PROBLEMS IN COMPLEX VARIABLE THEORY JAN G.KRZYZ
PROBLEMS IN COMPLEX VARIABLE THEORY JAN G.KRZYZ
MODERN ANALYTIC AND COMPUTATIONAL METHODS IN SCIENCE AND MATHEMATICS Number@@O
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Richard Bellman Editor
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PROBLEMS IN COMPLEX VARIABLE THEORY
by
JAN G. KRZYZ Maria Curie-Sklodowska University, Lublin (Poland) Institute of Mathematics, Polish Academy of Sciences
AMERICAN ELSEVIER PUBLISHING COMPANY, INC. NEW YORK
PWN-POLISH SCIENTIFIC PUBLISHERS WARSZAWA
197 1
AMERICAN ELSEVIER PUBLISHING COMPANY, INC. 52 Vanderbilt Avenue, New York, N.Y. 10017 ELSEVIER PUBLISHING COMPANY, LTD. Barking, Essex, England ELSEVIER PUBLISHING COMPANY 335 Jan Van Galenstraat, P.O. Box 211 Amsterdam, The Netherlands
International Standard Book Number 70-153071 Library of Congress Catalog Card Number 0-444-00098-4 COPYRIGHT 1971 BY PANSTWOWE WYDAWNICTWO NAUKOWE WARSZAWA (POLAND), MIODOWA 10
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, American Elsevier Publishing Company, Inc. 52 Vanderbilt Avenue, New York, N.Y. 10017.
I'RINTliIl IN (,OI.AND
To the memory of M ieczyslaw Biernacki
Contents Foreword
xiii
Notation
xv PROBLEMS
1. Complex Numbers. Linear Transformations 1.1. Sets and Sequences of Complex Numbers . 1.2. Spherical Representation . 1.3. Similarity Transformations 1.4. Linear Transformations . 1.5. Symmetry . . . . . . . 1.6. Conformal Mappings Realized by Linear Transformations 1.7. Invariant Points of Linear Transformations 1.8. Hyperbolic Geometry [3], [21] . . . . .
3 8 10 11
12 13 14 16
2. Regularity Conditions. Elementary Functions 2.1. Continuity. Differentiability . . . . . . 2.2. Harmonic Functions . . . . . . 2.3. Geometrical Interpretation of the Derivative. 2.4. Conformal Mappings Connected with w = Z2 2.5. The Mapping w = +(Z+Z-l) . . . . . . . . 2.6. The Exponential Function and the Logarithm 2.7. The Trigonometric and Hyperbolic Functions 2.8. Inverse Trigonometric and Hyperbolic Functions . 2.9. Conformal Mapping of Circular Wedges.
19 21 23 24 25 27 28 30 30
3. Complex Integration 3.1. Line Integrals. The Index [I], [10] 3.2. Cauchy's Theorem and Cauchy's Integral Formula . 3.3. Isolated Singularities 3.4. Evaluation of Residues. . . . . . . . . 3.5. The Residue Theorem . . . . . . . . . 3.6. Evaluation of Definite Integrals Involving Trigonometric Functions 3.7. Integrals over an Infinite Interval . . . . . 3.8. Integration of Many-Valued Functions [21] . 3.9. The Argument Principle. Rouche's Theorem.
33 38 41 43 45 48 49 52 54
4. Sequences and Series of Analytic Functions 4.1. Almost Uniform Convergence 4.2. Power Series . . . . . . . . . .
57 58
Ix
CONTENTS
x
4.3. 4.4. 4.5. 4.6. 4.7. 4.8.
Taylor Series .......... . Boundary Behavior of Power Series .. The Laurent Series . . . . . . . . . Summation of Series by Means of Contour Integration Integrals Containing a Complex Parameter. The Gamma Function . . . . Normal Families [I], [6], [16] . . . . . . . . . . . . . . . . . . . . .
60 65 66 69 72 75
5. Meromorphic and Entire Functions 5.1. Mittag-Leffler's Theorem [I] . 5.2. Partial Fractions Expansions of Meromorphic Functions [11] 5.3. Jensen's Formula. Nevanlinna's Characteristic [18] 5.4. Infinite Products [I], [10] . . . . . . . . . . . 5.5. Factorization of an Entire Function [I], [6], [10]. 5.6. Factorization of Elementary Functions [11] 5.7. Order of an Entire Function [6], [10], [13], [18] .
77 78 80 82 84 85 88
6. The 6.1. 6.2. 6.3. 6.4.
90 91 92 94
Maximum Principle The Maximum Principle for Analytic Functions Schwarz's Lemma [16] . . . . . . . . . . . Subordination [6], [15], [16], [22] . . . . . . The Maximum Principle for Harmonic Functions
7. Analytic Continuation. Elliptic Functions 7.1. Analytic Continuation [1], [2], [6], [10] 7.2. The Reflection Principle [I] . . . . . 7.3. The Monodromy Theorem [I], [10]. . 7.4. The Schwarz-Christoffel Formulae [I], [10] 7.5. Jacobian Elliptic Functions sn, en, dn [1], [6], [7], [10] 7.6. The Functions cr, 1;, tJ of Weierstrass [1], [6], [7], [10]. 7.7. Conformal Mappings Associated with Elliptic Functions [24] 8. The 8.1. 8.2. 8.3. 8.4. 8.5. 8.6.
Dirichlet Problem Riemann's Mapping Theorem [I], [6], [10], [16] Poisson's Formula [I], [26], [27] . . . . . . . The Dirichlet Problem [I], [14], [15], [19], [27] Harmonic Measure [I], [15], [16], [19], [25], [26], [27], Green's Function [15], [16], [17], [27j Bergman Kernel Function [6], [12], [24]
95 96 98 100 105 106 108
Ito III 113 114 116 118
9. Two-Dimensional Vector Fields 9.1. Stationary Two-Dimensional Flow of Incompressible Fluid [4], [9] 9.2. Two -Dimensional Electrostatic Field [4]. . . . . . . . .
121 124
10. Univalent Functions 10.1. Functions of Positive Real Part [26], [27] . 10.2. Starshaped and Convex Functions [16], [17] 10.3. Univalent Functions [16], [17], [20] . . . . 10.4. The Inner Radius. Circular and Steiner Symmetrization [171. [20]. 10.5. The Method or [nncr Radius Majorization [17]
127 129 130 133 136
CONTENTS
xi
SOLUTIONS 1. Complex Numbers. Linear Transformations . 2. Regularity Conditions. Elementary Functions 3. Complex Integration. . . . . . . . . . . 4. Sequences and Series of Analytic Functions 5. Meromorphic and Entire Functions . . . 6. The Maximum Principle ...... . 7. Analytic Continuation. Elliptic Functions 8. The Dirichlet Problem. . . . . 9. Two-Dimensional Vector Fields 10. Univalent Functions. Bibliography . Subject Index
141 155 168 191 212 224 229 245 256 260 277 279
Foreword This collection of exercises in analytic functions is an enlarged and revised English edition of a Polish version first published in 1962. The book is mainly intended for mathematics students who are completing a first course in complex analysis, and its subject matter roughly corresponds to the material covered by Ahlfors's book [1]. Some chapters, for example, evaluation of residues, determination of conformal mappings, and applications in the two-dimensional field theory may be, however, of interest to engineering students. Most exercises are just examples illustrating basic concepts and theorems, some are standard theorems contained in most textbooks. However, the 'author does believe that the reconstruction of certain proofs could be instructive and is possible for an average mathematics student. When the subject matter of a particular chapter is not covered by standard textbooks, the numbers in parantheses given in the contents indicate a corresponding bibliography position which may be consulted for further information. Some problems are due to the author, and some were adopted by the author from various sources. It was beyond the scope of author's possibility to trace the original sources and therefore the detailed references are omitted. The second part of the book contains solutions of problems. In most cases a complete solution is given; in some cases, where no difficulties could be expected, or when an analogous problem has been already solved in a detailed manner, only a final solution is given. The author is well aware that it was extremely hard to avoid mistakes in a book of this kind. He did his best, however, to reduce their number to a minimum. It is the author's pleasant duty to thank W. K. Hayman, Z. Lewandowski, and Q. I. Rahman, who suggested some problems included in this collection. Thanks are also due to Mrs. J. Zygmunt for her help in preparing the manuscript, as well as to M. Stark for his help and encouragement.
Lublin, July 1969
JAN
XIII
G.
KRZYZ
Notation 1. Set theory
a is an element of the set A a is not an element of A
aeA a¢A
BcA AnB
B is a subset of A
A u,B
A""-B A ""- a {a: W(a)} A
frA r)A
o
Intersection of sets A and B Union of sets A and B The complement of B with respect to A The set A with the element a removed The set of all a having a property W(a) Closure of the set A Boundary of the set A The boundary cycle of a domain A taken with positive orientation The empty set
2. Complex numbers I"CZ
Imz
:
1=1 largz IArgz
The real part of a complex number z = x+;y, i.e. the real number x The imaginary part of a complex number z = x+iy, i.e. the real number y The conjugate of z = x+iy, i.e. the complex number x-iy The absolute value of z = x+iy, i.e. yx2+y2 The argument of z =F 0, 00, i.e. any angle 0 satisfying the equations Izl cosO = rez, Izl sinO = imz. There exists a unique value of argz which satisfies - 7 t < argz ~ 7t. It is called the principal value of argument and is denoted Argz
3. Sets of complex numbers
I::, •• z 2] (Z •• Z2)
1= ••
Z2 • •••• Zn]
Closed line segment with end points z 1, Z 2 Open line segment with end points Zl' Z2 Polygonal line joining Zl' Z2' ••• , Zn in this order xv
xvi
NOTATION
[Zl' Z2) = [Zl' Z2]""'Z2, (Zl' Z2] = [Zl' Z2]""'Zl [0, + (0), (- 00, 0] Positive and negative real axis [0, +ioo), (-ioo, 0] Positive and negative imaginary axis ( - CXl, + (0) The set of all real numbers K(zo; r) The open disk with center at Zo and radius r K( 00; 1) The set of all z with Izl > 1 The circle with center at Zo and radius r C(zo; r) The upper (lower) half-plane H+ (H-)
C (C)
The finite (extended) plane
The open quadrants of C, e.g. (+; -) The convex hull of a set A dist(a; B) = inf{x: x = la-bl, b E B} dist(A; B) = inf{x: x = la-bl, a E A, bE B} (=f; =f)
=
{z: rez > 0, imz
< O}
convA
We use the same symbol C(zo; r) for frK(zo; r), as well as- for oK(zo; r); and similarly, [zo, z t1 denotes either a set, or an oriented segment. We hope that this does not cause any misunderstanding. 4. Functions and mappings 1: 1 f(A) p(a, b) 0, lei
z+i (. .){o z: <arg z-i < 4
7t }
11l
< a;
;
(iv) {z: 0 ~ reiz < I}; (v) {z: rez 2 > a}., lI. > 0; (vi)
{z: I ;~~ I < I};
(vii) {z: Izl+rez ~ I}; (viii) {z: Iz2-11 < I}; (ix) {z: re[z(z+i)(z-i)-l]
> O}.
1.1.22. Show that the set {z: arg(z-a)(z-b) = const} is an arc of an equilateral hyperbola whose center is located at 1-(a+b).
> 0 for which the set {z: Iz 2+az+bl < R}
1.1.23. Evaluate all R i~
connected. 1.1.24. Explain the geometrical meaning of the set
{z: AlzI2_Bz-Bz+C where A, C are real, A
=1=
0, IBI2
=
O},
> AC.
1.1.25. Find the radius and the center of Apollonius circle
Iz-allz-bl- 1 = k
(k
=1=
1, k
> 0).
