is such that all points of "tz' or of y." or of both. all its its frontierpoints frontierpoints are points both. This domain corresponds to a domain, domain, B R':] say, say, which which may have have points corresponds to the frontier of R., B,,, as as frontierpoints, frontierpoints, but but such a frontierpoint o(j) of B R~> of the cannot be at aa positive distance from the curve (y,,,' + positive distance from the curve (y.,' y.,''). For suppose that such a point. that ww is such point. There 'rhere isisaaneighbourhood neighbourhood N., of ww such that if w,, w,,W2, w2 , ... ••• is is a sequence sequence of of points points converging converging to to aa point point of of the the frontier frontier 1(z), say of R"j R::!within within14,. N.,,, then the inverse of the function functionf(z), say 4, q, (w), takes 4, (w) By § 133, values converging tor. By§ (w) is constant; butthis thisisisimpossible. impossible. to constant; but 133, cp has On the other other hand, hand, if B R':] property in in question, question, so so that, in On the property w", then rC' == '"· particular, w' This is is readily w' == w", readily proved proved if the the above above C". This argument is applied to the transformation w==1(z) and the domain transformation w f (z) and domain argument is applied to
!,
R'~,. z
137. Transformation of (23). 137. Transformation of one one Jordan Jordandomain domaininto intoanother another (23). assumethat that the frontiers of both the We now now assume the domains domains 14,, R., and 14, Rz are Jordan curves, denoted by by c,,, c., and and c,.. C2 • Let w w be an .Jordan curves, which which may may be denoted to,,, ... denote an arbitrary arbitrary point arbitrary pointofofc,,, c., and and let letto1, w1 , w arbitrary sequence sequence of of 2 , ••• points of R., tending tendingto tow. w. The corresponding pointsofofB,. R.are arez1, z 1 , z,, z;., .... corresponding points Since c., c,,.isisaa Jordan Jordan curve the points points w1, w,, w,,, w2 , ... ••• may be joined, joined, each to its itssuccessor, successor, by byaasequence sequenceof ofarcs arcswhose whoseaggregate aggregateforms forms aacut cuty,,, y 111 from ww into the the interior interiorofof11w. R.,. C of Cz into the By § 135, 135, yy,,, theimage image of of aa cut "t• from some point By§ point' 111 isisthe ...,,so so that interior of B,,. R •. Also, Also, y. contains the the sequence sequence of of points points z1, z1 , z,,, z2o ... of the sequence this sequence sequence must must converge converge to to C' '· The convergence convergence of sequence is simply simply aa consequence of the convergence {z,.} to C 'is consequence of convergence of {w,.} to w. w. It It follows that the C to converges depends only follows that the point point' to which which {C,.} converges only on on w, w, not choice of the thesequence sequence{w,.}. {w,.}. A A point point CCcorresponds uniquely to to on the choice each point since R. I?, and pointo,, w, and, since and B,,, R., may be interchanged without without modimodification corresponds uniquely uniquely to to each the reasoning, reasoning, aapoint pointww of c,,, c., corresponds fication of the point ofc,,. c•. pointCCof w22 of c.., c,,, have have two It is is seen seen at at once once that thattwo twodistinct distinctpoints pointsw1 w1 and w distinct corresponding andC2 C2 of ofc,.. c.. (This also follows follows from from distinct correspondingpoints pointsC1C, and
86
TRANSFORMATIONOF OF THE THE FRONTIER TRANSFORMATION
[cwtP. VI [cHAP. VI
§§ 136.) Further, Further, the transformation transformation of other is is of one one frontier frontier into into the other continuous. continuous. For Forlet letw1, ~, ~ •... to w0, w 0 , and let w2, ... be points of Cw converging to pointsofof;. ~~> '2> ... be be the corresponding corresponding points c•. that the If klc is aa positive If positive integer, integer, aa point pointwk wk can be be found in in Rw such that relations 1
1
are satisfied satisfied simultaneously simultaneously by by wk and and its its image image Zk. z,.. But the the points points w to w0. w 0 • It also follows that that Z1o ~ •... ...,, and therefore also w1, w2, ... converge to It follows 1, w 2 , ••• ...,, tend to a point the points ~I>~ •••• point~w2and whichare arepassed passedthrough through w1, w Finally, let let~. w 3 be three points points of ofc,,, cw which 2 and w3 in this this order orderwhen when the thecurve curveisisdescribed described in in the the positive positive sense. Suppose Suppose that cuts into the interior of Rw join Ow to the thatthree threecut<>intotheinteriorof the respective respectivepoints pointsw1, w~> "'s and w3, w 3 , and that these these cuts do do not not intersect intersect one oneanother anotherexcept exceptatatOs,,. Ow· We consider that the consider the the corresponding corresponding figure in the zplane and observe that the is conformal conformal at Ow· Then 'fhen it is clear dear that the the points points~~. transformation is occur on on c. c2inin this this order order when when the the curve curve isis described described in in the ~2 and ' 3 occur positive sense. These results results may may be be summarized summarized as as follows: follows: into another, another, THEOREM. coriftYrmally into THEOREM.IfIfone oneJtYrdan Jordan domain domain is traniftYrmed ccenformally the transformation isis oneone oneone and continuous continuo1ts in the the closed closed domain, domain, then the and and the the two tu'o frontiers frontiersare aredescribed describedininthe thesame samesense sense by by aa moving moving point point one and and the corresponding on one ctYrresponding point on on the the other. other.
of this this theorem theorem isis easily easily made. made. Let Rw 138. A slight generalization generalization of domain, and c,, Cw a a Jordan Jordan curve curve(with (with ororwithout withoutits itsend endpoints), points), be aa domain, conditionsare aresatisfied: satisfied: (a) every point and suppose suppose that the the following following conditions c,,,can canbe be joined joined w of c,, c,., is frontierpoint of Rw, (b) (b) every point point ww of Cw is a frontierpoint (c) every to any by aa cut into any interior interior point point 0,,, Ow by into the the interior interior of of Rw, (c) Jordan w1w2 c,, and and two .Jordan domain domain whose whose frontier frontier consists consistsofofa aportion portion w1 w2 of Cw two cuts into into the theinterior, interior,O,,w, O,w1and andO,,w2, O'"w2 , lies entirely within within Rw. Then Rw JtYrdan curve. curt•e. 14k,isissaid saidtotocontain containa/ree aires Jordan Suppose that w, § 136 showsthat that the cuts Suppose w1 =!= w Then § 136 shows cuts Ow~ and • (02. 2 two cuts cuts 0.'1 and O.Co, where ' 1=1= * ( 2 • The transO,w 2 are the images of two formation the interior of the formation ww==f(z) f (z) then transforms transforms the the Jordan Jordan domain domain ofof thethe Jordan domain OWWIW2OW. § 137, 0.,1 { 2 0. into into the theinterior interior Jordan domain Oww1w2 0 10 • By By§ 137, any arc any arc of of the the free free Jordan Jordan curve curve c.,., and hence hence the the whole whole curve, curve, is is continuous image image of of an an arc arc of of the frontier c. of B,,. a oneone oneone continuous R •. Just as not the image as in in §136 § 136 ititmay may be be shown shown that that c" .is not image of the whole frontier of R. except when c,,, the whole whole frontier frontier of B,,,, c, isisthe Rw, i.e. when when Ru is a Jordan Jordan domain. domain.
§§ 138, 139]
87
INVERSION
139. Inversion with respect to an analytic 139. anal:vtfc curve. A curve. in the xyplane :cyplane is given either by by an an equation equation A real real analytic curva F (x, y) = 0 ...... (139"1) (1391) or in parametric form form by two equations y=ifr(t). . ..... (139"2) (1392) X=
z=f(t)=4,(t)÷ii/í(t) z=f(t)=cfJ(t)+i.P(t)
.•.... (139"3)
as an analytic function function of of the the complex complex variable t in in the theneighbourhood neighbourhood real point point tt0. hypothesis f'(t,)0 ) *=1= 0, t — t, By hypothesisj'(t 0, so that that aa circle circle Itt 0 I
Substitution of of these these values values in in (139.1) (139'1) gives gives
F(z+w ~)=o ...... (1391) (139'7) 2 ' 2i ' and from from this relation w w can can be be calculated calculated as an an analytic analytic function function of of z. z. The relation (139'7) (1391) shows, in particular, particular, that that the shows, in the operation operation of of inversion depends only only on on the the form form of of the the curve curve (139"1), (1391), and and is independent independent of depends t in(1392) (139'2)(24). (24). choice of the parameter tin the choice G G
88
TRANSFORMATION OFTHE TEE FRONTIER TRANSFORMATION OF
[cn&p. (CHAP. VI VI
140. The equation equation (1391) (139"7) is is especially especially convenient convenient when algebraic 14.0. The when an algebraic curve (139"1) represents represents a straight straight line line or or a circle, circle, then the curve isisgiven. given. If If (1391) corresponding formulae formulae (139"7) (1391) are exactly the inversion inversion formulae given in Chapter i. 1. In the the case case of of the the ellipse ellipse
a'
...... (140"1) (1401)
b2
(1391) becomes (139"7) becomes 1 ) (w' 2 )zw+ (a' (a2— bb') (w2 + + z') z2 ) 2 2 (a (a22 ++ bb') zw + 4aW= 4a'b' = 0 ....... (140"2) (1402)
If this this isissolved solved for for ww aatwovalued twovalued function function of z is is obtained. obtained. This This the zplane zplane except except at atthe thefoci foci of of the the ellipse function is regular throughout throughout the (140"1 ). The corresponding transformation 2b2 = 0 ...... (1403) (a2 b2 )(z*2 + zt) 2 (a 2 + b2) z*z + 4a (140"3) 4a'b'=O can only be regarded regarded as an an inversion inversion with respect respect to to the the ellipse ellipse when when lie within within the ellipse to (140"1) (1401) and the points points transformed tranRformed lie ellipse confocal confocal to passing passing through the point a2 +
'.Ja2_.b2'
141. 141. The The inversion inversion principle. principle. Let B, R., be be aasimplyconnected simplyconnected domain domain in in the vplane, vplane, containing containing aa segment AA1B1 ofthe the real real axis, axis, and symmetrical with respect segment symmetrical with respect to this this 1 B 1 of axis. Suppose = 'Ji(t) (t) gives a conformal aXIs. Suppose that that the therelation relation v"'=.; conformal transformatransform&
A11
B2
vplane vpl&ne
Fig. 30 30
tplane tpla.ne
tion of R., B, into the the interior interior of of the the unitcircle unitcircle in the tplane, tplane, the the transtransformation being being such such that that the origin t ==00 corresponds 0, formation corresponds to some some point Ov of A1B1, A 1 Bh and that thatiJ/ .;· (0) >0. > 0. The Thefunction function if! (t) i'l uniquely determined Further, if the these conditions conditions (§ 112). Further, the two two figures figures are are inverted inverted by these with respect to A1B1 A 1 B 1 and A,B2 A 2 B 2 respectively, respectively; they are transformed into into themselves. From this itit follows that themselves. follows that ;jr (l) =.; (t), ...... (1411) (141"1) where ~ and Itare are the the numbers numbers conjugate conjugate to if! and t.t.
§§ 140143] fi 140—143]
INVERSION PRINCIPLE INVl!;.RBION
89
The relation (14r1) (141"1)shows shows that thatip.p (t) maybe may be written in the the form form of a powerseries coefficients, so thesegments segmentsA1 A 1B1 B 1 and and A2 A 1 B2 B2 powerseries with with real coefficients, so that that the correspond to one correspond to one another. another. Hence R~' and Rc' are corresponding corresponding domains.
142. To Toobtain obtainaaconformal conformal transformation transformation of of R~' into the interior interior of z <1, = 1/f (t) to transform a unitcircle unitcircle lzl < 1,we wemay mayfirst first use use the the relation relation vv= 14' R.,' into R/ and then a relation relation t = 4q, (z) to transform transform R/ into into Iz I<1. < 1. The function q, (z) already known (§54); (z) is already 54); it transforms the circular area. t I <1 < 1 into the splane zpla.ne cut cut along along an arc a.rc of IzI=1. Iz I = 1.Two Twopoints, points,z1 z1 area IItI correspond to which are are inverse inverse points points with with respect respect to JIz I = 1, correspond and z2, Zs, which two points which whichare aresymmetrical symmetricalwith withrespect respecttotothe thereal realaxis axisin in the two points tplane. t·plane. thereforetransforms transformslzlz <1 = iji (Sb (z)) therefore The transformation vv=l/f(f/l(z)) <1 into into Rv' in such a way that the points points z 1 and z 1 correspond to points points v1 v1 and and v2 v2 which which are symmetrically symmetrically placed with respect respect to to A1 A 1 B1. B 1 • The The segment segment A1B1 istransformed transformedinto intoan an arc arc of of the circle IzzI= A = 1, 1, and the function function 1 B 1 is 1/1 (4, (f/l (z)) is analytic on this arc. 143. Let Letc,,, Cw be be aa regular regular analytic analytic curve, curve, and suppose that the relation relation that the w = x (v) ...... (1431) (143"1) w=x(v) tra.nsforms it into a segment of the Letaa domain domain transforms the real axis in in the the vplane. vplane. Let be trans14,' wplane, having R 11,' in the thewplane, havingan anarc arcofofc,,, c.., as a free free Jordan curve, curve, be
("
w
wplane
Fig. 31 31
vpla.ne vplane
formed formed by by (143"1) into aa domain domain R1'. R.'. Let Letthe thedomain domain11w", R,/, which is transformed conobtained by inversion of 14,' with respect to obtained inversion R 11,' with c"', be conformally, by means means of of(143"1), (1431), into into the inverse R.," 14" of 14'. formally, by R.,'. In order to to obtain obtain aaconformal conformal transformation transformation of Rw' into the circle circle necessary to to substitute substitute the function v = '/' IzI z I <1, < 1, it is only necessary •/t (4, (q, (z)) of§ 142 in equation (143"1). We obtain the the transformation transformation ...... (143"2) w=f(z), I
90
TRANSFORMATION TRANSFORMATIONOF OFTHE THE FRONTIER
[CHAP. VI (CHAP.
which has the following properties: the arc B of c10 , which is part of which has following properties: are A AB of Re', of an an are arc of ofIzzl == 11 and on this arc the frontier frontier of R arcf(z) 10', is the image cf 1(z) is an analytic analytic function; function; ififz1 z1 and and z2 z2 are inverse points with with respect respect to to IzzI== 1, the corresponding corresponding points pointsw1 w1 and W2 w3 lie lie one in R 10' and and the other w2 are inverse inverse points with respect re~pect to c10 • in R 10", and w, and w. 144. Let Rv:' be be the domain domain just justconsidered, considered, and andlet letRL,' Rv' be domain be a domain circle and and having an arc A 3 B 3 of the circle as part of of lying within some circle arc A3B3 frontier. Suppose = F(v) ——— its frontier. Suppose that that the the relation relation w w=P(v) gives a conformal conformal transformation of Rv' into R 10' /"" ", "B3 in which in which the circular circular arc arcA3 A 3B3 B 3 corresponds corresponds to / B3 the arc AB of of C10 • Then F(v) F(v) has, has, on onA3B3, A3 B3 , / exactly the same as were exactly same properties properties as were found found to \ hold for 1(z) in for f(z) in§ 143. 143. For For P(v) F(v)isisobtained obtained by by \ R'v elimination of zz between equatiom w = f (z) ''... . between the equations =1(z) and v == gg(z), (z), the latter latter being being aa relation relation by bv which which A3 Rv' is is transformed transformed into Izz II <<1. 1. Fig. 39 32 11w' Now Now let R~ be an arbitrary arbitrary simplyconnected simplyconnected domain thewplane, domain in the wplane, having an arc MN MN of having of an analytic analytic curve curve c10 as curve. Let as a free Jordan curveS w=F(v) ...... (144"1) (1441)
t!)'
\
into the the circular circulararea areaIvI v <1. transform R~ into < 1. N In general itit isis not notpossible possible to to invert invertthe thewhole U'hole respect to to cce,, so that that the domain R~ with with respect 10 , so of§ used. We We know, know, method of § 143 cannot cannot be be used. however, that that the arc MN of c10 is transformed however, ~ircle IvII = 1, and that this into an arc arc of of the circle M oneone continuous continuous transformation(§138). transformation 138). is a oneone Fig. 33 arbitrary point point of of:J.IN. MN. We Now let let w w be an arbitrary We which can can be inverted with respect find a subdomain can find subdomain R 10' of R~, which which has, an arc arc AB ABofofca,, c10 , where where AB AB to c10 and which has, as as part of its frontier, an has was The relation relation (1441) (144"1) was an an interior interior point pointand andisisitself itselfpart partof of:J.:IN. MN. The transforms R 10' into a domain R v' which has all the properties mentioned the'beginning of this paragraph. paragraph. at the"beginning Thus F(t•) F(v) is is analytic analyticatat all all points pointsofofthe thearc arcofofIvi= v = 11 which corrc corrc · Thus Further, every sponds to c10 • Further, every interior point point of of this this arc arc has has aa neighbourneighbourhood that aapair pairofofinverse inversepoints pointsv1 t•1 and aud v2, v2 , both lying lying in the the hood such such that neighbourhood, are transformed into pair of of inverse points points with with respect respect into aa pair to ce,. c..,. These facts constitute 'rhese constitute the thefamous famous Principle Principle of ofSymmetry, Symmetry, or or Inversion Inversion Prin('iple, uamc of Sctiwarz Schwarz (25). l'ririeiple, associated with tltc the iiarne I
§§ 144146] 144—146]
TRANSFORMATION OF OF CORNERS TRANSFORMATION
91
LetBRbebeananarbitrary arbitrary simplyconnecteddomain domain whose whose frontier frontier 145. Let simplyconnected Suppose that the function function consists of consists of more morethan than one one point. point. Suppose that the ...... (145·1) w =f(z) gives aa conformal conformaltransformation transformationofofthe thecircle circleIzzI <1 < 1 into into the the domain domain fl. R. Chapter ii, rr, we now circle as a nonEuclidean nonEuclidean plane. As in Chapter now regard regard the the circle The nonEuclidean straightlines linesininIzz I<1 nonEuclidean straight < 1are aresuch suchthat thatififthe thecircular circular transformed into itself. one of of them them ititisis transformed area is inverted with respect to one
An analytic curve in the corresponds to to one one of of these these nonthewplane wplane which corresponds R into into two two subdomains subdomains R' Euclidean straight straightlines linesdivides divides B Eucidean B' and R", B", and these are interchanged when when they theyare are inverted inverted with with respect respect to to the curve. be regarded as lines lines of of symmetry in R. R. these curves curves may may be Thus all these
146. Transformation of corners,. 146. Let RIIJ be a domain frontier contains contains the the free free Jordan curve domain whose whose frontier curve and let A A be be an an interior point of which the MN, and of MN MN at at which the portions portions AM AM and AN we denote denote by by AP AP ANof of the the curve curve both bothpossess possess tangents, tangents, which which we A Q respectively. and A respectively. Then RVJ has aa corner cornerat at A, A, and this is measured by the angle angle ex between APand andAA.Q. Q. between the .two two directions AP If RVJ is transformed into the circular area I z I
p
M
Fig. 34
A
Fig. 35 35
AMand AMandANand AN a.ndlying lyingentirely entirelywithin within1?,,,. R •. The The function function z == .P 4 (w), which of the the unitunitI < 1,transforms transforms 8 111 into into aa. subdomain subdomain S1 Bz of transforms R. into Izz <1, subdomain has A1 A 1 as onlynecessary necessary circle, a.nd and this subdomain as aafrontierpoint. frontierpoint. ItItisisonly to show that that angles a.t at A A. are a.re transformed transformedproportionally proportionallyinto intoangles anglesat atA1. A.1.
92
TRANSFORMATION OF OP THE TRANSFORMATION TUE FRONTIER
..
[CRAP. VI (CHAP.
Now if A isis the thepoint pointw0, w 01 the the transformation transformation uu=(ww Now = (w — w0)0) ; represents 5,,, on on aa semicircle semicircle 8,., 5,,, and 5,, But the 8tD 8,. can then then be be transformed transformed into into5,,. 8.. But ratio of of the the angles angles in in question questionisisclearly clearly preserved preserved in the the transformation of 5,,, 8tD into 8,,, 8,., and the the second second transformation, transformation, 8,, 8. into into $,,, 8,, is is analytic analytic (§ 143) and therefore therefore conformaL conformaL
147. Our Ourproof proofof ofthe thegeneral generaltheorem theoremdepends depends upon uponthe thefollowing following lemma: Let the domain 8, but but suppose supposet.hat that RB and and 8S domain B R contain contain the domain domain 8, have the free Jordan GB as as part of have Jorda~ curve curve A AOB of their theirrespective respective frontiers. frontiers. B and S8 separately into the same circular area. area KK Let R sepa.T&tely be transformed transformed into same circular in such a way that both cases the ends A curve that in both A and B of the Jordan curve
lB1
B
Fig. 36
GB are transformed of the the circumAOB transformed into two two fixed fixed points points A1 A1 and B1 B 1 of circum· A ference of Now let 1 be aa circular arc within K, K, joining joining A1 A 1 and B1, Bu ference of K. K. Now within the and let let )'R 1a and y8be 1s be its its images. images. ItItwill willbe beshown shown that 1 8 lies within (YB and domain whose whose frontier frontierconsists consistsofofAAOB andYR• 'Ya· transformation,the the circle circleKK is is transformed transformed into a If, by a conformal conformal transformation, being made semicircle, the arc arc A1C1B1 A 1 0 1 B1 being to correspond correspond to the the diameter diameter A, A 2B,, B 2 of then the the curve the semicircle, semicircle, then curve y 1 is transformed into a circular arc passing transformed through We may may assume assume through A, A,,a~d and BB,. 2 • We that A,,B,, lies on onthe the real real axis. axis. From that A 2 B 2 lies From A this ititisis seen seen that that aatransformation transformation of Fin. 37 37 Rintoasemicircle, with correspondence Fig. Rintoasemicircle,withcorrespondence between Jordan curve curve A AOB diameter, transforms transformsYE 1a into between the Jordan GB And the diameter, passingthrough through AA,, andBs. B,,. The same transformation a circular arc arc YR' 1a' passing 2 and represents domain 8' S' lying lying within within the thesemicircle, semicircle, while while 1s represents 8S on aa. domain is is and joining A,, and curvey3' 1s' within the the semicircle semicircleandjoiningA, and B,,. B,. transformed into intoaa curve transformed lies between YE' and A,,B,,. between 1a' and AtBs. We have only to show that that 1; curve obtained obtained from from liz' 'YB' by means of be regarded regarded as the curve Now 'Ys' may be the transformation on 8' 8' transformation which which represents the semicircle semicircle conformally conformally on
§§ 147149] 147—149]
TRANSFORMATION OF CORNERS
93
into itself. and transforms transforms the diameter A2B2 A 2 B 2 into itself. The Theinversion inversion principle principle shows that the function which sets sets up this shows that function which this transformation transformation is analytic analytic throughout the circle whose whosediameter diameter isis AA2B2. real 2 B 2 • The function takes real onA2B2. · values on A 2 B 2 • ItItisisnow nowseen, seen,as asaadirect directconsequence consequence of the theorem of§ y; lies lies between between 7A' yl and and A2 A 2 B2. B2• that 'ye' of § 88, that 148. We We are are now now in a position position to prove prove the the theorem theorem of of §§ 146 in the whereRB has has aa corner corner A, A, one oneof ofwhose whosearms, arms,say sayAN, AN, is isaastraight straight line. case where before, let As before, let excxdenote denotethe themeasure measureofofthe thecorner corneratatA. A. Now construct two two new new domains, domains,R'Hand andR", H', as follows: follows: K H isis aa domain domain containing containing R. B. · Its frontier contains AN ANand andalso also aa segment segment AM' AM'which which makes makes an an angle angle (cx + e) with with AN. in R. B. Its (ex+~> AN. The Thedomain domainB" R"isis contained contained in Itsfrontier frontier contains contains AN and AN and also also aa segment segment AM" AM"which whichmakes makesan anangle angle(oc (ex~) with AN. — €) with threefunctions Take three functionsby bymeans means of of which the domains R, the three domains B, R' B' and and K' B" are transformed into a fixed fixed circle, circle, in in such aa way way that the the segment segment AN ANcorrecorresponds same circular sponds to to the same circular arc arc A A1N, 1N 1 R'' in each each case. case. Let 'Yo be a circular in circular arc arc N 1 , lying lying within within the joining A joining A,1 and N1, circle and and making circle making an angle angle 1r8 with Fig. arc A,N,. A 1 N 1 •Consider Consider the theimages images Fig. 38 38 the arc 147, y, jy'and andy" of this this arc arc in in the the three three domains domains R, R". By By§§ 14 7, ylies y, y" of B, R', B', B". y lies between between y'1' and and y"; j'; §§146 146 shows shows that the the curves curves y' y' and and y" y" both both have have (cx — €) re+ €) and at A, A,these thesetangents tangentsmaking makingangles angles0 (cx 8(ex+~) and0 8(ex~) tangents at spectively Since ~ is arbitrary arbitrary and andy always lies spectively with with AN. A N. Since y always lies between betweeny'j y", itit is is seen seen that that yy itself itself has has aa tangent tangent at at AA and and that this tangent and y", makes an angle withAN; AN; for for the contrary 8ex with contrary assumption assumption would would lead makes angle Ocx contradiction. to aa contradiction. This proves proves the the theorem theorem of§ of §146 146for for domains domainswhich whichhave have a.a straight line as one arm of the corner corner A. § 146 can be be made made to to depend depend on onthe the result result just just 149. The Thegeneral generalcase caseof of§ obtained. Let B with a corner obtained. R be be the the domain domain with corner at A. M Consider Consideraa domain domainKB' which which contains contains RB and has has a corner arms isis AM and the eorner at A, one of whose whose arms other a.a straight line line AK. AK. First Firsttransform transform F' R'1nto 1nto aa. halfplane at the the same same time time B R is is transformed transformed into halfplane;; at B, with with a.a corner corneratat Au A,, the the arm arm A1M1 A, M, A4 a.a domain domain Rt corresponds to to AM AM being now which corresponds now aa straight line. In this transformation transformation of and aa. further further of RB into R B,,1 , and ratio of of angles angles transformation of B1 R 1 into a circle, the ratio transformation of Fig. 39 at A is is unaltered, unaltered, as as isisshown shown by by §§ 148.