1.1.26. Find the equation of the circle through three not collinear points :I.ZZ,ZJ
(cf.Ex. 1.1.24).
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
6
1.1.27. Suppose IZII =/: e, 1 and 0, ZI' Z2 are not collinear. Show that the circle through ZI' Z2' zi i has [zl(1+lz212)-z2(1+lzI12)] (ZIZ2-ZIZ2)-1 as center and its radius is IZI-Z2111-ZIZ21IzIZ2-ZIZ21-I. Also show that Zi i is situated on this circle. 1.1.28. Suppose mI' m2' m3 are nonnegative, mi +m2-+ m3 = 1, and ZI, Z2' Z3 are not collinear. Show that (i) the point Zo = mizi +m2z2+m3z3 belongs to the closed triangle T with vertices ZI' Z2, Z3; (ii) conversely, for any Zo E T there exists a unique system of nonnegative mI' m2' m3 with mI+m2+m3 = 1 such that Zo = mIzI+m2z2+m3z3' The numbers mJ are called barycentric coordinates of Zo w.r.t. T.
1.1.29. The intersection of all closed and convex sets containing a given set A is called the convex hull of A and is denoted conv A. Show that
n
mk
~ 0, k
=
1, "', n; ~ mk =
I}.
1.1.30. Show that conv {ZI' Z2' ... , zn} = U Tk1m , where T k1m is the closed triangle with vertices Zt, ZI, Zm and the summation ranges over all triples {k, I, m} of positive integers ~ n. n
1.1.31. Prove that the equality
L (C -
Zk)-I = 0 implies: k=l CEconv {ZI' Z2' ... , zn}.
1.1.32. Prove following theorem (due to Gauss and Lucas): all zeros of the derivative of a polynomial are contained in the convex hull of zeros of the given polynomial. 1.1.33. Show that lim n .... oo
(1+ X+iy)n = eX(cosy+isiny). n
1.1.34. Discuss the behavior of the sequence {Zn} , Zn
= (1+i)
1.1.35. Show that, if {en} and is convergent, too.
(1+ ~) ... (1+ !).
L Ibnl
both converge, then the series
L
Cnb
1.1. SETS AND SEQUENCES
7
1.1.36. Suppose rez I ~ 0 (n = 1,2, ... ) and both that also IZnlz is convergent.
L
L Zn' L z~
converge. Show
1.1.37. Prove Toeplitz's theorem: Suppose (a"k) is an infinite matrix of complex numbers (n, k = 1, 2, ... ) which satisfies: 00
(i)
L la.kl ::::;; A
for n = 1, 2, ... ;
k=l
(ii) lim ank = 0
for n = 1,2, ... ;
k-+oo 00
(iii) lim (L ank) n--+oo
=
1.
k= 1
00
Then for any positive integer n and any convergent {en} the series
L ankCk k=l
is convergent. 00
Moreover, if Zn = Lank Cto then
lim Zn exists and is equal
lim Cn. n-+oo
k=l
P1 +Pz+ ... Pn MOl." ~ > lor n = 1,2, ... Ip11+IPzl+ ... +IPnl lim(!p11+lpzl+ ... +IPnD = +00. Show that for any convergent {zn} 1138 .. • Suppose
and
1.1.39. Suppose Zn = UO+U1 + '" +Un-1 +cun is a convergent sequence and rec > -t. Show that also Wn = UO+U1 + ... +Un -1 +un is convergent and has Ihe same limit. 1.1.40. Suppose {Pn} is a sequence of positive reals monotonically incr~asing 10 infinity. Show that for any convergent series Zn with complex terms we have:
L
1.1.41. Show that f
L ,unZn
converges and ,un
lim,un(z1+zz+ ... +zn)
--+
0 then
=
O.
n-+OO
1.1.42. Suppose {un}, {v n} converge to U and v resp. Show that Wn
rOllverges to
11'1).
1 =
-
n
(U1Vn+UZVn_1 +
'"
+unv 1)
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
8
1.2. SPHERICAL REPRESENTATION
1.2.1. Suppose OXlX2X3 is the system of rectangular coordinates whose axes OXl, OX 2 coincide with real and imaginary axes Ox, Oy of the complex plane C. Suppose, moreover, that the ray emanating from the north pole N(O; 0; 1) of the unit sphere s: xi + x~ + x~ = 1 and intersecting S at A (Xl; x 2 ; X3) intersects C at the point z. The point z = x+iy is called the stereographic projection of A(x l ; X2; x 3 ) whereas A is called the spherical image of z. Show that X
z+z 1+lzI 2 '
--~-=-
1 -
X
z-z 2 = iO+lzI2)'
X3
=
IZ12-1 1+lz1 2 '
and Xl +iX2
z=----~.
l- x3
Hence the points of the sphere S (also called Riemann sphere) can be used for geometrical representation of complex numbers. 1.2.2. Find the spherical images of eilZ , -1 +i, 3-4i. 1.2.3. Describe spherical images of northern and southern hemisphere. 1.2.4. Show that any straight line in C has a circle through N as its spherical image. 1.2.5. Show that the stereo graphic projection of any circle on S not containing N is also a circle. 1.2.6. Show that the spherical images of z, Z-l are points symmetric w.r.t. the plane OXlX 2 . 1.2.7. Find the relation between the spherical images of following points: (i) z, - z; (ii) z, z; (iii) Z, Z-l. 1.2.8. If cp, () denote the geographical latitude and longitude of A respectively, show that the stereographic pr~jection of A has the representation: z = e i9tan (t7t+tcp). 1.2.9. Show that
z, ,
correspond to antipodal points of S, iff
zC =
-1.
1.2.10. Prove that the circle AlzI 2+Bz+Bz+C = 0 (A, C real) has a great circle on S as its spherical image, iff A + C = O. 1.2.11. Show that C(zo; R) is the stereographic projection of a great circle on S, iff R2 = 1+lzoI2. 1.2.12. Find the stereographic projection of the great circle joining the points ( -] 3j ,• -
4 . 12) )-3 '-]3'
(
2 • 2 • I) 3 , f .
- -f'
1.2. SPHERICAL REPRESENTATION
1.2.13. The distance o"(Zl' Z2) between two points on S whose stereographic projections are Zl, Z2 is called the spherical distance between Zl and Z2' Show that o"(Zl' Z2) = 21z1-z21 [(1 + Iz112)(1+ Iz212)]-1/2. 1.2.14. Suppose dO", ds are lengths of infinitesimal arcs on Sand C resp. corresponding to each other under stereographic projection. Suppose, moreover, the arc of length ds emanates from the point Z E C. Show that
~; =
2(1
+ IzI2)-1.
Show, moreover, that the angle between any two regular arcs in C and the angle between their spherical images are equal. 1.2.1S. Suppose the sphere S is rotated by the angle rp round the diameter whose end points have a, _a- 1 (cf. 1.2.9) as stereographic projections. Suppose, moreover, z, Care stereographic projections of points corresponding to each other under this rotation. Show that
-C-a -- = l+aC
z-a
el'P---
l+az •
1.2.16. Suppose At, A2 E S and at, a2 are stereographic projections of At lind A 2 , resp. Find the set of all points a E C such that a is the stereographic r>rojection of a point A E S equidistant from At and A 2 • 1.2.17. Find the radius of the circle on S whose stereographic projection is C(a; r).
r
1.2.18. Suppose is a regular arc on Sand 'Y is its stereographic projection. Show that the length /(r) of r is equal to
~ 1+~ZI2 ds. y
011
1.2.19. Find the stereo graphic projection of a rhumb line on S, i.e. of a line S which cuts all meridians at the same angle.
r
1.2.20. Find the length /(F) of the rhumb line joining the points whose '\crcographic projection are Zl = r 1, Z2 = r2 eif%, 0 < ex < 2". Evaluate /(F) . 1 . 1 .• f i lor =1 =~, Z2 =-(3+IJl3). )13 2
III'
1.2.21. Evaluate the area of a spherical domain D being the spherical image II regular domain ;1 in C.
10
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.2.22. Show that the area ITI of a spherical triangle T with angles is equal a+{3+Y-7t.
(X,
{3, y
1.2.23. Evaluate the area of the spherical triangle T whose vertices are (0; 2- 1 / 2 ; 2- 1 / 2 ), (2- 1 /2; 2- 1 /2; 0), (0; I; 0).
1.3. SIMILARITY TRANSFORMATIONS
1.3.1. Show that each similarity transformation w = az+b (a i= 0) can be composed of a translation, a rotation and a homothety with center at the origin. 1.3.2. Prove that a similarity transformation (i) carries circles into circles and (ii) parallel straight lines into parallel straight lines; (iii) leaves the ratio (Z3-Z1)!(Z3-Z2) unchanged; (iv) leaves the angle between two curves unchanged. 1.3.3. Find a similarity transformation mapping the strip
{z: 0 < re z < I} onto the strip (w: limwl
so that (z
=
-t) +-+
(w
=
< +7tl
0).
1.3.4. Find the most general similarity transformation mapping (i) the upper half-plane onto itself; (ii) the upper half-plane onto the lower one; (iii) the upper half-plane onto the right half-plane. 1.3.5. Find the similarity transformation mapping the segment [a, b] onto [A, B] so that a +-+ A, b +-+ B.
1.3.6. Find the similarity transformation mapping the triangle with vertices 0, 1, i onto the triangle with vertices 0, 2, 1+ i. 1.3.7. Find the similarity transformation mapping the strip
{x+iy: kx+b l
:(;
y :(; kx+b 2 },
onto the strip
{IV: 0:(; re w :(; I} so that (z
=
ibJl)
+-+
(II' =-, 0).
bi
< b2 ,
11
1.4. LINEAR TRANSFORMATIONS
1.3.8. Show that for any similarity transformation w = az+b (a # 0, 1) there exists a unique invariant point Zo; show that the transformation can be . composed of rotation and a homothety center at Zo. 1.3.9. Find the invariant point zo, the angle of rotation and the ratio of homothety for the transformations in (i) Exercise 1.3.6; (ii) Exercise 1.3.7. 1.3.10. Show that for any similarity' transformation w = az+b (a # 0, 1) there exists a family of logarithmic spirals il}variant under the transformation. 1.4. LINEAR TRANSFORMATIONS
1.4.1. Show that any linear transformation w = (az+b)/(cz+d) (a, b, e, d are complex constants, ad-be # 0) is composed of a translation: Zl = z+/X, an inversion: Z2 = llz1 and a similarity transformation: w = Az2 +B (some of these transformati R, onto {w: p < Iwl < I}. Show that p=
~ - V(~r-1.
1.6.13. Find a linear transformation mapping the bounded domain whose boundary consists of C(O; 2), C(1; 1) onto a strip bounded by two straight lines parallel to the imaginary axis. 1.7. INVARIANT POINTS OF LINEAR TRANSFORMATIONS
1.7.1. If w = (az+b)/(ez+d), ad-be #- 0, and IX = (alX+b)/(elX+d), then IX is called an invariant point of the given linear trans/ormation. Find the general linear transformation with two different and finite invariant points IX, fJ. 1.7.2. Show that the general linear transformation with invariant points 0, ro is a similarity w = Az (A #- 0). 1.7.3. A transformation T whose inverse T- 1 is identical with T is called an involution. Show that a linear transformation (az+b)/(ez+d) different from identity is an involution, iff a+{[ = O. 1.7.4. Show that an involution different from identity has always two different invariant points. 1.7.5. Prove that any linear transformation with two different invariant points can be written in the standard form: (W-IX)/(W-(J)
1.7.6. Show that if L1 then
=
A(z-IX)/(z-fJ).