94
TRANSFORMATIONOF OFTHE THE FRONTIER FRONTIER TRANSFORMATION
(cHAP. VI [cHAP. vi
particular,ififthe thefrontier frontier of of the domain domain R 150. InInparticular, B has has a.a tangent at .A, A, = ir, then the so that A A is is aa corner comer with with ex=.,., the transformation transformation isis isogonal isogonal at at A. Iff(z) function and if' if is lff(z)isisthe the functionwhich whichgives gives the the transformation, transformation, and the point of of the the circumference circumference corresponding corresponding to A, then then isogonality isogonality is is expressed function expressedanalytically analyticallyby bythe thestatement statement that that the function arg
,_z
f(,)f(z)
...... (150•1) (1501)
tends to a constant as z tends to '· An analytical proof of isogonality isogonality at Lindelöf, who used used (150"1) (1501) and A has been given by Lindelof, and Poisson's Poisson's integral integral(23). (23). showed that that if the curve MAN is He further showed is smooth smooth (glatt) at A, then the function arg f' f' (z) which has has aa tangent (z) is continuous continuous at at '· (A curve which at A is is said said to to be be smooth smooth at A A ififevery every chord BG BC tends to the tangent as B and C Gtend tend to to AA simultaneously*.) simultaneously*.) It should be function f' (z) is not necessarily necessarily finite be noticed noticed that that the function /' (z) by Lindelof. Lindelöf. For in the the neighbourhood neighbourhood of ' even even in the case case considered considered by instance, the function instance, ...... (1502) (150"2) w=—zlogz, w=z logz, where z = r~, transforms the domain domain lying lying between between the the imaginary imaginary axis the the curve curve ir ir cos4 log r =  .p c?s .p f < .p < ?:2 and  '?2: < .P < sm¢'
]ltz=O ~ = 0 and
(o
o) ,
the domain domain lying lying between between the the convex convex into the
w=~r+it logr,
curv~
(O
— and and w = ~. But the and the the real real waxis waxis cut cut between between the the points points w w= = ~ derivative of (150"2) is w'=—(l÷logz), ...... (150•3) w' =(1 + logz), 1
and tends to oo oo as as zz tends tends to tozero; zero;arg argw', w',however, however, tends tends to to zero. zero.
151. The Thetheorem theoremofof§§93 93 may maybe be completed completed and and can can then thenbe beapplied applied in many and leads leads to to a.a more resultthan than that that just in many cases, cases, and more precise precise result given. * The "The
curve y=f(x) with withf(O)=o,f(x)=x2sin!: forz*O,hasy=Oastangent CUl'VP y=f(z) /(0)=0, j(z)=z2sin!: for z*O, hasy=O as tangent
at the the point point xz = =0, but is not smooth at this this point. point.
z
§§ 150, 151]
APPROACH TO APPROACH TO THE THE FRONTIER
95
functionregular regularininIz I <1 and such that (z) be a function < 1 and that 11(z) I/ (z) Ij <<1 1 in Let I1(z) this domain. domain. Consider Consider the function.f(z) function/(z) at points within within the triangle ABC, where A is where A is to the the point point z == 1, 1, BC is perpendicular perpendicular to unitcircle, real axis, lies entirely within the unitcircle, A and is at at aadistance distance Iik from from A. There There is is aa positive positive number M such that, that, for for all all values values of z within the triangle triangle ABC, <M(1 — Izi), ... (151"1) 11zl Ii —zJ<M(1Izl), and hence Fig. 40 40 1—Lf(z)l 11/(z)l < M ~1/(z) ...... (151•2) 1lzl 1z · Thus either 1/(z) 11(z)
I
1z 1 —z
as zz tends to A tends uniformly to infinity as A by by values values within the triangle, the triangle and sequence of or there is a sequence of points points z~> z,, z~o z2, ... ...,, lying within the
I
f(z) such that j 1 ~~ tending to z = 1, such ~z) and, by (1(151"2), 1lf(z,.)l 1 — 1lz,.l are bounded; the the theorem theorem of of §§93 93 can can then then be be used. used. let zz1, ;, ••• In the latter case, case, let arbitrary sequence sequence of of points points ... be an arbitrary 1 , z2o within ABC = ':1.. Let two ABC tending tending to to zz= two sequences sequences of of numbers, numbers,rr,, r,, ... 1 , r~o defined by by the relations and Pi, Pt, P2, ..., ... , be defined and cLr,, 1~z.. <Xr,. ( ) . r,.=k', Pn=1 p,.=1,·,.(ltX)' ······ (1513) 151"3 where tX has the same same meaning meaning as in (93"10). Further, let
1z=r,.(1t), ...... (1514) (151"4) 1 —1(z) = p,. (11/(z)
....
96
(CHAP. vi VI
TRANSFORMATION OF THE THE FRONTIER TRANSFORMATION OF FRONTIER
Also these. these functions functions form form aa normal Also normal family family within I t I<1. < 1. ItItfollows follows that (15r6) (151"6)holds holdsuniformly uniformly in in any anyclosed closed set set of ofpoints points lying lying within within Itl < t. ItI
1z,.
and this this gives gives
r,.
1t,. i—tj'
lim 1f(z,.) =«. (151"8) urn ,......., 1z,. we observe observe that that since In conclusion, we since {4,, {4>,. (t)} (t)} is a normal family, so also (1515) with (t)}, and (t,.) = 1. Now Now differentiate differentiate (151"5) with is {4>,.' (t)}, and that lim tj>,.' (ta) for t. This respect to t,t, and, and, taking taking account account of of (151 (151"4), substitute t,. for respect 4), substitute givesf'(Zn) (z,.) r,. = p,.tj>,.' (t,.). Thus (ta). Thus gives!' limf' (z,.) =ex. . ..... (15r9) (151"9) 0
0
0
•••

,._..
theorem:: We have proved the the following following theorem thefunetionf (z)is regular and 11(Z) THEOREM. 1, tkefunctionf(z)isregularand lf(z) I<<1,1, and ThEOREM.Q,if,ininIZz I< <1, of numbers numberslying lyinguithin withintke thetriangle triangle ABC ABC z~o ... is any sequence sequence of if z1 , z9, tending to toz= and tending z = 1, 1, then tken 1 lim 1f(z,.) lim ,.._., 1z,. e:z:ists. This either + <x:> or it it isis aa number number aex > 0. In the tke latter This limit isis either case a. limf lim f (zn) (z,.) = =ex.

152. Conformal tranaformation traniforma.tiou on the frontier. domainand andPP aa point of its frontier Let if R be be aa simplyconnected simplyconnected domain frontier through twocircles circlesKKandK' K' lying entirely entirely outside outside andK'can canbe be drawn, .K'lying through which two entirely inside inside B. R. We Wenow nowmake makeaaconformal conformal transformation transformation of and K entirely if areaIzz <1. will R into into the circular circular area I < 1.It It willbebeshown shown that thatthere there is is aa point point A on 1 such that, if triangle ABC, on I zz I= = 1 if z approaches A from from within aa triangle whose base BC BC lies lies within within the the circle, circle, the the corresponding correspondingpointf(z) point 1(z) in in RB whose base approaches P, F, and and also also the derivative!' derivativef'(z) (z)tends tendstotoaaunique, unique,finite, finite, nonnonzero limit. limit. ItItisis therefore therefore legitimate legitimate to to speak speak of of aa conformal conformal transformazero tion of frontierpoints.
§§ 152, Io2, Io3] 153]
97
FRONTIERPOINTS
By means means of ofaa Mobius Möbiustransformation, transformation,let letthe the interier interier of of the the circle circle K' K' be represented representedon onthe the exterior exteriorofofthe theunitcircle unitcircleI w I = 1, be 1, in such such a way way that PP isistransformed transformed into into the the point w = 1. 1. Then B R isis transformed transformed into wI<1. a domain B, R 1 lying lying within within II w < 1. The circle transformed into a circle KK is transformed circle K ouly show circle K11 with with centre centre 0C1 lyingwithin withinRB,. show that that if 1 lying 1 • We need only the function function giving the transformation is suitably suitably chosen the number «. of§§ 151 is finite. finite. Take Take that that transformation transformation w =1(z) =f(z) which which makes z = 0 of correspond to to the centre 0 1 of~. the radius of K,. correspond and let p denote K1. of K,, let Pi1
42 Fig. 42
Fig. 41 41
By Schwarz's lemma, every circle (0<< 1), is transformed into circle II zz I=0, = 6, (6 a curve Thus this curve curve surrounding surrounding the circle circle I w —  (1 —  PI) = p,O. PI6. Thus curve Pi)II= on the real passes through at least least one one point point to' w' which lies lies on real axis axis and and passes through satisfies
§§ 151,
(X~ PI·
153. Attention should be be called calledto to the the fact fact that that P may 153. Attention should may be be aa frontierfrontierpoint at at which which the the transformation transformation is is conformal conformal and yet yet not notlie lieon onaafree free <1, Jordan curve. curve. For Forexample, example, consider consider the the unitcircle unitcircle I to w I< 1, and in it · la.r arcs JOIDIDg · · · 1 to les ± circular joining—i + 11and angles = 1,2, 1,2,...) ClrcU to+ andmaking mak"mgang ± .,.2 n  1 ((an= ··· )
n'
with the real real axis axisat at these these points. points. Now with the Now cut the the circle circle along along each each arc a.rc between the the point w = — between  11 and the first first intersection of the arc with the —(1— with these cuts, is aa domain circle ( 1 PI) domain to to circle II wPi)II =PI. = Pi The unitcircle, with which our our theorem theoremapplies. applies. The functionf(z), which functionf(z),considered considered in in the thewhole whole circle I z I < 1, is not even continuous at z = 1. But, within an angle at z = 1 whose whosearms armsare are chords chordsof of the the circle, circle, both both f(z) 1(z) and f (z) are are continuous andbounded bounded(26) (26) (27). tinuous and
f
CHAPTER VII VII
TRANSFORMATION TRANSFORMA'riON OF OF CLOSED CLOSED SURFACES SURFACES
154. BlendIng 1M. Blending of domains. In aauplane passing through uplane let let three three Jordan Jordan arcs arcs a, a, b and c, c, all all passing through PP Q but with common points, domains with no no other common points, define define three three Jordan domains and Q and (A A, B Band (A + +B). vplane let three three circular circular arcs arcs b', c' and d', all B). In a vplane crescentsBB' and and C'; C'; the passing through P' F' and passing through and Q', ([, define two two crescents the sum sum of of these domains, (B' C'), is aa circle circle whose whose circumference circumference isis made made up of of ( B' + 0'), the two arcs b' and d'. d:. Further, the the angle angle between between the the arcs arcs b' b' and and c' c' is is ;.. , where where nn is is aa positive positive integer. integer. Suppose known function function Suppose that that a known if' (v) u ==if! (v) represents B' on on B B in insuch suchaaway way that thatLib corresponds corresponds to to b' b' and to c'. c'. It will be shown shown that two functions c to It will functions zz =f(u) =f(u) and z = g (v) can be + B") found, such such that that the former found, former transforms (A + +B) (A"+ B) into a domain (A" and the latter C') into a domain (B" + C"). latter transforms transforms the the circle circle (B' + 0') 0"). (A"+ ;l'he transformations Here (A" + B" + 0") C") isis aa circular circulararea. area. The transformations are are to to be such that that AA and andA", A",C'0'and andC" 0"are arecorresponding correspondingareas, areas,while while B" B" correcorresponds and to to B'. B'. Given Givenany anypoint pointofofB" B"the thecorresponding corresponding sponds both both to B and (v). points in the uu and vplanes vplanes are to satisfy the relation relation uu == i,li 1{1 (v).
QI,
a p
uplane
vplane Fig. 43 43
z plane
B) into the First take aa function function a1 Ut = transforms (A ++B) = 4>I (u) which transforms u 1 I <<1, 1, in in such such a way way that that the the centre centre of of the the circle circle is the the circular area I u1 image of of an an interior point of A. Let image Letthe thenew newdomains domains corresponding corresponding to A and respectively. The The frontier consists of of aa and B B be be A1 A 1 and B1 B 1 respectively. frontier of B1 B 1 consists Jordan curve andaa circular Jordan curve b1 b1 and circulararc arcc1. c1 • The function u1 = 4>I (1/J function~= transforms the domain domain B' B into into B1 B1 (if' (v)) (v)) = I/J1 (v) transforms correspond. The The inin such a way way that the the two two circular circular arcs arcs c' c' and cc1 1 correspond.
154—156] §§ 154156]
REPRESENTATION OF OF' A A SURFACE SURFACE
99
version version principle principle shows showsthat that 1{1fr'1 (v) (v) isis defined defined not not in B' alone, but in a crescent which which has has angles angles 7r/2"' 7r/2"' at P' crescent P and and Q'. Q'. We Wemay may suppose suppose that is bounded bounded by by arcs arcsb'b'and andc'.ë. The function this crescent crescent is function u = = 1{1i/ti1 ( v) transforms forms this this new newcrescent crescentinto intoaa domain domain made made up up of B B1 and its its inverse Jj, 1 and with respect to the the circle circlea1 a 1 ++ c1. c,. Our problem has thus been been reduced to replaced by another of the same same kind which the number number nn has been replaced kind in which (n1). (n—i). By process we we obtain in succession succession the functions By repetition repetition of this process the functions U2 ==
155. The 'fhe special domains B' B' and and C' C'in inthe thevplane vplanecan cannow now be be replaced by much much more more general general ones. ones. We We assume assume that that the frontiers frontiers of B' and C' C' consist of curves joining Q' and having consist of three Jordan curves joining the the points points P' F' and Q' be a function these points. points. Let vv = x function which which distinct tangents at these x (w) be 'fhen B' is transformed into a domain transforms transforms II w I <<11 into (B' + C'). Then whosefrontier frontierconsists consistsofofaa circular circular arc arc /3' and a Jordan whose {3' and Jordan curve curve which which meets either end of {3' at at an an angle anglewhich which differs differs from from zero. zero. This Thisshows shows of /3' that the the domain domain just justdescribed describedcontains containsaacrescent crescent {3' y' whose that whose angle angle is 7r/2". The function u == l{t (x ((w)) w)) transforms y' into a curve in the uplane c, our problem we cut cut away away the the domain domain that that lies lies between between y and c, and, if if we that of of§ § 154. i54. is the same as that I
transformation of aa threedimensional 156. Conformal transformation threedimensional surface (28). surface (28). X YZspace Let a surface surface 8S in threedimensional threedimensional X Y Zspace be represented, in some point point of of 8, 8, by means of the equations the neighbourhood neighbourhood of some equations /3) ...... (i561) Y= Y(or., /3), Z=Z(ca, X=X(oc, /3), Y= X=X(1X,{3), Y(1X,{3), Z=Z(a.,{3), (156·1) are parameters. parameters. The where IX and /3 {3 are 'fhe same same surface surface can be represented in in terms of other parameters of the and v~' by by substitution in (156'1) Of terms parameters u and = IX (u, (u, v), v), {3/3= (u,v).v).We Weassume assumethat that the functions a. (u, v) functions IX= = {3/3(u, (u,v) v) can can be be so so chosen chosenthat that the new equations for 8, and /3 {3 (u, v) Y=77(u, v), Z=,(u,v), (1562) X=~(u,v), ...... (156'2) X=E(u, v), Y=lJ(u,v), have continuous first partial derivatives have continuous first deriv11tives in in some some domain domain Rw of of the uvplane, and and at the same same time uvplane, ds2=A(u,v)(du2+dv2), di' =A (u, v) (du2 + dv'), ...... (156·3) where A* A *00at at all all points B,,,. where Rw. It then then appears appears that that the thedomain domain B,,, Rw isis R 8 of ofSS by by a oneone continuous correspondence, represented on on aa portion portionB5 represented that any any two two curves curves in inB,,, Rw which which have have continuously continuously turning tangents tangents
100
vii (CHAP. VII
TRANSFORMATION OF CLOSED TRANSFORMATION CLOSED SURFACES
and cut cut at atan anangle angleaotcorrespond correspond to totwo two curves curves which which cut at an angle that the the scale scale of of the the representation !epresentation at the the point point of of interintera on R8 , and that section isis independent independentof ofdirection. direction. In In other words section words the representation representation is conformal. R 8 is of RfiJ on 118 It isis known It known that thatititisisalways alwayspossible possible to to introduce introduce parameters parameters u and vv such that (156"3) holds in in aacertain certainneighbourhood neighbourhood a point (ot0 /3k,) , Po) of ofof a point (oc0, such (1561) provided (156·1) provided that in in some some domain domain of of the the a/3plane otpplane the the functions functions .X (a, p), /3), Y (ex, (a, /3) and Z (ot, (a, P) /3) have have continuous continuous first partial X (ot, p) and partial derivatives derivatives three Jacobians, Jacobians, which satisfy which satisfy aa Lipschitz Lipschitz condition, condition,and and that that the three a(X,Y) a(Z,X) o(X, Y) o(Y, Z) o(Z, X) a(Y,Z) a (a, /3) 0 0 (a, a(cx, P) ' a((a, ot, /3) P) ' a (ot, /3) p) ' do not all vanish at (oto, Po). It is that the equation (156·3) is conceivable conceivable that may be be satisfied, satisfied, for for aa suitable suitable choice choice of parameters, may parameters, even even on on surfaces surfaces conditions just just mentioned. mentioned. which which do do not not satisfy alJ all the conditions
157. 157. Conformal Conformalrepresentation representationofofaa closed closed surface surface on on a sphere. Suppose that that SS isisaasurface Suppose surfacewhich whichcan, can, by byaaoneone oneone continuous transformation, be made formation, made to to correspond correspond to to aa closed closed sphere ~ Further let every point P of of SSlie liewithin withinaaportion portionofofSSwhich whichcan canbe beconformally conformally reprereprewill be shown shown that the the whole whole sented portion of a plane plane(§ 156). It will sented on on aa portion surface S can be be represented represented conformally conformally on on ~Consider an arbitrary arbitrary oneone oneone continuous transformation of S into into ~ The north north pole correspondstotoaa point point N N of S. pole N1 N 1 of ~ corresponds S. Stereographic projection from N 1 represents ~ on on a plane plane T, T, and andthere thereisisaaoneone oneone projection from 1V continuous correspondence between betweenTand Tand the pricked prickedsurface 5* obtained surfaceS* obtained omitting the point N from from S. by omitting
158. Now Nowconsider consider in in TTan an infinite infinitesequence sequenceofoftriangles trianglesT1', 1'/,T2', 1'2', ••• arranged spirally arranged spirally (like (like the the peel peel of of a peeled peeled apple) and and covering plane T. apple) covering the whole whole plane 14 13 12 ... + Tn' of the 'rhe sumoS,,' u,.' == T1' 1'1' ++ T2' 1'2' ++ ... The sum 15 first n triangles triangles always always covers covers aa simplysimply2 16 10 connected domain. The The figure gives a rJ,___ connected domain. figure gives 17 simple exampleofofwhat whatisis meant. meant. The simple example The 18 6 8 triangles may have triangles may have curvilinear curvilinear sides; sides; 19 are drawn drawn with with straight straight sides purely they are convenience. The triangles as a matter of convenience. triangles Fig. 44 Tn' are, however, so chosen Fig. 44 however,totobebéso chosenthat thattheir their images Tn on on S* 5* satisfy conditions: (a) the sides of satisfy the two two following following conditions: 1
§§ 157—161] 157161]
REPRESENTATION ON A SPHERE SPHERE
101 T,. are curves on 8 which which have endpoints, and two two havetangents tangentsat at their endpoints, (b) any any two two triangles triangles with a intersecting sides sides have have distinct tangents; tangents; (b) intersecting common side lie liewithin withinaaportion portionof ofSSthat that can be represented represented conformally on a plane domain. domain.