=
(d-a)2+4be and the sign of .
A
-=
(a jd+}lIi)/(a j-d--}lt1).
}/Lf is suitably chosen,
1.7. INVARIANT POINTS
15
1.7.7. Bring the linear transformation w of Exercise 1.7.5.
=
(z+i)/(z-i) to the standard form
1.7.S. Prove that a linear transformation with only one invariant point 00 is a translation. Also prove that a linear transformation with only one finite invariant point IX (or the parabolic transformation) has the form (W-IX)-l
=
(z-IX)-l+h
(h i= 0).
1.7.9. Find the parabolic transformation mapping ceO; R) onto itself whose only invariant point is z = R. 1.7.10. If IX, {3 are invariant points and A = IAle i9 (cf. Ex. 1.7.5) then the circle Cz with diameter [IX, {3] is carried into a circle Cw with radius R -= +IIX- {3llcosOI- 1 . 1.7.11. The sequence {zn} is defined by the recurrence formula: Zn+1 =/(zn), wherefis a linear transformation with at most 2 invariant points and Zo is given. Discuss the convergence of {z,,}. 1.7.12. Find the points of accumulation of the sequence {zn}: Zo =
0,
1.7.13. If A in Exercise 1.7.5 is real, the corresponding linear transformation is called hyperbolic, if A = eiq> (with real q;) it is called elliptic. Prove that in both cases there exists a family of circles such that any' circle of the family is mapped onto itself under the transformation. 1.7.14. Prove that for any parabolic transformation with an invariant point rx there exists a family of circles tangent to each other at IX and such that each d rde of the family is mapped onto itself under the given transformation. 1.7.15. Suppose IX, {3 are invariant points of a linear transformation which is not an identity and carries a circle C into itself. Show that either IX, f3 are ,itllated on C, or are symmetric w.r.t. C. 1.7.16. Suppose a linear transformation which is neither elliptic, nor hyperholic, has two finite and different invariant points. Show that no circle can be IIlIlpped onto itself by this transformation. 1.7.17. Suppose w = (az+b)/(cz+d), ad-bc = 1 and a+d is real. Prove thai the transformation is elliptic if la+dl < 2, hyperbolic if la+dl > 2 and (lilraholic if la+dl = 2.
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
16
1.7.18. Show that the rotations of the Riemann sphere correspond to elliptic transformations in the plane after stereo graphic projection. 1.7.19. Suppose IX, P are invariant points of an elliptic transformation and ~ 2. Prove that this transformation corresponds to a rotation of the Riemann sphere followed by a translation.
IIX-PI
1.7.20. Show that the linear transformation
w = (az+b)/(-bz+a),
lal+lbl
> 0,
corresponds to a rotation of the Riemann sphere. Evaluate stereographic projections of the end-points of the diameter being the axis of rotation, as well as the angle of rotation in terms of a and b. 1.7.21. Find the linear transformation representing the rotation of the Rieround the diameter with end-points (f; f; t), mann sphere by an angle (-f; -f; -t)·
t'lt
1.7.22. Find the general linear transformation representing a rotation of the Riemann sphere by an angle 'It. Show that this is an involution. 1.7.23. Show that for any involution with two finite invariant points IX, P which is different from identity the factor A in Exercise 1.7. 5 equal~ -1. Prove that the straight line through IX, fJ is mapped onto itself. 1.7.24. Find all
str~ight
lines remaining invariant under the involution
2wz+i(w+z)-2
=
o.
1.8. HYPERBOLIC GEOMETRY
In hyperbolic (Lobachevski-Bolyai) geometry the axioms of Euclid are valid except for the parallel axiom: there are at least two different straight lines in the plane through a given point not on the straight line L which do not meet L. There is a very simple and elegant way essentially due to Poincare of satisfying the axioms of non-Euclidean geometry by a suitable choice of cert~ configurations in Euclidean space. The points in the hyperbolic plane or h-points are the points of the unit disk K(O; 1). The straight lines (hyperbolic straight lines, or h-lines) are the arcs of circles, or straight line segments orthogonal to the unit circle and interior to it. Hyperbolic motions are linear transformations mapping K(O; 1) onto itself. Two sets of h-points are congruent if there exists an h-motion carrying one set intoanother one.
I.s. HYPERBOLIC GEOMETRY
17
We can also introduce in a natural way h-distance in the hyperbolic plane which is invariant under h-motion. Complex numbers and linear transformations are very convenient tools in analytic treatment of hyperbolic geometry.
1.8.1. Prove that there exists a unique h-line through any two h-points represented by Zl' Z2 (Zl"# Zz, IZll < 1, IZzl < 1). 1.8.2. Prove that there exists a unique h-line through a given point direction ei!l (i.e. meeting C(O; 1) at ei!l).
Zl
in a given
1.8.3. The unit circle C(O; 1) is called the h-line at infinity. Two h-lines meeting at infinity (i.e. two circular arcs orthogonal to C(O; 1) intersecting each other at a point on C(O; I) are called h-parallels. Show that there are two h-parallels to a given h-line L through a given h-point Zl not on L, as well as infinitely many h-lines through Zl not meeting L. 1.8.4. Find the general form of an h-rotation (i.e. an h-motion with a unique invariant h-point zo). 1.8.5. Find a general h-translation, i.e. an h-motion with two invariant points the h-line at infinity.
011
1.8.6. Find a general h-boundary rotation, i.e. an h-motion with a unique invariant point on C(O; 1). 1.8.7. Find a general h-motion. Verify the group property for h-motions. 1.8.8. Write parametric equation of an h-segment [Zl' ZZ]h' i.e. a subarc of h-Iine with end-points Zl' Z2. 1.8.9. Suppose the h-segments [a, Z]h' [b, W]h are congruent in the sense of hyperbolic geometry. Verify that
l(z-a)/(I-az)1 = l(w-b)/(l-bw)l· 1.8.10. Suppose C and r are two regular curves situated in the unit disk and carried into each other under an h-motion. Show that
1.8.11. Consider all regular curves situated inside K(O; 1) and joining two Ji1lcd points 0, R (0 < R < I) of K(O; 1). Show that
f. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
18
has a minimum for y being the segment [0, R], the minimum being equal 1 I+R Tlog l-R = artanhR. Hint: Verify first that we can restrict ourselves to regular curves with parametric representation 0 = O(r), where 0, r are polar coordinates.
1.S.12. If C is a regular curve situated inside K(O; 1) then
C Idzi
J l-lz[2
c
is called hyperbolic (or h-) length of C. Show that the h-segment with end points Zl' Z2 is the curve with shortest h-Iength among all regular curves in K(O; 1) joining Zl to Z2' The h-Iength of [Zl' Z2]h is called hyperbolic (or h-) distance P(Zl' Z2) of points Zl' Z2' Also show that P(Zl' Z2)
=
artanhlzl-z2111-Z1Z21-1.
1.S.13. Find the h-circle with h-radius R, i.e. {z: p(zo, z)
=
R}.
Also find its h-Iength lh' 1.S.14. Verify the usual properties of a metric for P(Zl' Z2)' 1.S.lS. Show that P(Zl, Z2) = tlog (Zl , Z2' ei/l, ei"),
where e i", Zl, Z2' ei/l are successive points of a circle orthogonal to C(O; 1). 1.S.16. Suppose a regular domain D, into [j. Prove that
CC
JJD
Dc
dxdy (l-x 2_y2)2 =:'
rc
JJ
K(O; 1), is carried under h-motion d~dYJ (l_~2-172)2 .
Q
The integral on the left is called hyperbolic (or h-) area of D and will be denoted
IDlh' 1.S.17. Find l[jlh for Q
=
-, {z: Izl < R}.
1.S.lS. Consider an h-triangle T, i.e. a domain bounded by three h-segments with angles rx, (:1, y. Show that
ITlh
=
t[1t-(rx+(:1+y)].
Hint: Take the origin as one of the vertices.
1.8.19. Evaluate the h-area
or
an h-triunglc with Vl:rti~cs
ZI' =2'
=.\.
CHAPTER 2
Regularity Conditions. Elementary Functions A complex function w = fez) defined on a set of complex numbers A is actually defined ~y a pair of real-valued functions u(x, y), vex, y) of two real variables x, y (x+iy = z) with a common domain A. In a formally identical manner as in real analysis we can introduce the notions of limit, continuity and differentiability. If fez) = u(x, y)+iv(x, y) is differentiable at zo = xo+iyo, then u, v have partial derivatives at zo satisfying Cauchy-Riemann equations at this point: Ux = vy, uy = -Vx ' On the other hand, if all the four partials of first order of u, v exist in some neighborhood of zo, are continuous and satisfy Cauchy-Riemann equations at zo, then f = u+ iv is differentiable at Zo, i.e. f(zo+h)
=
f(zo) +ah +h'fJ (h) ,
where
lim 'fJ(h) = 0; h .... O
the constant a is called the derivative off at zo. The most interesting and most important case occurs when f is defined and has a derivative at every point of some domain (or open, connected set) D in the plane. Thenfis called analytic, Iw/omorphic or regular in D. Regularity has far-reaching consequences that go much beyond what one can obtain from differentiability in the real case. The theory of analytic functions has as its central theme just the investigation or these consequences. So, for example, regularity in a domain D implies the existence of derivatives of all orders at all points of D. Since the definitions of Ihe derivative in real and complex domain are formally identical, the usual I"lIlcs of differentiation as the formulas concerning the derivative of a sum, a prodnct or a quotient, as well as the chain rule, remain the same in complex case.
2.1. CONTINUITY. DIFFERENTIABILITY
2.1. t. Discuss the continuity at z = 0 offunctions defined at z :1= 0 as follows: (i)
/~~Zl; (ii) z-lrez; (iii) z-2 rez2; (iv) z-2(rez 2)2 and equal 0 at z 19
=
O.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
20
2.1.2. Suppose J is defined and uniformly continuous in K(O; I). Prove that for any sequence {zn}, Zn E K(O; I), convergent to C(lCJ = I) there exists a limit €p(C) depending only on C, and not on a particular choice of {zn}. Also prove that F: F(z) = J(z) for z E K(O; I), F(C) = €p(C) for CE C(O; I), is continuous in K(O; 1). 2.1.3. Verify that the function J:
J(O)
=
0,
J(z)
Izl- 2 (I+i)imz2
=
satisfies Cauchy-Riemann equations at z 2.1.4. Verify that J(z) 2.1.5. Suppose J(z)
=
=
0. Is
J
differentiable at z
zrez is differentiable at z
= u(x, y)+iv(x, y)
z i= 0,
for
=
°
=
O?
only.
and the limit
limreh-1 [f(zo+h)-J(zo)] h-+O
exists. Show that the partials
u x , Vy
at Zo both exist and are equal.
2.1.6. Suppose u(x, y), v(x, y) are continuous and have continuous partials of first order at Zo = xo+iyo' If J = u+iv and the limit
lim Ihl-1If(zo+h)-J(zo)1 h-+O
exists, then either f, or 1 = u- iv has a derivative at Zo. 2.1.7. Verify that the following functions fulfill the Cauchy-Riemann equations in the whole plane: (i)J(z) = Z3; (ii)J(z) = eXcosy+iexsiny. 2.1.8. Verify that J(z) 2.1.9. If J a constant.
=
=
x(x2+y2)-1_iy(x2+y2rl is analytic in C"-O.
u+iv is analytic and satisfies u 2 = v in a domain D, then J is
2.1.10. Suppose Llu
a2 u a2 u
=
ax2
+ ay2 .