159. Itwillbeshownthatforn=1,2, It will be shown that for n = 1, 2, ... the sum rr,. = T 1 + T 2 + ... + T,. can be represented represented conformally c~mformallyon onaacircle circleII z,. I < r,. in a complex znplane. z,.planeI
Assume represented on on a circle circle I z,._ 1 I< r,._~> Assume that that rr,._ 1 has already been represented and that that the the figure figure ( T,._ 1 + T,.) has been been represented represented on on aa domain domain of of a appears in both representations, and if if the vplane. triangle T,._1 appears representations, and vplane. The triangle image of a point of T,._ 1 in one is made to correspond to the im&ke of the same point The same point in in the other, a new conformal conformalrepresentation representationisisset set up. up. The is thus reduced representation representation of rr,. on on Izz I
transformation of of the 160. The transformation the whole whole pricked pricked surface surface S* has now now The results results there there been reduced to to the the problem problem considered in §§ 125—129. 125129. The showthat that s• 5* can be transformed either into the obtained show the circle circle IIw I <
into the surface 161. Conformal Conformal transformation transformation of of polyhedral polyhedral surfaces surfaces into of a sphere of the the theory sphere was was one one of the earliest earliest applications applications of theory to be be
102
TRANSFORMATION OF CLOSED CLOSED SURFACES SURFACES [CHAP. vrz, VII,§ § 161
attempted. Schwarz that both both the general general tetrahedron tetrahedron and and the the Schwarz showed showed that attempted. cube can be transformed into the sphere. cube can transformed into sphere. The The transformation transformation of the the general closed polyhedron into the the closed closed sphere is a problem which most general We cannot cannot require require prepreis almost almost identical identicalwith withthe theone onejust just treated. treated. We at edges edges and and corners corners of of the thefigure, figure, and andto tospecify specify our servation of angles at following conventions intentions intentions we we make make the following conventions as as to to representation of neighbourhoods of of these singular points. neighbourhoods points. Along an edge meet. We cut them them from from edge two two faces meet. We suppose suppose them rigid, cut spread t:Uem them out out on on aa plane. plane. Thus the figure figure and spread Thus aa neighbourhood neighbourhood of any point of the edge edge is is represented represented as as aaplane planeneighbourhood. neighbourhood. is a corner we make aa quasiconformal quasiconformal representation of a neighIf C If Cis we make bourhood of of C onto a plane plane domain, domain, as as follows: follows : By means means of a small bourhood small neighbourhood neighbourhood of of C is cut off. sphere, sphere, centre C, C, a small Cis off. The The pyramidpyramidlike figure figure so so obtained is cut cut open open along along one one of of its its edges edges and spread out on the uplane, uplane, the the point C corresponding to u = 0. IfIfnow now 1rct is the the sum sum 2
of the superficial angles angles of of the the polyhedron polyhedron at C, = u~ C, then the function v = gives the required quasiconformal gives quasiconformal representation of the corner corner C. C. The method used used above above now now gives a solution solution of ofour our problem problem (29).
CHAPTER VU! VIII
THE GENERAL THEOREM THEOREM OF OF UNIFORMISATION UNIFORMISATION THE GENERAL
162. 162. Abstract surfaces. We finite or or infinite infinite collection of triangles T1, T1 , T2, T 2 , ..., ••• ,and We take a finite and supwith each each side side of of each triangle Ti is associated exactly one one side pose that that with of some may picture associated some other triangle triangle Tk. Tk. We may associated sides sides as being being will belong belong to welded welded together; together; when when this this has has been been done, done, any any vertex PP will several triangles. We We assume assume that that the several triangles. the triangles triangles adjoining adjoining aa common common of adjacent adjacent triangles. finite cycle cycle of vertex form form aa finite The welding theabstracl abstractpolypolyThe welding operation need only be be done done mentally; mentally;the hedral surfaceS surface S consists consists merely merely of the the triangles T;, the pairs pairs of of associated edges of vertices. sides sides (the edges ofS), 8), and and the vertices.
163. We directed from Weshall shallconsider consider polygonal polygonal paths paths PP1 PP1 ... ..• Pn_1Q, directed P to of S. S. We to Q, Q, containing a finite finite number of adjacent edges edges of We call call S connected anytwo two of of its its vertices vertices P and connected ififany and Q Q can be joined by such a path. A connected surface isis closed closedififitit contains contains only only aa finite A connected surface finite number number of of triangles, open open if it contains an an infinite infinite number. number. Given aa path ex from from PP to Q, Given Q, we shall denote by ex1 the same same path we shall shall denote denote by ex/3 described from Q Q to P. If Iffi {3 is a path path from from Q Q toR, to R, we from P to Q the path PQR obtained by describing ex Q followed by {3 from a from
QtOR. QtoR. deformaa path path exa by making We deform making a finite finite number number of of operations operations of the following following two kinds: which corresponds corresponds to to aa side side AB of a triangle ABC is (a) An edge of ex which (a) replaced by AC + CB, CB, or or AC AC ++ GB CB is is replaced replaced by by AB. AB. replaced (b) A pair of edges corresponding corresponding to to AB + BA is (b) of successive successive edges is inserted inserted or deleted. If aex can can be he deformed deformed into {3 in this way, way, fi {3 is called hornotopic homotopic to a. IX. Clearly this isis an an equivalence equivalencerelation, relation,sosothat that the the paths joining P tp Clearly this tp Q Q are divided into into classes classes {PQ} {PQ} of homotopic paths. paths. For most most surfaces, surfaces, there there are areseveral severalclasses classes {PQ}, {PQ}, generally an infinite number. IfIfthere i.e. if all all paths joining P to thereisisonly onlyone oneclass class {PQ}, {PQ}, i.e. to Q Q number. S is honiotopic (for any choice of P and are homotopic and Q), Q), the surface surfaceS is called called simply simply connected. H
104 104
THE GENERAL GENERAL THEOREM THEOREM OF OF UNIFORMISATION UNIFORMISATION
(CHAP. VU! VIII [CHAP.
164. Now fixedvertex vertex00 of of S, S, and two 164. Now consider consider aa fixed two paths paths a oc and and fJ joining to P. The and ends endsat at 0, 0, and oca and fJ joining 00 to The path oc{J1 then begins begins and are homotopic if and only if oc{J1 is homotopic homotopic to the path path consisting consisting of the single point 0. It foll?ws easily easily from from this that that we we can can choose chooRe a set set of of closed closed paths It foll'ws containing 0 ...... (164.1) (164·1) such that, that, given the paths paths given any paths paths aoc and and fifJ from 0 to F, P, exactly exactly one of the oc, wllloc, w"'oc, .. . . ••••. (164·2) a, is homotopic to fl. 165. 165. The universal covering covering surface.
Consider two two surfaces surfaces SS and 8, formed with triangles Consider triangles 1', T; and and T/•' qi,(2), 1, ••• respectively. Suppose ... is Suppose that that each of the triangles T;'i', TP is related to the the same same triangle triangle1',, T;,which whichisiscalled calledthe theprcvjection projection of T/• 1• Suppose further that, if T/• 1 and Tklp.l are adjacent adjacent on on 8, their projections 1', T; and adjacent on on S. S. Then T 1e are adjacent Then 8 is called aa covering covering surface of S. A covering surface 8 of S is is called called unbranched unlnanched if if every every cycle cycle of triangles adjoining adjoining aa vertex vertex PP of 8 projects on to aa similar similar cycle cycle on S, both both cycles having the same number of of triangles. triangles. cycles
166. We We shall shall now now show show that thatany anysurface surfaceSShas hasa simply a simplyconnected, connected, surface of S. unbranched covering surface 8, called unbranched called the universal universal covering covering surface To construct 8, we first first choose chooseaa vertex vertex00 of of SS and and a set (164·1) of 5, with closed paths on on S containing 0. Next, closed paths containing 0. Next,given given any any triangle triangle T., of S, P, Q Q and R, R, we we associate associate with it triangles triangles vertices P.
T"' T.,m, T.,121, ... ... ...... (166·1) (or any any path path joining joining00 toP to P corresponds to the path w<•loc (or
where homotopic to w<•loc). This involves involves selecting selecting aa definite definitevertex vertex PP from each triangle T n; we we restore symmetry by by associating associating with with T ..1• 1 not only a class {,8} of paths from 0 to toP, butalso also classes classes {,8'} and {,8} of paths paths from from P, but 0 to to Q Q and and 00 to toR. R.The Theclasses classes are related in pairs by the rule:
Tn 1• 1
homotopic to ,8 + PR. A path ,8' is homotopic PQ, a path path fl,e homotopic homotopic to ,8 + PQ, Finilly, we Fin~lly, we arrange arrange that the the vertices vertices pv, p!vl,Q(v), Qlvl, Rlvl of T., 1• 1 project on P, Q, R, and similarly similarly for the the sides. sides. F, Q, decide how how to to associate pairs To complete the definition of S, we muEt muet decide sides of triangles triangles on S. Let T n and betriangles triangles S with a common and TTm m be ofof S with a common of sides edge PQ, and let let oc be any path path from from 0 to to P. P.This Thispath pathgives givesrise rise to to well well
§§ 164167] 164—167]
UNIVERSAL COVERING SURFACE
105
defined triangles T,. 1• 1, T m ,,., , and we we aBBociate associate those those two sides of T,. '"' defined which and which project on PQ. PQ. This This makes makes the thecollection collection of of triangles triangles which is is clearly an unbranched covering T,. '"' into an abstract abstract surface surface S which covering surface of S.
'I'm'"'
To prove prove that S is simply simply connected, connected,take take any any two two vertices vertices PP and Q of S projecting on P and and Q, Q, and let ~$ and y be any two paths joining P to Q projecting on to fJ andy to Q. Q. ToP corresponds aa path pathccrx and y joining P to To P corresponds from joining00 to to Q 9 must correfrom 0 to to F, P,and andboth boththe thepaths pathsixfi rx{J and and ccy rxy joining if = PP1 ... ... P,._1Q, the paths spond to Q. (For if~= ••• P,,_1Q P,._1Q and fi fJ = PP1 PP1 ••• P and cc belong to related classes rx + PP1 .•• P; rx + classes as defined defined + PP1 ••. Pi+ 1 belong above. follows easily easily by induction induction that that cc rx + + P ... P; P. corresponds corresponds to above. ItItfollows P;, and in particular particularthat thatccf3 rx{J corresponds since ccfl rx{J and and ccy rxy corresponds to Q.) Now since both correspond to Q, they must be be homotopic, homotopic, and hence so are rx1rx{J and y are homotopic. and therefore fi and cc'ccy, rx1rxy, and fJ andy homotopic. But the the operations operations (a) (a) and and (b) (b) which which deform deform fJ into y will will deform deform ~ ~ and y are homotopic, and S is simply
into y if transferred to S. Thus connected.
167. Domains and their boundaries. 167. Domains and ...,, T ".t of triangles on S, 8, their Given any flmte finite collection Ta,, T ,., T ns• ... their sum sum taken modulo modulo22will willbe becalled calledaadomain domainI:Eof ofSSand and denoted denoted by by taken = T,., ± I:= t Tns t ... ... ±t T,.,. . ..... (1671) Any finite number of domains I:1 , I: 2 , ••. ,:Em can also be added modulo 2, s,,, has a meaning. so that that the symbol symbolI:E1 E2 ..... t :Em so meaning. 1 t± I: 2 t± .
bourulary !J ((T) T) of a triangle is the sum of its sides; the boundary The boundary of the domain (167.1) is then then defined defined by by the equation (167·1) is S(I:) == S(T,.)
tf
!J(Tns)
tf ... ... t
!J(T,.,)•...... (167·2)
This very convenient convenient formalism formalism for proving proving the the theorems theorems which which This is is a very follow: If lfrx cc and and fi fJ are homotopic homotopic paths, and and ifif[rx] [cc] and and [{J] are the sums inodulo modulo 2 of their sides, (E) of sides, then then [cc] [rx] t± [fi] [{J] is is the the boundary boundary !J (I:) of some some domain I:. For if a triangle is used § 163, triangle TTis usedfor foroperation operation(a) (a)ofof§ 163,[cc] [rx] is transformed transformed unchanged. Hence [rx) t± S (T), whilst whilst operation operation (b) (b) leaves leaves [cc] [rx] unchanged. Hence if if into [cc] fi fJ is homotopic homotopic to rxa we can write [cc] t + S (T,,1) [{J] = [rx] (T,.,) t+ ... ··· t± S (Tflk) (Tn,.) = [cc] [rx] t± S (I:), [fi] (s), and therefore (1673) . ..... (167·3) [rx] t± [fi] [{J] = !J (I:). [cc]
106 106
THE GENERAL GENERAL THEOREM OF UNIFORMISATION UNIFORMISATION
[CHAP. [CHAP. VIII
Every an open open connected conmcted suiface sur,face SS has a nonnonEvery rumempty nonempty domain~ damain on an empty boundary 8 (~). (s). Since SS is open there exist triangles T of S which which do not belong belong to ~.
since SS isisconnected connectedat at least least one one such such triangle triangle isisadjacent adjacent to to aa triangle and since of~. of L
Two different different domains domains ~ and Two have the the same same boundary. have
~· on
an open open connected conmcted surface S cannot cannot
E') = 0, and and this For if '8 (E) For (~) = 8 (~'), this contradicts contradicts (s'), then 8 (~ t± ~') the previous theorem since since ~ t ~· is not empty. ItItshould shouldbe beobserved observed that on on aaclosed closed surface surface there there are arealways always pairs pairsofof(complementary) (complementary) domains domains with the same boundary. pie closed On an open open simply simplyconnected connected surface, surface,every every sun simple closed polygon is the the boundary of of exactly one domain.
Two distinct distinct vertices vertices PP and Q divide itit intv into paths Two Q of the polygon polygon TTir divide ex to Q. Q. Since Sis f1 are homotopic. S is simply connected, exa and fi a and f1 from P to Also (since the the polygon is simple) polygon is simple) exa = = [a] [ex] and f1 == [a], [{1], and and therefore Also (since by (167·3) Ir=[a] TT = [ex] t ± [/1] = 8 (~).
(30). 168. The Theorem Theoremof ofvan vander derWaerden Waerden (30). Given Given a simply connected conmcted open open surface surfaceS, 5, its triangles can be be arranged arranged in in aasequence sequence T1, T2, ..., ... ...... (1681) (168·1) in in such such aa way way thai th.at each each domain domain (n = 1, ~ .. = T 1 + + T2 T 2 + ... + T, (n 1, 2, 2, ... ...)) ...... (168·2) has ltas aa simple simple closed closed polygon polygon as boundary. Suppose Suppose that that T1, T 1 , ... ...,, T,, T, have have already been been chosen. chosen. We We must then find find such that that S (~, T n+I adjacent to ~ .. such simple closed closed polygon. polygon. (s,, + T n+I) is a simple common with with the polygon This will be so if T n+I has one or two sides in common but 1wt not one one side sideand and the theopposite oppositevertex. vertex.We Wemust mustalso alsoensure ensure that that S % (~,)but every triangle of S occurs occurs in the the sequence sequence (168·1).
Choose any triangle ~ of SS adjacent to ~ ... The sides sides p, p, qq and and r of Choose any of~ cannot all all lie on cannot on !8 (~ .. ), because if they did did S (s,,) (~,) would not be a simple closed closed polygon. polygon.Therefore Thereforeone oneofofthree threethings things must must happen: simple
I. Two Two sides sides
of~
lie on !8
(~,).
II. One side of of~ lies on S (~,). opposite vertex does not. (s,,), the opposite III. One (E,,). Oneside side p of of~ and the opposite opposite vertex P both both lie lie on on 8 (~,).
§ 168]
SIMPLY CONNECTED SURPACES SURFACES 107 If satisfies II or If A satisfies or II, II,we wetake takeT,,÷1 T n+1 == & A. In case case III, III, the thepolygon polygon splits into three S (E,,) (~ .. ) splits three parts partsp,p,cc ot and p, and ex T q, p ± q, P T r are simple
II* (see (see Fig. Fig. 45). closed polygons, polygons, bounding bounding domains domains II H and H5 Thus S (II) =ex t q, S (II*) = p t r and
S (II
t II*)
t
/J) tt (p (p t (ptt ex t P) ± q t t r)
11*) == (p =
It follows It follows that that
S
cc
(~.. )
t S (A). ±
j= .. + & . ..... (1683) (168·3) t 11* II*=~ A. The triangles ...,, T 7',,.. belong triangles T1, T 1 , ••• them belongs belongs belong to to ~n• E,,, and so each of them either to II or to II*. But But ifif T1 T 1 belongs to II*, say, then so do T2 , ••• ...,, T., (11*) = p + rr belongs to more and A, because no side of the polygon polygonS (II*)= more than one one of of these these (it (n + Thus H II censists consists of of triangles triangles outside Olltside t 1) triangles. Thus now try to ~.. + A ; we we now to choose choose T n+1 from one of these triangles. triangles. H ~
p
Fig. 45 45
Let A' be a triangle in argument above, in H II adjacent adjacenttotocc. ex. We repeat the argument yeplaced by A', and find find that: that: either eitherwe we may maychoose choose T.,+l = =A', with A..replaced bounding aa domain domainII'. H'. But or there is a polygon polygon q' +ex' But all the vertices cc' bounding ex' lie in cc, ex, and least one one of q' ++cc' and so ex' cc'isisobtained obtained from ot ccby by removing removing at at least side. We We now now repeat repeat the the argument, argument, starting starting with with aa triangle triangleA' iY in in II' H' side. adjacent to After aa finite finite number of steps, we must must arrive at a suitable adjacent to cc'. ex'. After triangle T ..+l which, being adjacent adjacent to tocc, ex, belongs belongs to H. II. 7',,÷1 and II H by Now replace, in the original figure, by = the original figure, ~.. ~ ..+1 = ~ .. + +T n+1 and H II —  T n+l• and repeat the argument. argument. After After aa finite finite number number of of steps, steps, we we obtain ~ .. + H II = ~.. + T.,+l + ... ··· • Tn+m = 1:n+m and are now ready to adjoin the triangle triangle A= = Tn+m+l·
To show show that SS can can be be exhausted exhausted by by this this process, process, we we start with with any triangle T 1 ' of ,S, adjoin all all adjacent adjacent triangles, triangles, then then adjoin adjoin all triangles ~. adjoin obtained already, In this this way way we we obtain obtain adjacent to those obtained already, and and so so on. on. In
108
THE GENERAL GENERAL THEOREM THEOREM OF OF UNIFORMISATION UNIFORMISATION
[CHAP. VU! VIII [CHAP.
exhausting S. S. If a sequence {T,.'} of triangles triangles exhausting If we we n;:,w arrange, arrange, at each stage of the process above, that d is stage process described described above, is chosen chosen to to be the first first suitable triangle suitable triangle T,,', T ,.', it is is clear clear that thatevery every triangle triangle Ti,' T ,.' will will appear appear somewhere in the the sequence sequence (168.1). (168·1).
169. 169. Riemann surfaces. surfaces. So far, far, our arguments have not involved any individual So arguments have individual points on S edges of SS may be be regarded regarded as ordered pairs of except the vertices; the edges vertices, the triangles triangles as as triples triples of ofvertices. vertices. In In order order to to obtain obtainaaRiemann Riemannsurface, surface, we we must must convert convert this thisframework framework into aa twodimensional space which which carries carries the the local local angular twodimensional topological topological space of the ordinary ordinaryplane(31) plane(3l)(32). (32). metric of
This This is done done most most simply simply by by assigning assigning to each each triangle 1', T; a Jordan Jordan curve;a.;lying On;a.iwe curve lyingininthe theplane planeof ofaa complex complex variable t;. On we mark mark three points P;, P,, C), and R;, I?,,and andwe weregard regardthe theinterior interiorofof;a.; asasaaconformal points Q; and conformal representation of 1',. Since Since any any two two Jordan Jordan domains transformed ofT;. domains can be transformed conformally conformally into into each each other, other, the transformation transformation being being continuous continuous if extended to their frontiers frontiers (§ 137), shape and size size extended 137), we we see seethat that both the shape of;, of a.;,and andthe thepositions positionsofofP,, P;,C), Q; and R,, R;, are irrelevant. Nevertheless, Nevertheless, we we transform we can can now now identify identify individual individual points points of of T;: T,: we transform the interior of;a.;conformally of conformally into into aa circle, circle, so so that that any any point point of of T, T; isis mapped mapped into into a point pointA, A;of ofthe thecircle circle and and isi~ characterised characterised by by the the crossratio crossratio (P;, C),, Q;, R;, A,) A;) (P,,
u). (see§ (see § i:i).
The angular angular metric metric which which we we want want to to set set up is then well defined at at all The well defined inner points of i.e. those for which (P,, C),, R1, A1) is not real. ofT;, those for which (P;, Q;, R;, A;) real. We must still define :;ides and vertices vertices of all define the angular metric on the sides our triangles. To assumptions which which are our triangles. To do do this, we need further assumptions are best dealt with with by by considering considering aa special special case. case. Suppose that that the Suppose the triangle triangle T1 T 1 with with sides sides p, p, 1, l, q is to be welded welded along q to T2 sides q, q, m, in,rr (Fig. (Fig. 46 46a,a, b). b).We Weassume assumethat that there is at least one T2 with sides conformal representationofofTT1 suchthat that the the angular conformal representation angular metric metric of of the the 1 such (1planeisisvalid validon onthe theside sideqqof ofTT1, andthat that aa strip oa between y lcplane between q and andy 1 , and (Fig. (Fig. 46a) 46a) is the conformal conformal image image of aa part partofofT2 T2(Fig. (Fig. 4Gb). 46b). Then Then the the welding of T1 and T2 along qq can can iJe ie performed § 154 welding T2 along performed by by the the method method of of§ 154 (Fig. 46c). 46c). This process process is is uniquely defined defined if we we know know the the images images in in the (Fig. 4of aa particular point tc and and t2planes t 2 planes of point M M of ofq. q. Suppose further further that around thatthree threetriangles trianglesT1, T1 ,T2 T 2 and and T3 T3 form a cycle arour.d ci). We apply a vertex vertex A, A, anL am~ that thataccordingly accordinglyT3 T3 has hfl.ssides sidesp, p,r,r,an(Fig. (Fig.46 46d).
§ 169]
RIEMANN SURFACES SURFACES
109 the preceding preceding method, + p) p) method, with with the the difference differencethat thatnow nowthe thepart part (r(r + +A A+ must be welded to the corresponding part of the of the frontier frontier of of T3 must welded to corresponding part frontier of T1 and that that A T1 + 2'2, T 2 , and A takes takes the the place place of of M. M. The The common vertex vertex A lies inside inside the the resulting circle circle (Fig. (Fig. 46e), 46€),sosothat that the angular metric is certainly defined there. case when when the the cycle cycle around around A A contains contains more more than than three triangles The case triangles is treated tr~ted by bythe theobvious obvious extension extension of of the the method method above. above.
,,
::!...,
I
I
A
c
(a)
B
(b)
D
, ,. /
 ... ',s \
\
I
I
I I
B B (c)
(d)
(e) Fig. 46
llO 110
THE GENERAL THEOREM OF UNIFORMISATION UXIFORMISATION
(CHAP. [CRAP. VIII
170. The Uniformisation 170. Uniformisation Theorem. Let S be is closed LetS be aa Riemann Riemann surface. surface. IfIfSSis closed and and simply simply connected, connected, the apply. developments of Ch. Ch. vn vu apply. In all all other other cases, cases, the the universal universal covering covering surface surface .cy of S is is an an open, open, aerden holds simply connected surface for which which the the theorem theorem of van van
T1 + T 2 + ... + Tn
(n
=
2, 3, ... )
a circle circle I tn I ~ rri,, n• and find find on this this circle circle the the image image of of done this for for all n, we may use the results 125—130 Having done results of of§§ 125130 to establish establish the conformally on on ~onformally
I
~nl·
Unlformiiation Theorem. Uniformiaation The universal universal covering covering surface S of any open or closed closed Riemann su,face su1jace S can can always always be be represented represented conformally on (1) aa closed (1) closed sphere, (2) an Euclidean Euclidean plane, pla·ne,
(3) the the interior interior of ofaacircle circleI itt
pattern in in the tplane In this way way we we obtain obtain aa complete complete conformal conformal pattern tplane of the triangulation of S. Its edges edges are are Jordan curves and do do not not necesnecessarily have tangents at at their their endpoints, endpoints, so that that in in general general the the angles angles of the triangles not defined. defined. However, However, the the given triangulation triangulation can can the triangles are are not always be replaced by an equivalent equivalent one one for for which the edges are analytic arcs the particular particular metric metric arcs or, or, even even more more specially, specially, are are geodesics geodesics in the involved.