If J is analytic and does not vanish in a domain D, then
2.1.11. Prove that for an analytic function
J:
2.2. HARMONIC FUNCTIONS
21
2.1.12. Write Cauchy-Riemann equations for fez)
=
U(r, O)+iV(r, 0),
where
z
=
re i8 •
Express l' in terms of partials of U, V. 2.1.13. Prove thatf(z) = z" (n is a positive integer) satisfies Cauchy-Riemann equations and f'(z) = nz,,-l. 2.2. HARMONIC FUNCTIONS
A real-valued function u of two real variables x, y (resp. of one complex variable z = x+ iy) defined in a domain D is said to be harmonic in D, if it has continuous partial derivatives of second order that satisfy in D Laplace's equation: Llu
=
u.u+u yy
=
O.
Notice that continuity of partial derivatives of second order -implies continuity and U y, as well as continuity of u. Two functions u, v harmonic in a domain D and satisfying Cauchy-Riemann equations in D: UX = V y , Uy = -Vx are called conjugate harmonic functions. Any pair of conjugate harmonic functions u,v determines an analytic function u+iv. fix
2.2.1. Find all the functions harmonic 10 C"", (- 00, 0] which are cOllstant one the rays argz = const. 2.2.2. Find all the functions harmonic in C"", 0 which are constant on the circles ceO; r). 2.2.3. Verify that the functions u = log Izl, v = argz are conjugate harmonic functions in C"", (-00,0] and Logz = log Izl+iArgz, where Argz is the principal value of argument: -7t < Argz < 7t, is analytic in C"",(- 00, 0]. 2.2.4. Verify that ~cosy, ~siny are conjugate harmonic functions in C. Also verify that the analytic function expz
=
eXcosy+iexsiny
flllfills the identity: expLogz III
C "'-, ( 00 , 0] . 2.2.5. Show that
!
Logz
= Z_l
=
Logexpz
=
z
in C"", (00, 0] and
!
expz
=
expz in C.
2.2.6. Suppose w = fez) is analytic in a domain D and feD) II (- 00,0] Show that F(z) CO" loglf(z)l+i Argf(z) is analytic in D. Evaluate P'.
=
0.
22
1. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.2.7. Suppose u is harmonic in a domain D. Verify that! = u,,-iuy is analytic in D. 2.2.8. Suppose V1' V2 are harmonic and conjugate with u in a domain D. Verify that V1-V2 is a constant in D. 2.2.9. Suppose u is harmonic in {z: Izl > O} and homogeneous of degree m, m =/: 0, i.e. for any t > 0, u(tz) = tmu(z). Verify that v = m-1(yux-xUy) is a conjugate harmonic function. 2.2.10. Find a conjugate harmonic function v for u equal: (i) x 2_y2+xy; (ii) x3+6x2y-3xy2_2y3; (iii) X(X 2+y2)-1; (iv) (X 2_y2) X
(X 2+y2)-2. Evaluate in each case a corresponding analytic function u+iv as depending on z = x+iy. X
2.2.11. Find a conjugate harmonic function v for (1 +x 2+y2) X u(x, y) = 1+2(x2_ y 2)+(x2+y2)2 .
Write u+iv as depending on z = x+iy. 2.2.12. Show that a function u harmonic in a domain D has a conjugate harmonic function v in D, iff! = ux-iuy admits a primitive in D. 2.2.13. Find a conjugate harmonic function v for
u(x, y)
=
eX(xcosy-ysinx).
2.2.14. Write Laplace's equation in polar coordinates r, O. Verify that r"cosnO, r"sinnO are harmonic for any positive integer n. 2.2.15. Discuss the existence of nonconstant harmonic functions having the form: (i) lP(xy); (ii) !p(x+f'x 2+J2); (iii) tp(x 2+y), where tp is a suitable, realvalued function of one real variable. Fisd a corresponding conjugate harmonic function in case it does exist. 2.2.16. Given a real-valued function F with continuous partial derivatives of second order in a domain D such that F;+F; > 0 in D. Suppose a < F(x, y) < b for x+iy ED and 1p is a real-valued, continuous function of t E (a, b) such
that (F",,+Fyy)(F;+F;)-l = 1pO F. Then there exists a real-valued function tp defined in (a, b) and ~uch that tp 0 F is harmonic in D. Evaluate tp as depending on 1p. 2.2.17. Find an analogue of Exercise 2.2.16 in case F is given in polar coordinates r, O.
2.3. GEOMETRICAL INTERPRETATION
23
2.2.18. Verify the existence of functions u harmonic in C'\, (- 00, 0] and constant on confocal parabolas with foci at the origin and vertices on (0, +(0). Find all these functions. 2.2.19. Find all the functions I analytic in C '\,0 such that III has a constant value on circles x 2+y2-ax = O. 2.2.20. Find all the functions I analytic in C'\, (-IX), 0] such that argl has a constant value on circles C(O; r). 2.2.21. Verify the existence of functions u(r, 0) harmonic in C '\, ( - 00, 0] having a constant value on arcs of logarithmic spirals r = ke)./J, where r, 0 are polar coordinates, A. is fixed for all the spirals and k is a parameter determining the individual arcs. 2.2.22. Find all the functions regular in C '" 0 whose absolute value is constant = a2 sin20.
on lemniscates r2
2.3. GEOMETRICAL INTERPRETATION OF THE DERIVATIVE
If I is analytic in a domain D, Zo ED and J'(zo) =F 0, then I(zo+h)
=
l(zo)+hJ'(zo)+O(h 2 )
as
h -+ O.
This means that locally I is a similarity transformation composed of a rotation hy the angle argf'(zo) , a homothety with the ratio 1f'(zo)1 followed by a translation (=0)' The angle between any regular arcs intersecting at Zo and the angle between Ihe image arcs are equal, therefore the mapping realized by an analytic functionl withf'(z) =F 0 is said to be conformal. An analytic function realizing a conformal lind homeomorphic mapping of a domain D is said to be univalent in D. 2.3.1. The linear transformation w = (z+ I)/(z-I) carries the boundary of Ihe upper half-disk of K(O; 1) into two rays emanating from the origin (why?). Iii nd the angle (l( between the image ray of (-I, I) and the positive real axis liN well as the local length distortion A. at z = -I. 2.3.2. Given a linear transformation w = (az+b)/(cz+d) with c =F 0, find I he sets of all z for which (i) the length of infinitesimal segments is preserved; (ii) the direction of infinitesimal segments is preserved. 2.3.3. Suppose z = z(t) is a differentiable, complex-valued function of a real \1Il'inhle t E (a, b) such that z'(t) =F 0 and IV =/(z) is a conformal mapping
24
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
defined in a domain D containing all the points z(t), t E Ca, b). Show that if arg[f'(z(t))z'(t)] is constant in (a, b), then the image of the arc z = z(t), a < t< b, is a straight line segment (not necessarily bounded). 2.3.4. Show that for any linear mapping w = (az+b)/(cz+d), ad-be:F 0, c :F 0, some straight line has as its image a parallel straight line. 2.3.5. Find the sets of all z where an infinitesimal segment is expanded, contracted or preserved under the given transformation: (i) w = Z2; (ii) w = z2+2z; (iii) w = Z-l. 2.3.6. Find local magnification and the angle of local rotation at Zo under the mapping w = Z3. 2.3.7: Show that the Jacobian
~~::;]
of the mapping
= -3+4
f= u+iv, where
f is analytic in a domain D, is equal 11'12. Give a geometrical interpretation. 2.3.8. Verify that if f = u+iv is analytic and !'(zo) :F 0, then the lines u = const, v = const, intersect at Zo at the right angle.
2.3.9. Find the lines u = const, v = const for the mappings (i) w = Z2; (ii) w = Logz. 2.3.10. Find the length of the image arc under the univalent mappingf: D - C of the arc given by the equation z = z(t), a:« t b. Also find the area of the image domain of Q, QeD.
:«
2.3.11. Show that under the mapping w = Z2, the image curve of C(1; 1) is the cardioid w(If!) = 2(1 +cos If!) eiq>. Find its length and the area enclosed. 2.3.12. Evaluate the integral ~~(x2+y2)dxdy, where D is a domain situated D
in {z: rez > 0, imz > O} whose boundary consists of the segment [1,2] and three arcs of hyperbolas x 2- y2 = 1, x 2- y2 = 4, xy ~ 1. Hint: Cf. Ex. 2.3.9 (i). 2.3.13. Evaluate the length of the image arc of the segment [0, i] under the mapping w = z(1_Z)-2. 2.4. CONFORMAL MAPPINGS CONNECTED WITH w
= z'-
2.4.1. Evaluate the maxiq1al error in K(i; +0) if the mapping w = Z2 of this disk is replaced by its differential at z = i. 2.4.2. Find the image domain of the square: 0 < x < 1, 0 < y < 1 and the length of the boundary of the image domain under the mapping w Z2. Z -=, x+ iy. ",c
25
1.S. THE MAPPING w = t(z+z-l)
2.4.3. Find the univalent mapping of the domain {z: rez onto K(O; 1) such that Zo = l+i corresponds to the center.
> 0, imz > O}
2.4.4. Find the univalent mapping of K(O; 1) onto the inside of w(O)
=
2(1+cosO)eif1 ,
0::::;; 0::::;; 2".
2.4.5. Find the univalent mapping of the domain situated on the right-hand side branch of the hyperbola x2_y2 = a 2 onto K(O; 1) which carries the focus of the hyperbola into w = 0 and the vertex into w = -1. 2.4.6. Find the univalent conformal mapping of the domain
< +00, 2px < y2}, p > 0 onto the unit disk such that the points z = -t and z = 0 correspond to w = 0 {z
=
x+iy: -00
0, and the rays argz = 1=+" onto the strip {w: limzl < I}. 2.4.8. Show that the mapping z = ay2w(1+w 2)-1/2 carries the unit disk {w: Iwl < I} into the domain bounded by the branches of the hyperbola
x 2 _yZ = a2 • 2.4.9. Map conformally the inside of the right half of the lemniscate < p ::::;; a, onto the unit disk.
11I'2-a 21= p2, 0
2.4.10. Map conformally the inside of lemniscate Iw 2-a 21 = p2, P > a, onlo the unit disk. I,l
2.4.11. Map conformally the strip domain between 4(x+l), y2 = 8(x+2) onto the strip {w: lim wi < I}.
the
patabolas:
2.5. THE MAPPING w = t(Z+Z-l)
2.5.1. Suppose C is an arbitrary circle through -1, 1 and Zl' Z2 do not lie = 1. Show that one of these points lies inside C and an-
tin C and satisfy ZlZ2 t II hl'r one outside C.
2,5.2. Show that the mapping w = t(Z+Z-l) carries both the inside and the oiliside of any circle C through the points z = 1=1 in a 1: 1 manner onto the ~lIlIll' domain in the w-plane. Find the image domain. /lint: Show that (w-l)j(w+ 1) = [(z-l)j(z+ lW. 2.S.3. Show that the image domain of the upper half-plane under the mapping II'
~(Z·I-z-l) is
C,,",-{(-oo, -1] u [1, +oo)}.
26
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.5.4. Show that the image domain of the unit disk under the mapping t(Z+Z-l) is C",[-I, 1].
W=
2.5.5. Suppose C is a circle through z = =f 1 and r is a circle having a common tangent with C at z = 1 and situated in the outside of C. Describe the image under the mapping w = t(Z+Z-l). curve of
r
2.5.6. Find the image curves of: (i) circles C(O; R); (ii) rays argz
=
(1
under the mapping w = t(Z+Z-l).
2.5.7. Map conformally the ellipse {w: Iw-21+lw+21 < l00/7} slit along [-2,2] onto the annulus {z: 1 < Izl < R}. Evaluate R. 2.5.8. Map conformally the outside of the unit disk onto the outside of the ellipse: {w: Iw-cl+lw+cl > 2a} (c 2 = a2 _b 2 , a, b, c > 0). (a
2.5.9. Map conformally the domain C",{K(O; 1) u [-a, -1] u [1, b]} > 1, b > 1) onto the outside of the unit disk.
2.5.10. Map conformally the outside of the unit disk slit along (- 00, -1) onto: (i) w-plane slit along the negative real axis;
(ii) the right half-plane.
2.5.11. Map conformally the domain whose boundary consists of three rays: (-00, -1], [1, +00), [2i, +ioo) and of the upper half of C(O; 1) onto a halfplane. 2.5.12. Map conformally the domain bounded by two confocal ellipses:
{w: onto an annulus {z: Rl
2V5
0, im w < O} is carried 1: 1 onto the complementary domain of the set {W: re W ;;;::, 0, im W ;;;::, O} u [- 2, 0]. 2.5.16. Map conformally the part of the z-plane to the left of the right-hand branch of the hyperbola x 2 - y2 = 1 on a half-plane. Hint: Map the upper half of the given domain by the mapping W = Z2. 2.6. THE EXPONENTIAL FUNCTION AND THE LOGARITHM
2.6.1. Use the identity:
expz
=
e%
=
eXcosy+iexsiny,
z
=
x+iy,
to verify that (i) le%1 = eX; (ii) exp(z+2rci) = expz; (iii) exp(zl+Z2) = (expzl)(expz2)'
"-
2.6.2. Show that for any complex w =f. 0 and any real ct the equation e% has exactly one solution z satisfying ct < imz < ct+2re.
=
w
2.6.3. Find the image domain of the strip -re < imz < re under the mapping II' = e%. Also find the images of segments (xo-rei, xo+rei) and of straight lines y =Yo·
2.6.4. Find the image of the straight line y e% (m =f. 0).
=
mx+n under the mapping
II' .•. '"
2.6.5. Find the image domain of the strip mx-re llIapping w = eZ •
0
[cotzl ~ [I+(sinho)-2]1/2.
2.7.8. Verify the identity sin 2z+cos 2z
=
1 for all complex z.
2.7.9. Verify the identity sinz = sinz and its analogues for cos, tan, cot. 2.7.10. Find all z for which sinz, cosz, tanz are (i) real; (ii) purely imaginary. 2.7.11. Evaluate cos(5 - i), sin(1- 5i) up to 4 decimals. Show Zo
=
fTC+ilog(4+ ]115) then sinzo
=
that
if
4.
2.7.12. Verify for complex Zl' Z2 the addition formulas: COS(Zl+Z2)
=
COSZ1COsz2-sinz1 sinz2'
sin(zl+z2) = sinz1cosz2+Cosz1sinz2'
2.7.13. Write coshz +iv(x, y), x+iy = z.
=
t(ez+e- Z), sinhz
=
t(eZ-e- z)- in the form u(x, y)+
2.7.14. Express [sinhz[2, [coshz[2 as functions of x, y. 2.7.15. Find the relation between corresponding trigonometric and hyperholic functions and give a geometric interpretation. 2.7.16. Verify the identity: 'R) (1 .) 'R) (1+1.) cot (IX+If' + - I cot (IX-If'
=
2 sin2IX+sinh2f3 h2 2R . cos IX-COS f'
2.7.17. Find the image domain of the strip [rezJ < tTC under the mapping II' sinz. Find the image arcs of segments (-tTC+iyo, tTC+iyO) and of straight lincs x = xo and verify the univalence of sine in the strip considered. 2.7.18. Find the image domain of the rectangle: 0
IInder the mapping w
=
0, imz > 0 under
2.8.10. Show that w = arcsinez maps the z-plane slit along the rays y = - 00 < x ~ 0 (k = 0, =f 1, =f2, ... ) onto the upper half-plane.
k7t,
1.9. CONFORMAL MAPPING OF CIRCULAR WEDGES
Elementary functions such as logarithm, exponential function, linear functions, power w = z" = exp(cdogz), carry some families of rays, circular arcs, or straight lines, into a similar family, e.g., Logz carries the family of rays argz = const into the family of parallel straight lines. A suitable superposition of such transformations enables us to map any circular wedge, i.e. a simply connected domain whose boundary consists of two circular arcs (not necessarily different, either arc can be replaced by a straight line segment), onto a disk, or half-plane. In particular the mapping w = z" for Ot: suitably chosen carries an angular sector with vertex at the origin into a half-plane. On the other hand the exponential
1.9. CONFORMAL MAPPING OF CIRCULAR WEDGES
31
function transforms zero angle into a half-plane. It may even happen that a particular circular triangle with two right angles can be mapped onto a circular wedge and hence onto a half-plane. 2.9.1. Map 1: 1 conformally a circular wedge in the z-plane whose sides intersect at a, b and make an angle IX orito a half-plane. Hint: Consider first a linear mapping carrying a, b into 0, 00 respectivel.y. 2.9.2. Map 1: 1 conformally the angle D
{z: IX < argz < {3},
=
0
< {3-1X
1.
a
C(.:a)
3.2.11. Evaluate
~ z2~a2 dz,
a> O.
C(O:2a) (
3.2.12. Using Cauchy's formula for f"(a) evaluate 21 . \ ( zez )3 dz, where C 'Ttl J z-a c is a contour containing a inside. 3.3.13. Evaluate in a similar way the integral
_1_\ ez dz 21ti J z(1-z)3 C in case: (i) C = C(O; t); (ii) C = C(l; t). 3.2.14. Evaluate
\ J C(O:R)
fez) dz in case a, be K(O; R) for (z-a)(z-b)
f analytic
in some domain containing K(O; R). Prove Liouville's theorem: Any function analytic in C and bounded is a constant. 3.2.15. Using Cauchy's formula for the derivative evaluate \
J
dz (z-b) (z-a)m '
lal
< r < Ibl.
C(O;r)
3.2.16. Suppose C is a contour containing 0 inside and leaving z outside and analytic in ~0 and bounded outside C. Prove that
J is
fez) = ~ \ zf(C) dC. 2m ~ (z-C) C
Evaluate the right-hand side term for z situated inside C and 3.2.17. Let Pn be the polynomial j
~ k.
(Z-Zl)(Z-Z2) •••
How many values can take the integral
~ P~~Z) C
with positive orientation omitting all
Zk?
~
O.
(z-zn) with
Zj ~ Zk
for
for C being contours
3.3. ISOLATED SINGULARITIES
41 3.3. ISOLATED SINGULARITIES
If I is analytic in K(a; r)",a, then the point a is called an isolated singularity 01f. If the limit lim/(z) exists, then the singularity is called removable: we can z ...a
extend the domain of I putting I(a)
=
lim I(z) and obtain in this way always z...a
a function analytic in the whole disk K(a; r). If lim/(z)
=
00,
then the point a
z ... a
is said to be a pole. In this case there exists a finite A #0 and a positive integer n such that lim (z-a)n/(z) = A. The number n is called the order 01 the z ...a
pole. Otherwise a is called an essential isolated singularity. If I is analytic in {z: Izl > R}, then 1(1/z) has an isolated singularity at 0 and the character of singularity of/at 00 is defined to be the same as that of/(1fz) at z = O.
z_
3.3.1. Suppose I is analytic in K(a; R)",a and lim (z-a)/(z)
=
O. Prove
that for any closed, regular curve in K(a; R)""a we have ~/(z)dz
=
O.
,.
3.3.2. Show that under the assumptions of Excercise 3.3.1 we have
1m = _1_. 2m
r
I(z) dz,
where
J z-C
O I, show that the equation z+e-~ positive real part. 3.9.14. Prove that the polynomial I+z+az", n at least one root in R(O; 2).
=
z+2 are contained
= A has one solution with
~
2, has for any complex a
3.9.15. Suppose 0 < lal < 1 and p is a positive integer. Show that the equation (;=-I)P = ae- Z has exactly p simple roots with positive real part and all of these arc located inside K(I; I). 3.9.16. Prove that all roots of the equation tanz-z = 0 are real and each ill tcrval «n - -}) 'It, (n +~-) 'It), n =1= 0, contains exactly one root All'
3. COMPLEX INTEGRATION
S6
3.9.17. If
laml < 1
(m = 1,2, ... ,
n), Ibl < 1 and n
F(z)
=
IT l-a z-a z
m ,
in=1
show that the equation F(z)
=
m
b has exactly n roots in the unit disk.
3.9.18. If F is analytic in K(O; a) and continuous in .K(O; a), IF(z) I > m on C(O; a) and IF(O) I < m, show that F has at least one zero in K(O; a). 3.9.19. If I is analytic in the annulus A z E A and Yt =
=
{z: r
I( C(O; t)), show that for any t E
< Izi < R}, I(z)
=1=
a for all
(r, R) the index n(y" a) has
the same value. 3.9.20. If I is analytic in a domain D except for one simple pole
tinuous in l5"",zo, I/(z) I = every value a with lal > 1.
and con1 on D",D, show that I takes in D exactly once Zo
CHAPTER 4
Sequences and Series of .Analytic Functions 4.1. ALMOST UNIFORM CONVERGENCE
Following Saks and Zygmund [10] we shall call a sequence of functions {In} ddi'iied in an open set G almost uniformly (a.u.) convergent on G to alunction!, if {In} tends to I uniformly on each compact subset of G. In what follows w~ use the notation: In =l GI. If all functions!.. are analytic in a domain D and!.. =l'DI, then also lis analytic in D. Moreover, for any fixed, positive integer k, JC:) =l DJC k). We can also consider a.u. convergent series of functions. The so-called M-test of Weierstrass yields a quite convenient, sufficient condition for a.u. convergence of functional series. Let L Un be a series of functions defined on an open set G. If for each compact subset F of G there exists a convergent series L Mn with positive terms such that Iun(z) I ~ Mn for all z E F, then L Un is a.u. convergent on G. 4.1.1. Show that the sequence {zexp (_-}n 2 z 2 )} is uniformly convergent on the real axis and at the same time it is not a.u. convergent in any disk K(O; r). 4.1.2. Prove that the series co
00
n=1
n=1
l>n(Z) = I: zII[(1-zII)(1- zn+1)]-1 is a.u. convergent in K(O; 1) to z(1-Z)-2 and also a.u. convergent in C,,-K(O; 1) to (1-z)-2. co
4.1.3. Prove that the ~eries
l: 3-nsinnz
is a.u. convergent in the strip
n=O
lil11zl < log3, its sum I being analytic in this domain. Evaluate /,(0). co
4.1.4. If I is analytic in K(O; 1) and 1(0) = 0, show that the series l:/(z") is !l.U. convergent in K(O; 1) and its sum is analytic. n=1 57
4. SEQUENCES AND SERIES
58
4.1.5. Prove that in any closed disk K(zo; r) leaving outside all negative in00
tegers the series
L
•
(_1)n+1(z+n)-1 is uniformly convergent and its sum is ana-
n=1
lytic in C""{-l; -2; -3; ... }. 4.1.6. Suppose y is a closed, regular curve not meeting any negative integer and I is the analytic function of Exercise 4.1.5. Evaluate ~ I(z) dz. l'
4.1.7. If Izl
4.3.12. Verify that
q;(z)
= -. /
V
z Arctan'" / z l-z l-z
is analytic in some neighborhood of z of its Taylor series at the origin.
V
=
O. Evaluate the radius of convergence
4. SEQUENCES AND SERIES
64
4.3.13. (cont.) Verify that rp satisfies the differential equation
2z(l-z)rp'(z)
with the initial condition rp(O)
=
o.