Remark. Care Care must be taken always Remark. always to weld weld adjacent adjacent triangles triangles of S together in in the same together same way way as as their their projections projections on S. S. The The resulting resulting tritriangulation angulation of S in the the tplane tplane then then admits admits aagroup groupofofdisplacements displacements transforming equivalent triangles into into one one another. another. 171. The Theuniformisation uniformisationtheorem theoremshows shows that thatwe we can can always always introduce introduce a local which is spherical spherical in the first first case, case, euclidean euclidean in the the local metric on on S which second and (Lobalschewskyan)ininthe thethird. third. We can even define and hyperbolic hyperbolic (Lobatscheu·skyan) the distance between any any two two points of S as the distance between distance between between their tplane. images in the tplane.
Since two triangles of S which which have have the same projection 8ince two projection on S have have points, we we can can transfer the local the same same metric at at equival equival ~nt points, local metric metric
§§ 170, 172]
REPRESENTATiON OF REPRESENTATION OF A A TORUS TORUS
111 Ill
from S to S. S. But Butwe wecannot cannotdo dothis thisfor forthe thedistance distancebetween between two two points, since general any any point of of SB has has an aninfinity infinityof ofcorresponding corresponding points points since in general on onS.
172. Conformal repre1entatlon 172. Conformal representation of of aa toru1. torus. It is often important to determine which kind It is often important to determine which kind of of metric metric is is introduced introduced given Riemaiin Riemarin surface surface by by the themethod methoddescribed described above. above. This This can can on aa given sometimes be be done done with very little sometimes little effort. effort. Considerfor forinstance instanceaa torus torus S in threedimensional space. By threedimensional space. By comcomConsider bining the method method of of§ of the present present chapter, we hnd that § 156 with that that of that the universal is represented either on ·the universal covering covering surface surface S is represented either on the infinite infinite 1plane or or on on a finite The metric metric induced induced on on Sis S is either tplane finite circle circle I t I < R. R. The I
euclidean or or hyperbolic. hyperbolic.
Fig. 47 47
We We now now consider considertwo two closed closedpaths paths r:x.' a' and fi' fJ' on S which are homotopic nor to to each each other, other, and and which which intersect intersectat at aa point 0 of of S. neither to zero zero nor of r:x.' a' and fJ' in the 1plane are arcs arcs joining joining aa point 0 1 to two The images images of tplane are differentpoints 0 2 and 003. We replace replace r:x.' a' and fi' fJ' by curves r:x.a and fifJ which differentpoints 3 • We are mapped on the segments segments (eucidean (euclideanor ornoneuclidean noneuclidean as as the the case case may may and 0103. be) O1O2 0 10 2 and 0 10 3 • Since r:x.a and fi fJ are the shortest shortest closed closed curves curves on on S passing through 00 and homotopic to r:x.' a' and passing through homotopic to and {J', they cannot intersect intersect except at 0. on SS except
If If we we cut S along r:x.a and {J, we obtain a surface S' which which is mapped on in the the tplane. 1plane.This Thismapping mappingisisconformal conformalatat0,0,so sothat that the quadrangle in a quadrangle quadrangle has has anglesuni anglesum 2ir. 211. But the the anglesum anglesum of aa noneucidean noneuclidean quadrangle thanIT, 11, hence the anglesum anglesum of a quadrangle is less than triangle is is less less than 2IT. 211. Therefore the metric must must be beeuclidean. euclidean. Moreover, opposite oppositesides sidesof ofthe thequadrangle quadrangle have have equal equal length, length, so so that Moreover, the quadrangle quadrangle is is in in fact fact aaparallelogram. parallelogram. Any Any segment segment parallel parallel to to one one on S. S. joining two two opposite oppositesides sidesisisthe theimage imageofofaaclosed closed geodesic geodesic on side and joining which form form aa net on Hence S is covered by two two systems systems of of geodesics geodesics which on S. S.
112
THE GENERAL GENERAL THEOREM THEOREll OP OF UNIFORMISATION UNIFORMISATION
(CHAP. (CHAP. VIII
It is well known It is known that that an infinity of such nets can be constructed on on S. S. Among them ·isisa a"reduced" the Among them• "reduced" net, net, characterised characterisedby bythe the fact fact that the triangle 0/)/)3 is acuteangled. acuteangled. The shape of of the the reduced reduced parallelogram parallelogram is a conformal conformal invariant S: two two such such surfaces surfaces can can be be transformed transformed conconjrnjarjang of 5: formally only ifif their their reduced reduced parallelograms parallelograms are formally into each other if and only similar. Since any given point of the tplane tpla.ne can be be moved moved to another another similar. Since translation, it follows follows that can be be transtransgiven point given point by by a translation, that any torus S can formed is formed conformally conformallyinto intoitself itselfininsuch sucha away waythat that aa given given point point P is transformed into aa second second given given point point Q. Q.
BIBLIOGRAPHICAL NOTES BIBLIOGRAPHICAL NOTES I. BOOKS I. BOOKS
Encykiopadie der mathematischen Encyklopii.die mathema.tischen Wissenechaften: \Vissenschaften: L. Lichknstein. Lichtenstein. Neuere Entwickelung der Potentialtheorie. Potentialtheorie. Konforme Neuere Entwickelung Abbildung. n, c. 3. (Vol. (Vol. n, II, 3, 3, 1.) 1.) Abbildung. ii, L. Bieberback. Bieberbach. Neuere Neuere Untersuchungen Untersuchungen uber iiber Funktionen Funktionen von von komplexen komplexen Variablen. II, n, c, 4. (Vol. Variablen. (Vol. ii, n, 3, 1.) H. A. Schwarz. Gesammelte Abhandlungen. Abhandlungen. II. ii. H. Schwarz. Gesammelte iv). sur la tht\orie theorie generale genérale des dessurfaces. surfaces. I,r, Livre Livre IIii (ch. IV). G. 0. Darboux. Darboux. Leçons ~ODS sur E. Traite d'Analyse. E. Picard. Pward. Traité d'Analyse. II. n. G. Fubini. Fubini. Introduzione Introduzionealla aliaTeoria Teoriadei deiGruppi Gruppidiscontmui discontinuieedelle delle Funzioni Funzioni 0. (Pisa, 1908.) 1908.) Automorfe. Automorfe. (Pisa, 0. G. Julia. Julia.Leçons Le~nssur surles lesfonctions fonctions uniformes uniformes a point singulier singulier essentiel essentiel isolé. isole. (1923.)   Principes Principes gbométriques geometriques d'Analyse. iI (1930). J. L. Coolidge. Coolidge. Treatise on the the Circle Circle and the the Sphere. Sphere. (Oxford, (Oxford, 1915.) 1915.) J. L. HurwilzCourant. HurwitzCourant. Funktionentheorie. (3rd ed. ed. 1929.) 1929.) H. Weyl. Weyl. Die Die Ides Idee der der Riemannschen Riemannschen Flii.che. (2nd ed. ed. 1922.) 1922.) (Paris, 1927). P. Montel. Montel. Leçons ~ODS sur les families normales. 19271. C. Carathiodory. Carathiodory. Funktionentheorie. (Basie, C. (Basle, 1950.) 1950.)
II. NOTES II. NOTES AND AND PAPERS (1) p. p. 1. L.Lagrange. Lagrange. Sur Surlalaconstruction constructiondes descartes cartesgeographiques. goographiques. 1. J. L. (1) iv, pp. pp. 63792. (1779.) <Euvres, IV, 637—92. C. F. Gause. die Theile (2) 2. C. Gauss. Ailgemeine Allgemeine Auflosung Auftoaung der Aufgabe Aufgabe die Theile einer (2) p. 2. gegebenen Flii.che Fiäche so so abzubilden, abzubiiden, dass die Abbildung dem abgebildeten in gegebenen den kieinsten den. kleinsten Theilen Theilen ãhnlich ii.hnlil.'h wird. wird. (1822.) (1822.)Werke, Werke,Iv,IV,pp. pp.189—216. 189216. (3) Riemann. sine aligerneine (:J) p.2. p. 2.B.B. Riemmm.Grundiagen Grundlagenfür fiireine allgemeineTheorie Theorie der der Functionen Functionen einer compiexen (Dies. Gottingen, Gottingen, 1851.) verii.nderlichen GrOsse. Grosse. (Diss. 1851.) Werke Werke einer complexen veränderlichen pp.3—41 341 (§ 21, p. 39). (2nd ed.), ed.), pp. (4) p. See e.g. e.g. C. C. Carathéodory. Caratheodory. Variationsrechnung und partielle pa.rtielle DifferenDifferen· (4) p. 2. See tiaigleichungen tialgleichungen erster erster Ordnung Ordnung(Leipzig, (Leipzig, 1935), 1935), Pp. pp.305, 305,309, 309, 333. 333. · (5) p. p. 2. D. D. Hilbert. I)as Dirichietsche Prinzip. (Gott. (5) Hilbert. Das Dirichletsche Prinzip. (Gott. Festschrift, Festschrift, 1901.) 1901.) Gesammelte Abhandlungen (Berlin, (Berlin,1935), 1935), in, III,pp. pp.15—37. 1537. Gesammeite Abhandiungen (6) p.2. H.A.A.Schwarz. Schwarz.Gesamineite GesammelteAbhandiungen Abhandlungen(Berlin, (Berlin,1890), 1~90), ii, II, p. p. 145. (6) p. 2. H. E. Picard, Picard, Traité Traite d'Analyse, d'Analyse, T. ii, II, Chap. x. (7) p. 4. A. (7) A. F. F.Möbius. MObius. Werke, Werke, H, 11, pp. 205—43. pp. 20543. (8) p. 18. (8) 18. F. F.Klein. Klein.Vorlesungen VorlesungenOber tibernichteuklidische nichteuklidische Geometric Geometric (Berlin, (Berlin, 1928). (9) p. 18. 18. H. Poincare. n. Poincare. (Euvres, <Euvres, rr. (9) (10) (tO) p. 22. P. P.Fin,qkr. Firurler. Uber tiberKurven Kurvenund undFlaehen Fliiehenininaligemeinen allgemeinen Räu.' Rii.u.· en. en. (Dies. GOttingen, Gottingen, 1918.) reprint has (Diss. 1918.) AA reprint has been been published published(Basle, (Basle, 1951). 1951). (11) (ll) p. 34. 34. The Themonodromy monodromy theorem theoremwas wasoriginally originallygiven given by byWeierstrass. Weierstrass. gave in in his his lectures lectures is is to to be found found in 0. 0. &olz J. Orneiner, The proof he gave Stolz and J. Gmeiner, Einleitung in die die Funktionentheorie Funktionentheorie(Leipzig, (Leipzig,1910). 1910).