=
rp(z)+z
Prove the identity:
_ 2 3 2·4 5 2·4·6 7 rp(z) - z+3 z +TI z +T5.'7z +
... ,
Izl < 1.
4.3.14. Prove that . 2 _ 2 1 2 4 1 2.4 6 (Arcsmz) - z +"2·3 z +3·TIz +
... ,
Izl
< 1.
4.3.15. If f is analytic in K(O; R) and Rn is the remainder of Taylor series center at the origin, i.e.
Rn(z)
=
z"
f(z)- f(O)-z!, (0)- ... - - , j b.
< Ibl,
4.5.6. Find the Laurent development of (1 + Z2)-1 (2+ Z2)-1 for (i) 1 < Izl
< V2; (ii) Izl > V2.
4.5.7. Find the Laurent expansion of Log [Z2 j(z2-l)] for Izl
>
1.
for
4. SEQUENCES AND SERIES
68
4.5.8. Express the Laurent coefficients of exp(z+z-l) at the origin in terms of (i) integrals involving trigonometric functions; (ii) sums of infinite series by using the identity exp(z+z-l)
=
e Z e l/z •
4.5.9. The Bessel function In(z) is defined as the nth coefficient of Laurent expansion at the origin (n ~ 0): co [a,b]
a
+co
4.7.1. The integral ~ W(z, t)dt is said to be almost uniformly (a.u.) cona
vergent in a domain G, if for any compact subset F c G~nd any exists A such that for any b, B with A < b < B we have
B
> 0 there
73
4.7. THE GAMMA FUNCTION B
I~ W(Z, t)dt! < e
for all z in F.
b
Ifboth W(z, t), W;(z, t) are continuous on Gx [a, +(0) and H(z) =
+00
ia W(z, t)dt
is a.u. convergent in G, show that H is analytic in G. Hint: Use Weierstrass theorem on a.u. convergent series of analytic functions. +00
4.7.2. Prove that ~ e-Iltdt is a.u. convergent
m
the
right
half-plane
o
1
+ 00
{z:rez>O}and ~ e-ztdt--=Ofor all z with rez>O. By separating o z
real and imaginary parts in both terms find the values of two real integrals. +00
4.7.3. Prove that H(z) = ~ e-'t,,-ldt is analytic in the open plane and 1
+00
H'(O)
=
~ t- 1 e-:,}dt.
1
4.7.4. If +00
and
l
~
is a complex-valued function of a real variable x
1~(x)ldx
O.
a
4.8. NORMAL FAMILIES
A family ff of functions I analytic in a domain D is said to be normal, if every sequence {J,,} of functions In E ff contains a subsequence {Ink} which either converges a.u. in D, or diverges to 00 a.u. in D. The limit function I = lim Ink ,
k~oo
is analytic, unless it reduces to the con~tant 00. If any sequence {[,.} of functions fn E ff contains an a.u. convergent subsequence, then ff is said to be compact. A necessary and sufficient condition for compactness of the family ff is the existence of a common, finite upper bound of the absolute values of all IE ff on each compact subset of D (Stieltjes-Osgood theorem, also called Montel's compactness condition). Clearly each compact family is normal. The real-valued !'unction p(z,f) = 2 If'(z) 1(1+I/(zW)-l is called the spherical derivative of I and has an obvious geometrical meaning (cf. Ex. 4.8.1). Now, a family ff of functions I analytic in a domain D is normal, if there exists in every compact ~ubset of D a common finite upper bound for the spherical rlerivative of all !'unctions of the family. This criterion is due to F. Marty. Another sufficient condition for normality is due to Monte! and its proof is based on the properties of the modular function: if all functions I of a family :iF are analytic in a domain D and every IE ff does not take in D two fixed, lillite values ct, {3 (ct =f. {3), then I is normal. The concept of normal family is very important in the existence questions for solutions of extremal problems. 4.8.1. Explain the geometrical meaning of the spherical derivative. 4.8.2. Verify that the family of all similarity transformations az+b is not II
normal family in the finite plane.
4.8.3. If ff is normal in a domain D and there exists a point Zo E D and a real I"llllstant Mo < + 00 such that I/(zo)1 ~ Mo for all I E ff, show that ff is a complld family. ..
4. SEQUENCES AND SERIES
76
4.8.4. If ff is a compact family of functions analytic in a domain D, show
that also ffl = {g: g = 1', IE ff} is compact. By considering the sequence {n(z2 _n 2)} verify that the derivatives of functions of a normal family not necessarily form a normal family. 4.8.5. Suppose F(w) is an entire function (i.e. a function analytic in the finite plane C) and ff is a compact family of functions analytic in a domain D. Verify that the functions F 0 I, IE ff also form a compact family in D. 4.8.6. Suppose ff is the family of all functions I(z) = az, where a is a complex constant and F(w) = eWsinw. Verify that ff is normal in C""K(O; 1), whereas the functions Fol do not form a normal family. 4.8.7. Prove Hurwitz's theorem: If {In} is an a.u. convergent sequence in a domain D and/n(z) 1= 0 for all ZED and all In, then the limiting function I is either identically 0 in D, or does not vanish in D. Hint: Verify that p(z,J) == p(z, Iff). Also prove that the terms of an a. u. sequence form a normal family. 4.8.8. Suppose {In} is an a.u. convergent sequence of univalent functions in a domain D and g = limfn. Show that either g is univalent in D, or is a constant. Hint: Consider I(z)-/(a) in D~a. 4.8.9. Show that the family of all functions I analytic in the unit disk and such that 1(0) = 0,1'(0) = 1, is not a normal family. 4.8.10. Let To be the family of all functions I analytic in the unit disk and such that 1(0) = 0,1'(0) = 1, I(z) 1= 0 for z 1= o. Show that To is compact. Hint: Consider log(f(z)/z). 4.8.11. (cont.) Show that there exists a constant ct > 0 such that any I E To takes in the unit disk any value Wo E K(O; ct). Hint: If In does not take the value ctn and ctn ~ 0, consider the sequence gn(z) = log (1-/n(z)/ct n). 4.8.12. Show that the family So of ail functions analytic and univalent in the domain D omitting one fixed value ct is normal. 4.8.13. Suppose G(M) is the family of functions analytic in a domain D and
such that ~ ~ If(z)1 2dxdy ~ M. Show that G(M) is compact in D. D
Hint: If K(zo; R) cD, verify that 2",
If(zoW
~
217':
~
o and deduce that 7':R21/(zo)12 ~ M.
If(zo+re i6WdO,
0< r
~ R,
CHAPTER 5
Meromorphic and Entire Functions 5.1. MITTAG-LEFFLER'S THEOREM
Let {a"} be an arbitrary sequence of complex numbers such that ao = 0 = 00 and let {G"(w)} be an arbitrary sequence of polynomials with vanishing constant terms. There exists a meromorphic function F which is analytic in the finite plane except for the poles ao, a1' a 2, ... and has G"(I/(z-a")) as singular parts at a".
< lall < la 2 1< ... and lima"
00
Let
L Un be an arbitrary, convergent series with positive terms and let {H"(z)}
"=1
be a sequence of polynomials such that
IG" (z 1aJ -
H"(z)
I~
U"
(n
=
1,2, ... )
for z eK(O; r"), where rn < la"1 and limr" = +00. In particular H" can be a suitable partial sum of Taylor's expansion of the singular part which is analytic In the open disk K(O; la"D. Then the function
where H is analytic in the finite plane (i.e. H is an entire function), has all the required properties. This representation due to Mittag-Leffler enables us to lind a meromorphic function with given .poles and given singular parts at these poles up to an entire function. 5.1.1. Find the most general meromorphic function F whose only singularilics are: (i) poles 1, 2, 3, ... of first order with res(n; F)
= n;
a" of first order (lal > 1, n = 1, 2, ... ) with res (a" ; F) = a"; (iii) poles l/n of first order (n = 0, ],2, ... ) with residues 1; ' p the series 2..: lanl-
IX
n=l
is convergent. 5.7.9. Show that an entire function J of finite order p has a representation of the form J(z)
= t'exP(g(z))"U
E( :n ,m)
where g is an entire function and m :::; p does not depend on n. 5.7.10. Determine the order of the following entire functions:
(ii)
f eo;n r a
zn
n=l
2: e-" zn; 00
(iii)
2
n=O
(iv)
cosh lin n , z. n.
(a>
0);
CHAPTER 6
The Maximum Principle 6.1. THE MAXIMUM PRINCIPLE FOR ANALYTIC FUNCTIONS
6.1.1. If I is analytic and not a constant in a domain D, show that III cannot have in D a local maximum. Hint: Cf. Exercise 4.2.5. 6.1.2. If I is analytic and not a constant in a domain D, show that III cannot have a local minimum at Zo ED, unless I(zo) = O. 6.1.3. Suppose I is analytic and not a constant in a domain D and III has a con-
15. If I(z) i= 0 in D and m = inflf(z),l, M = sup If(z) I, < I/(z) I < M in D. 6.1.4. Suppose I is analytic in a domain D and If I has a continuous extension on J5 without being a constant. If III has a constant value on the boundary of D, show that I(zo) = 0 at
tinuous extension on z E frD, show that m
some point Zo ED. 6.1.5. If P is a polynomial of degree n, show that {z: IP(z)1 = A} cannot have more than n components for any fixed A. 6.1.6. If I is a nonconstant, analytic function in K(O; R), show that M(r,!) sup I/(z) I is a strictly increasing function of r E (0, R). \zj=r
6.1.7. Prove Hadamard's three circles theorem: .
Suppose I is analytic in B = {z: r1 < Izi < r2}, continuous in Band Mk = sup I/(z)l, k = 1,2. Izl=rk
If M(r)
=
sup I/(z)l, then 10gM(r) is a convex function of logr, i.e. Izl=r
Iog M() r
~
logr2-logr I M logr-Iogr 1 I M og 1 + - - - - - og 2' logr 2-logr1 logr 2-logr1
Hint: Consider [/(z)]pz-q, where p, q are suitably chosen integers. 90
6.2. SCHWARZ'S LEMMA
91
6.1.8. Iffis analytic in K(oo; 1), continuous in K(oo; 1) and has a finite limit at 00, show that If I attains a maximum at a point of C(O; 1). Also prove that M(r,!) = sup If(z) I strictly decreases, unless f is a constant. 6.1.9. If P is a polynomial of degree nand IP(z) I :0:;; M in K(O; 1), show that IP(z)1 :O:;;Mlzl n in K(oo; 1). 6.1.10. If P is a polynomial of degree nand IP(z) I :0:;; M for z E [-1, 1], show that for all z situated inside an ellipse with semiaxes a, b and foci -1, 1 we have: IP(z)1 :0:;; M(a+b)". Hint: Consider the mapping z = t(w+w- l ). 6.1.11. Suppose f is analytic and bounded by 1 in absolute value in K(O; 1) and tends uniformly to 0 in the angle lI. :0:;; argz :0:;; f3 as Izl -+ 1. Show thatf = O. Hint: Consider the product rp(z) =f(z)f(wz)f(w 2z) .. .f(wn-lz), where (II = exp (27ti/n) and n is a suitably chosen integer. 6.1.12. Suppose f is analytic in S = {z: Irezl < a} and there exist two real constants C, A such that If(z)1 :0:;; expAIYI for z = x+iy, Ixl < a, and lim If(z) I :0:;; C for z -+ a+iyo, resp. z -+ -a+iyo with arbitrary Yo. Show that for all z E S we have: If(z) I :0:;; C. Hint: Consider rp(z) = f(z)exp(ez2) , e > O. 6.2. SCHWARZ'S LEMMA
6.2.1. Iffis analytic and iess than 1 in absolute value in K(O; 1) and if/CO) = 0, show that either If(z) I < Izl in K(O; 1)",0, or elsef(z) = ei<x z with some real lI.. Hint: Consider z-lf(z). 6.2.2. If f is analytic in K(O; 1) and If(z) I < 1 for all z
E
K(O; 1), show that
If(Z)~f(O) I:0:;; 11-f(0)f(z)l. 6.2.3. Under the assumptions of Exercise 6.2.2 show that either 11'(0)1 = e i l7.z (0( is real).
< 1,
or fez)
6.2.4. If f maps K(O; 1) 1: 1 conformally onto a domain D containing the IInit disk and f(O) = 0, show that 11'(0)1 ~ 1 with the sign of equality for fez) l,la z only. co
6.2.5. If fez) =
2: anz"
is analytic in K(O; 1) and If(z) I :0:;; M in K(O; 1),
n=O
',how that Mlall
0, show that the set of all values taken by £0 in R(O; r) is contained in K(zo; p), where
and U·
flint:
Consl·der :'4 rI() z
=
1£O(z)-oc ( ). -oc£O Z
6.2.8. (cont.) Show that any point of K(zo; p) is a value of a certain function £0 taken at a point of K(O; r). 6.2.9. (cont.) Find the set D, of all possible values taken in K(O; r) by functions £0 analytic in K(O; 1) and such that £0(0) ;;;: 0, 1£O(z) I ~ 1. 6.2.10. If Q is a polynomial of degree n with complex coefficients and IxQ(x)1 ~ M for x E [-1, 1], show that
IQ(z)1 ~ M(I+v'2)n+\
z EK(O; 1).
Hint: Cf. Exercise 6.1.10. 6.2.11. If Q is a polynomial of degree nand
1(z-'I]) Q(z) I ~ M,
z
E
('I] is a constant, 1'1]1
ceO; 1)
=
1),
show that
IQ(z)1 ~ t(n+2)2 M,
z E K(O; 1).
6.3. SUBORDINATION
Suppose J, F are analytic in K(O; R). The function I is said to be subordinate to Fin K(O; r), r ~ R, if there exists a function £0 analytic in K(O; r) such that £0(0) = 0, 1£O(z) 1 < r in K(O; r) and 1= Fo£O in K(O; r). If I is subordinate to F in K(O; r), we write 1- 0,
q
> 0)
function.
7.4.3. Map the upper half-plane onto a rhombus with angles 7tIX, 7t(I-IX) so that its vertices correspond'to z = 0, =FI, 00. Find the side-length. 7.4.4. (cont.) Verify that the upper semicircle and the positive imaginary axis correspond to the diagonals of the rhombus. Also find the preimage of its center. 7.4.5. Map 1: 1 conformally the upper half-plane {z: imz > O} onto the part of the upper half-plane lying outside the rectangle: - k re w 0, o ~ im w « h, so that the points =Fa, =F~ correspond to the vertices. Discuss the limiting case: ~ -+ O.
«
«
7.4.6. Map I: 1 conformally the upper half-plane imz > 0 onto the domain containing the first quadrant whose boundary consists of two rays: {w: rew ~ 0, imlv = I}, {w: rew ~ 0, imw = -I} and a segment [-i, i]. The points : ·co -1, 1 should correspond to w = i, - i. 7.4.7. Map 1: 1 conformally the upper half-plane onto a part of it with the boundary consisting of two rays: {w: rew « 0, imw = 7t}, [-7tcot~, +00) II nd a segment [-7t cot ~, 7ti], where 0 < ~ < 1-7t. The points z = -1, 0 should wrrespond to w = 7ti, -7tcot~. Discuss the limiting case ~ -+ O. 7.4.8. (cont.) Find the mapping of the right half-plane reC > 0 onto the II'-plane slit along the rays: im w = =F7t, re w 0, which carries the points , '=Fi into w = =F7ti.
«
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
102
7.4.9. Find the mapping of the upper half-plane imz > 0 onto the strip domain 0< imw < 7t slit along the ray imw = Vi' rew ~ 0 (0 < Vi < 7t). 7.4.10. Find the mapping of the upper half-plane imz > 0 onto the first quadrant slit along the ray: im w = 7t, re w ;;:: h, which carries the points z = 0, a 2 into w=O, h+7ti (a>l,h>O). 7.4.11. Suppose F is a function analytic and non-vanishing in K(O; r5) which has a constant argument on (- r5, r5). Prove that the function Z
W(z) = ~ C- 1F(t;)dt; Zo
where zoEK(O; r5), imzo >0, is analytic in K(O; r5)",,-(-ir5,O] and maps the radii (- r5, 0), (0, r5) onto two paralle} rays the distance between them being equal to 7tIF(O)I. 7.4.12. Find the image domain of the upper half-plane under the mapping
-
z
~ ()-
w-Hz-
dt; t;(t;-b)}f·t;-a
(0
< a < b, imzo >0).
Zo
Find the distance between two parallel rays in each pair of boundary rays and evaluate b so as to make both distances equal. 7.4.13. Show that the mapping Z
W
= ~ t;-1(t; -a)- lj3 dt; o
carries the upper half-plane imz > 0 into a domain bounded by a polygonal line consisting of two parallel rays and a segment such that interior angles are equal -t7t, T7t. Evaluate the distance between the rays. D
7.4.14. Find the mapping of the upp~r half-plane imz = C""-(K(i; 1) u Ql) where Ql is the first quadrant. Hint: Map first D under W = w- 1 •
>
7.4.15. Find the mapping of the upper half-plane imz Dil
=
{w: 0 < imw < I} u {w: 0 < arg(w-i) < fm}
Assume that z = -1, 0 correspond to w = i,
00
0 onto the domain
> 0 onto where
0
0 onto Dil U (-00, +(0) U D!, where D: is a reflection of Dil w.r.t. the rea] axis. 7.4.17. Show that the function mapping the inside of the unit disk onto the
7.4. THE SCHWARZ-CHRISTOFFEL FORMULAE
103
inside of a simple, closed polygonal line has the same form as in (7.4A), replaced by the points on the unit circle.
Xk
being
7.4.18. Show that the mapping
carries the unit disk into the inside of a regular n-angle. Evaluate its perimeter. 7.4.19. Find the mapping of the unit disk It I < 1 onto the n-pointed star (i.e. 2n-angle whose all sides and alternate angles are equal). 7.4.20. Map 1: 1 conformally the unit disk It I < 1 (i) onto a pentagram (a five-pointed star obtained by extending the sides of a regular pentagon); (ii) onto a Solomon's seal (a six-pointed star formed of two congruent equilateral triangles placed one upon another). 7.4.21. Show that the mapping
A
W =
where Zk
=
expicpk,
Ck
[ l-zkC Jrrrn k=I (l-CkC)1+ k P
]
dC+B
expiVJk> 0:;:;;; CPI < VJI < CP2 < VJ2 < ...
0,
L (h =
2, carries the unit disk Izi < 1 into the w-plane slit along
k=l rays 11 , 12 , ... , In.
I he are angles between
Verify that Zk correspond to the end-points of Ik and and Ik (/n+l = 'i)·
fh
Ik+l
7.4.22. Show that the mapping z
W
r~_~_d_t===--
=
~ (1-t 4 )Jh+t 4
l"nrries the unit disk Izi < 1 into the domain
(w: Irewl < ta} u {w: limwl < ta}. I'valuate a. 7.4.23. Prove that the function J mapping 1: 1 conformally the unit disk 1.:1 ..:: 1 onto a convex domain bounded by a polygonal line and such thatJ(O) = 0 hilS the form z
J(z)
=
n
A ~rr (l-texpiOk)-a.kdt+B o k= 1
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
104
n
where 0 ~ 01 < O2 < ... < On < 21t,
OCk>O,
L
OCk
~ 2.
k=1
Verify that If'(z) I ~ 1f'(0)1(1-lzl)-2,
If(z) I ~ 1f'(O)llzl(1-lzl)-l.
7.4.24. A function f analytic in a disk K is said to be close-to-convex, if there exists in K a univalent, analytic function (/) mapping K onto a convex domain and such that re!'(z)/(/)'(z) > 0 in K. (i) Show that each close-to-convex function is univalent in K. Hint: Consider the mapping fo (/)-1 and cf. Exercise 3.1.14. n
(ii) If
L
(1Jlk-rpk)
=
1t in Exercise 7.4.21, show that the mapping function
k=1
is close-to-convex. [Remark: Exercise 7.4.24 (i), as well as its counterpart for a half-plane is a convenient tool in proving the univalence of mappings obtained by a formal application of the Schwarz-Christoffel formulas.] 7.4.25. If IC1 1= IC21 = ... = ICnl = 1 n
n
k=l
k=l
L OCk = 2, L OCkCk = 0,
and
OC k
> -1
(k = 1,2, ... , n),
show that "
IT (1-Cfk)"' dC
1
k=1
W(z) = ~ C- 2
n
1c
is continuous on C(O; 1) and maps it onto a closed polygonal line L with interior angles (1- OCk) 1t. If L has no self-intersections, show that W is meromorphic and univalent in K(O; 1) and maps it onto the outside of L. 7.4.26. Find the 1: 1 conformal mapping of the outside of the unit disk onto the outside of a triangle with exterior angles OCk 1t (k = 1, 2, 3) such that the points 'fJk on C(O; 1) correspond to the 'vertices and the points at infinity to each other. Find the relation between OCk and '17k. 7.4.27. Iff maps 1: 1 conformally the outside of the unit disk onto the outside of an equilateral triangle T so that f( (0) = 00, show that the preimages of vertices of T also form an equilateral triangle. ,
7.4.28. Find the 1: 1 conformal mapping f of the outside of the unit disk onto the outside of a regular n-angle such that f(oo) = 00 and f(r/) = 'YJk ('YJ = exp(21ti/n), k = 0, 1, ... , n-l).
7.5. FUNCTIONS sn, cn, an
105
7.5. JACOBIAN ELLIPTIC FUNCTIONS sn, en, dn
If 0 < k < 1, then the Schwarz-Christoffel integral z
U = u(z)
~ [(1-t 2)(1-k 2t 2)tl/2dt
=
o
maps 1: 1 conformally the upper half-plane imz > 0 onto the inside of ~he rectangle with vertices =fK, =fK+iK'. We have: u(O) = 0, u(=f1) = =fK, u(=fk- 1) = =fK+iK', u(oo) = iK'. The inverse function z = sn(u, k) can be continued analytically by reflections all over the open plane and becomes a meromorphic function with two periods 4K,2iK'. We have: ~2
1
K
=
K(k)
~ [(1-t 2) (1-k 2t 2)tl/2dt
=
=
o
~ [1-k 2sin2q?rl/2dq? 0
F being the complete Legendre elliptic integral. The functions en, dn are defined by the equations: sn 2u+cn 2u
=
1,
k 2sn 2u+dn 2u
=
=
.
F(k, 7t/2)
1.
All the roots of the equations sn 2u = 1, sn 2u = k- 2 are double, hence en, dn have no branch points and are merom orphic in the finite plane. z
7.5.1. If u
=
(6- 5t 2+t4)-1/2dt, show that
~
o
z z
7.5.2. If u
=
~
V 2 sn(uV3,
=
vt).
(1-t 4)-1/2dt, show that
o
z
=
cn[}i2 (u-K), 2- 1 / 2 ].