114
BIBLIOGRAPHICAL NOTES
(12) p. 35. analytisehenKurven. Kurven. GOtt. Gott. (12) p. 35. Uber Uberdie dieUniformisierung Uniformisierung beliebiger beliebiger a.nalytischen Nachr. Paper presented presented on on 12 12 April.) April.) See See especially especially p. 13. 13. Nachr. (1907. (1907. Paper (13) H.A. A.Schwarz. Schwarz. Zur ZurTheorie Theorie der der Abbildung. Abbildung. (1869.) (1869.) Gesammeite Gesammelte (13) p. P. 39. H. Abha.ndlungen, ii, n, p.p.108 108(especially (especially §§ 1). 1). Abhandlungen, The was first pubThe proof proof given givenininthe the text text is is due due to to Erhard Erhard Schmidt. Schmidt. It It was lished lished by C. C. Carathéodory. CaratModory. Sur Sur queiques quelques généralisations generalisations du du theorême theoreme de de M. M. Picard. Picard. C.R. C.R. 26 26 Dec. Dec. 1905. 1905. A A similar similar proof proof had had already already been been given given by by
Poincare. Sur Sur les les groupes groupes des des equations equations lineaires. lineaires. Acta Math. Vol. Vol. 44 H. Poincaré. (1884) especially especially p. 231. (1884) (14) 0. Pick. (14) p. 41. 41. G. Pick. Uber Ober eine eine Eigenschaft Eigenschaft cler der konformen konformen Abbildungen Abbildungen kreisformiger 77 (1916), (1916), p. 1. l. kreisformiger Bereiche. Bereiche. l\:lat.h. Math. Annalen, 77 (15) (15) p. 48. P. P.Koebe Koebe made the conjecture conjecture that (0) I ~ 4. The first complete that If' f' (0) proof was was !{iven given by 0. G. Faber. Faber,Neuer Xeu£'rBeweis Beweiselnes eines KoebeBieberbachschen K(){'beBieberbachschen uber konforme Abbildung. Munch. Satzes tiber Munch. Sitzungsber., Math. }lath. Phys. Kiasse (1916), Encykiopadie, i.e. (1916), p. 39. 39. (Compare Bieberbach, Bieberbach, Encyklopadie, I.e. p. p. 511). 511). Klasse (16) G. Julia. Julia.Extension Extensionnouvelle nouvelled'un d'unlenmie Iemmede deSehwarz. Schwarz. Acta Acta Math. Math. (16) p. 53. 53. 0. (1918), p. 349. 42 (1918), The J. Wolff Wolff and a.nd was was pub. pubThe general general theorem theorem given givenininthe the text text is due to J. Rendus, 13 The theorelll theoretn is here presented lished in Comptes Rendus, 13 Sept. 1926. 1926. The in the the form form in in which which itit was was proved proved by by the the author: author:Uber Uberdie die Winkel. Winkelvon beschrankten beschränkten analytischen analytisehen Funktionen. Funktionen. Ben. derivierte von Berl. Sitzungsber., Sitzungsber., 39, where, in ignorance of Wolff's priority, Phys.Mathem. Klasse (1929), p. p.39, priority, his result was proved afresh. For the definition definition of of continuous continuous convergence convergence for meromorphic meromorphic functions, functions, cf. ef. (17) p. 61. (17) 61. A. .4.Ostrow8ki. Ostrowski. Uber Uber Folgen Folgen analytischer analytischer Funktionen Funkt.ionen und undeinige einige Verscharfungen Picardschen Satzes. Satzes. Math. Math. Zeitschrift, Zeitschrift, 24 24 (1926), (1926), Verseharfungen des des Picardschen
p. 215. 215.
(18) 62. C. C. Carathéodery. Caratheodvry. Stetige Konvergenz Konvergenz und und normale normale Familien Famiien (18) pp. 61, 62. Funktionen. Math. Math.Annalen, Annalen,101 101(1929), (1929), p. p. 315 315 and and I.e. I.e. FunktionenFunktionenvon Funktionen. theorie, Book IV. iv. (19) p. 61. 61. Cf. (19) Cf. P. Montel. MonteZ. i.e. I.e. (20) p. 62. Cf. (20) Cf. M. Fréchet. Frechet. Les Les espaces espaees abstraits. (Paris, (Paris, 1928.) 1928.)
(21) p. p. 66. 66. Three attack have (21) Three distinct distinct methods methods of attack have been been used used to prove prove the fundamental theorem fundamental theorem of of conformal conformal representation: representation: Dirichiet's Dirichlet's Principle, Principle, the methods methods developed developed in in Potential PotentialTheory Theoryfor forproblems problemsconcerning concerningvalues values on the boundary, from the the Theory of on boundary, and a.nd methods methods taken taken exclusively exclusively from Theory of Functions. Proofs all, after after the the others others Proofs on on this this latter latterbasis basis were were given last of all, The history of the the problem problem up to 1918 isexhaustively exhaustively had been disposed of. The 191~ is dealt with by by Lichten8tein Lichlensl$in in Encyklopadie. in his article in the Encyldopadie. The proof presented presented in in this this book book goes goes back back to to L. L. Fejer Fejir and a.nd F. F. Riesz. Riesz. It was It was published published by T. T. Radó. Rad6. Uber i:berdie die Fundameutalabbildung Fundamentalabbildung schlichter schlichter Gebiete. Acta Acta Szeged, Szeged,I 1(1923). (1923).Rad6 Radódraws drawsattention attentiont{) to the the important important Gebiete. fact that thatsimple simple connectivity connectivityisisnot notmade madeuse use of ofin in the the course course of of the the proof. proof. Carathéodory,AAProof Proofofofthe the first first prinprin.For another type of of proof, proof, see C. C. Caratheodvry, cipal Theorem of Conformal Anniversary Volume Volume Conformal Representation, Representation, Courant Anniversary (New 1948), and Funktionentheorie, Funktionentheorie, l.c. l.c.Book BookVI, VI, Chap. ii. n. (New York, 1948), also the interesting paper by 0. Cf. aJao interesting paper G. Faber, Faber. Uber Ober den den Hauptsatz Hauptsa.tz den der
BIBLIOGRAPHICAL NOTES
115
Theorie der konformen konformen Abbildungen. Abbildungen. Munch. Miinch. Sitzungsber., Sitzungsber., Math..Phys. Math.Phys. Klasse (1922), Klasse (1922), p. 91. transformations ponctuelles et et leurs applicaapplicaSee also P. P. iSimonart. Simnnart. Surles Sur les tr&DSformations géométriques. (2e tions geometriques. (2" partie) Ia la representation representation conforme. conforme. Ann. de la la Soc. p. 81. Scient. de Bruxelles, Bruxelles, 50 (1930), (1930), Mémoires, Memoires, p. The theory theory also alsoapplies applies totosequences tiequences of ofRiemann Riemann surfaces, surfaces, cf. cf. (22) (22) p. P. 76. 76. The Untersuchungen über C. Caratheodory. Untersuchungen iiber die konformen konformen Abbildungen Abbildungen von Math. Ann. festen und Ann.72 72(1912), (1912),pp. pp.107—144. 107144. mid verii.nderlichen veränderlichen Gebieten. Math. (23) pp. 85 and 94. The (23) The method method here here used used isis aa simplification simplification of a proof proof given by Lindewf, Sur representation conforme conforme d'une a.ire simplement simplement by E. Lind,elaf, Sur Ia. la representation d'une aire 4e Congres Congrès dPs des ma.them. mathém. scandinaves connexe sur J'a.ire d'un d'un cercie. cercle. 4e sca.ndina.ves a connexe sur l'aire Stockholm (1916), pp. pp.59—90. 5990. (See (Seeespecially especiallypp. pp.75—84.) 7584.) Stockholm (1916), The modifications modifications made made were were suggested suggested in in the course of as. conversation with T. Radó. Rad6. (24) p. 87. For Forextensions extensions see: see: E. E.Lummer. Lummer. Uber Uberdie diekonforme konforme Abbildung Abbildung (24) p. 87. bizirkularer "urven vierter bizirkularer .S:urven vierterOrdnung. Ordnung.Diss. Diss.Leipzig Leipzig(1920). (1920). For extensions extensions see: tr. Seidel. On the the distribution distribution of values values of (25) (25) p.p. 90. 90. For see: W. &idel. On of bounded analytic functions. functions. Trans. Trans.Amer. Amer.Math. Math.Soc. Soc.3636(1934), (1934),pp. pp.201— 201Zurn Schwarzschen 226. C. C. Carathéodory. Caratheodory. Zum Schwarzschen Spiegelungsprinzip. Spiegelungsprinzip. Comm. Comm. Hclvetici, Vol. 19 19(1947), (1947),pp. pp.263—278. 263278. Helvetici, Vol. (26) For extensions extensions see: see: W. Jr. Seidel. Seidel. Uber Uberdie dieRanderzuordnung Randerzuordnung bei bei (26) p. p. 97. 97. For konformen Math.Annalen, Anna.len, 104 104 (1931), (1931), p. p. 182. 182. konformen Abbildungen. Abbildungen. Math. (27) (27) p. 97. 97. L. Ahlfor8. Ahlfora. Untersuchungen Untersuchungen zur zur Theorie Theorie der der konformen konformen AbbilAbbil9 (1930). dungen und der ganzen ganzen Funktionen. Funktionen. Acta ActaFennicae, Fennicae,A, A,x,I, No. No.9 (1930). (28) (28) p. 99. Cf. Cf. 0. G.Darboux, Darbvux, i.e. i, I, Livre 2, Chap. 3. (29) p. p. 102. (29) 102. For For the the application application of this this theorem theorem to Plateau's Plateau's problem problem see: T. Radó, The problem problem of ofthe theleast leastarea areaand and the the problem problemof ofPlateau. Plateau. Math. RadO, The Zeitschrift, 32 32 (1930), (1930), p. p. 763. 763. Zeitschrift, (30) p. 106. (30) 106. B. B.L.L.van vander derJvaerden. Waerden. Topologie Topologie und undUniformisierung Uniformisierung der Rie. Riemaunsehen Flii.chen. Flächen. Leipzig. mannschen Leipzig. Sitzungsber. Sitzungsber. Math.Phys. Math.Phys. Kiasse. K.lasse. Vol. Vol. 93 93 (1941}. (1941). (31) p. p. 108. T. Radó. Uber den Begriff (31) 108. T. RadO. Uber Begriff der der Riemannschen Riemannschen Fiache. Fliiche. Acta Szeged, 2 (1925), (1925), pp. pp.101—121. IOI121. (32) 108. C. C. C'arathéodory. Caratheodory. Bemerkung Riemannschen (32) p. 108. Bemerkung iiber über die Theorie der Riernannschen Flachen. Flii.chen. Math. Math.Zeitschrift, Zeitsehrift,5252(1950), (1950),pp. pp.703—8. 7038. 131—136 contain contain in principle all that The ~§ 131136 principle all that isisnecessary necessary for for the study study of the tre representation representation of the frontier frontier in in conformal conformal transformation transformation of the most genera; genera: simplyconnected domain. The relevant relevant literature literatureis is mentioned by Lichgen8tein hisarticle article in in the the Encyklopii.die Encykiopedie I§ 48). Lichtenstein ininhis 120—123ononthe thekernel kernelof ofaa sequence sequence of of domains domains can be The results of§§ 120123 be extended to to the the case case of ofrepresentation representationof of2ndimensional 2ndimensional domains domains on one one another by systems of nn analytic functions functions of na variables. variables. See: See: C. Uber die Abbildungen C. Carathéodory. Caratlteodory. Cber Abbildungen die durch Systerne Systeme von analyanalytischen Funktionen mit erzeugt werden. werden. Math. tischen mit mehreren mehreren Veränderlichen Verii.nderlichen erzeugt Zeit.schrift, 37 37 (1932), (1932), p. 758. Zeitschrift,
There is an extensive literature literature on on the the representation representation of of multiplyconnected multiplyconnected domains and Riemann Riemann surfaces. surfaces. (Cf. (Cf.Lichten8tein, Lichtenstein, I.e. i.e. §§ 41, 44 and and 45.) 45.)