7.5.3. Let k' be the complementary modulus w.r.t. k, i.e. k' that K(k') = K'(k) and K(k) = K'(k').
=
y1-k2. Show
7.5.4. Derive. the expansions: (i) sn(u, k) = u-(1+e)u 3 /3!+(1+14k 2+e)u 5 /5!- ... ;
(ii) cn(u, k) (iii) dn(u, k)
=
=
1-u2/2!+(1+4k 2)u4/4!+ ... ; 1-k2U2/2!+k2(4+k2)U4/4!+ ... ;
lind find the radius of convergence in each case. IL
7.5.5. Express K(k) as the sum of a power series. Verify that K'(k)/K(k) is strictly decreasing function of k E (0, 1) which decreases from +00 to O.
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
106
7.5.6. Show that sn(u+iK') = (ksnu)-l. Find the initial terms of the Laurent expansion of snu in the neighborhood of u = iK'. 7.5.7. Determine the periods, zeros and poles of sn(iv, k') as depending on k. 7.5.8. Verify the identity sn- 2 (u, k)+sn- 2(iu, k') == I. 7.5.9. Find an elliptic function f(u) with periods 2, 2i which has simple zeros at points whose coordinates are either both odd, or both even and has simple poles at points whose one coordinate is even and the other one odd. Discuss the uniqueness. 7.5.10. If z moves round the rectangle with vertices iK', 0, K, K+iK', show that w = cnzj(1 +snz) moves round a quadrant of the unit disk.
•
7.6. THE FUNCTIONS a, C, f.J OF WEIERSTRASS
The function O'\:z) = a(z; £01' as the infinite ptoduct
(0
2) of Weierstrass is an entire function defined " .
a(z)=a(Z;W1,W2)=ZI)'(I-
~k)exp[~k + ~ (~kr],
where Dk = mkwl +nkw2, (mk; nk) is the sequence of all pairs of integers and the prime after the product sign indicates,that the pair (0; 0) should be omitted; £01' £02 are two complex numbers with im(wtfw2) ¥= o. The logarithmic derivative a'(z)ja(z) is the merom orphic function C(z) of Weierstrass. The function f.J(z) = -C'{z) is a meromorphic, doubly-periodic (or elliptic) function with periods £01' £02 whose only singularities are double poles mWl +nw2. We have: f.J(z) = f.J(z; £01'
(0
2) = z-~+
2:' [(Z- Dk)-2_ Dk"2]. k
7.6.1. Show that a(z)
=
z+CSZS+C7Z7+ ...
7.6.2. If £01 is real and £0 2 purely imaginary, show that a is real on the real axis and purely imaginary on the imaginary axis. 7.6.3. If imwl
~
rew 2 = 0, show that a(z) = a(z).
7.6.4. Verify that a(u+wk)ja(u)
=
-exp[(2u+Wk)C(twk)]
(k
=
),2).
go
7.6. FUNCTIONS a, "
107
7.6.5. Show that (.) cr(2u) _ _ 1
0'4(U) -
,(). fIJ u,
..
fIJ"(u) fIJ'(u)'
(n) 2C(2u)-4C(u) =
7.6.6. If C(u) has poles at 2mw+2nw' and 7:=£0'/£0, h=eiT.., 'f}=C(w), im 7: > 0, show that 2'f}w =
~~(1-24
I
k=l
7.6.7. If 1 < Izl < q, show that fIJ(logz; 2logq, 27ti)
2:
1 (q-znz+qznz-1_2)-1-12-2
0 0 ,
=
n=-oo
L: 00
(qZn+q-Zn_2)-1.
n=l
7.6.S. Suppose fIJ has periods 2a, 2bi, where a, b are real and positive. Show that (i) fIJ is real on both axes; (ii) fIJ is also real on straight lines rez = an, imz = bn (n = 0, =fl, =f2, ... ); (iii) fIJ maps 1:1 conformally any rectangle R: na O}"", K(O; 1) such that the points z = 0, ib, ib-a, -a and w = 00, -h, -1, 1 (h > 1 depends on a, b) correspond to each other. Use the reflection principle to show thatfhas a double pole at the origin and is doubly-periodic with periods 4a, 2ib. If ~(z) = ~(z; 4a, 2ib) and e1 = ~(2a), e2 = ~(ib),e3 = ~(2a+ib), show that
7.7.2. Find the function mapping 1: 1 conformally the annulus 1 < lei < Q onto K(oo; 1)"",(-00, -h] where h> 1. Also find the relation between Q and h. 7.7.3. Show that the mapping w = yksn(u, k) carries the open segment (-K++iK', K++iK') into the upper semicircle of C(O; 1). 7.7.4. (cont.) Show that the image domain of the rectangle with corners =fK=f~ iK' is the unit disk Iwl < 1 slit" along the radial segments (-1, -Vf], [Vk, I). . K'(k) 1 a+b 7.7.5. If 0 < b < a and k IS such that 2K(k) = --;-log a-b show that the function
w = w(z)
=
V k sn [2K(k) 7t arcsin 1-
Va z_b 2
2
2
is analytic in H, where H is the ellipse with boundarY;i-
,
k] 2
+ -t-2-
=
1.
Also prove that w = w(z) maps HI: 1 conformally onto the unit disk.
7.7. CONFORMAL MAPPINGS
7.7.6. Given k, 0
109
< k < I, show that the function w = w(z) =
yIC sn ( 2:i
is analytic in the annulus A = {z: 1 < Izl 1:1 onto K(O; I) jlkj.
logz,
< R}, R
k)
= exp(7tK'/4K), and maps it
"'[-Yk,
7.7.7. If A = {z: 1 < Izl
b > 0) such that u(z) -+ A as z approaches the ellipse and u(z)-loglzl = 0(1) as Z -+ 00. Discuss the uniqueness. Find the value A for which u(z)-loglzl = 0(1) as z -+ 00. Hint: Consider first an analogous problem for the unit disk. 8.3.12. Suppose u is harmonic and bounded in the upper half-plane slit along [i, + ioo) and has the boundary values 0 on the real axis and boundary values 1 on the ray [i,
+ioo).
Find the derivative
~:
on the real axis.
8.4. HARMONIC MEASURE
r
Let G be a domain such that the spherical image of = fr G is a finite systcm of simple, closed curves. If y is a finite system of open arcs on r, then the function w(z; y, G) harmonic and bounded in G which tends to 1 as z -+ CE Y and lcnds to 0 as z -+ CE r~y is said to be harmonic measure of y at z.
8.4. HARMONIC MEASURE,
115
8.4.1. Find a function harmonic in K(O; 1) and tending to 0 as z -+ !;, 1)",1 which is not identically O. Verify that the boundedness condition in the definition of harmonic measure is essential for its uniqueness. ~ E C(O;
8.4.2. Determine the lines of constant harmonic measure w(z; y, G), if: (i) Y = (a, b), G = {z: imz > O}; (ii) Y = {z: Izl = 1, fJ 1 < argz < fJ 2 }, G = K(O; 1). 8.4.3. Find the harmonic measure of each of the boundary rays 11 , 12 of the angular domain G= {z: O<argz(z) = [1-f(zo)f(z)]f[f(z)-f(zo)], show that the interior normal derivative of g(z, Zo; G) is equal to Ic!>' (C) I, C E frG. 8.5.5. Determine g(z, Zo; K(O; 1» and derive Poisson's formula from (8.5A). 8.5.6. Determine Green's function for the upper half-plane and derive the formula of Exercise 8.3.8 from (8.5A).
B.S. GREEN'S FUNCTION
117
8.5.7. Find the Green's function g(z, CXl; G) for (i) G
=
K( CXl; 1);
(ii) G
=
C", [-1, 1];
(iii) G =
{z = x+iy:
::
+ ~:
> I}.
8.5.8. Suppose 0 < h < c < 1 and take the constant T determining the function {} of Exercise 7.6.13 equal to 10gh/7ti. Verify that the function F(w) = {}
(2~i log ;
) : {}
(2~i 10gCW)
has the following properties: (i) F is single-valued and analytic in A = {w: h (ii) F has exactly one simple zero in A; (iii)
IFI
< Iwl < I};
has constant values on each of two components of fr A .
8.5.9. (cont.) Find the Green's function g(w, c; A). 8.5.10. (cont.) Show that the interior normal derivative
J3g an Express
~~ in
= \'
([J'(w) \ ([J(w) ,
loge
where
([J(w) = w logh /F(w).
terms of Weierstrass C-function.
8 5.11. Suppose G is a simply connec,ted domain in the finite plane whose complement contains at least 2 different points and p( w, Wo; G) i~ the hyperbolic distance of two points w, Wo E G (which is defined as hyperbdlic distance of their image points after a univalent mapping onto the unit disk). Find the relation between pew, Wo; G) and the Green's function of G. 8.5.12. Let G be a domain whose boundary r is a finite system of analytic Jordan curves. Show that g(z, Zl; G) is a harmonic function of Zl in G"'z. Also verify the symmetry of the Green's function: g(z, Zl; G) == g(Zl' z; G). Hint: Cf. formula (8.5A) and Exercise 8.5.1.
8.5.13. If Go c G, show that g(z, Zl; Go) :S;; g(z, Zl; G) for any z, Zl
E
Go.
n
8.5.14. Let G be a domain whose boundary G =
U rb where r k are
analytic
k= 1
Jordan curves with positive orientation w.r.t. G. If h(z, zo) is the (multi-valued) conjugate of g(z, zo; G), show that the increment i1h (C, zo) as C describes r k
118
B. THE DIRICHLET PROBLEM
is equal to - 27tWk(ZO) , where Wk(ZO) = w(zo; Fk> G) is the harmonic measure of Fk w.r.t. G. Hint: Express Wk by means of (8.5A). 8.5.15. Find the increment of 10gF(w), where F is the function of Exercise 8.5.8 and w moves over vA. 8.5.16. Suppose a, b, c are complex numbers different from each other and such that c ¢:. [a, b]. Determine the domain yielding the maximal value of g(a, b; D), among all convex domains D such that a, bED, C E C"'-D. 8.5.17. Let G be a domain such that H = C~ G is a finite system of bounded continua. Given two real constants A, B, show that there exists a unique function cp harmonic in G, continuous in C, which takes the value B on H and has the form Aloglzl+O(1) as Z -+ 00.
8.6. BERGM,AN KERNEL FUNCTION
In this chapter G denotes a domain in the finite plane whose complementary set is a finite system of disjoint continua and L 2 (G) is the family of all functions
I analytic in
G and such that the integral
HIf(x+iy)1
2 dxdy
(taken in the sense
G
of Lebesgue) is finite. 8.6.1. Verify that L2(G) is a complex, linear space with the usual definitions of addition and multiplication by numbers. 8.6.2. If f, g EL 2 (G), show that the integral
Hfgdxdy =
(f, g)
is finite
G
and has usual properties of an inner product. 8.6.3. Show that for ~ny 1; E G there ~xists f E L2(G) such that 1m = 1. Hint: If is the unbounded component of C"'-G consider the univalent function cp mapping c",-r onto a disk.
r
8.6.4. If A(M) is the family of all (f,f) = IIfll2 :;:;:; M2
f
E
L2(G) such that
and
K(a; p)
c:
show that there exists a uniform bound of If(z) I for f Hint: Cf. Exercise 4.8.13.
G, E
A(M),
Z E
K(a; p).
8.6.5. Evaluate a common bound for If(z) I in terms of M, a, p (Ial+p < I), when G = K(O; 1).
8.6. BERGMAN KERNEL FUNCTION
119
8.6.6. show that for any ~ E G there exists fo E L 2 (G) which has the minimal norm (f,f)1/2 among all f E L 2(G) taking the value 1 at the point 1;. 8.6.7. (cont.) Prove that for any g E L2(G) with g(1;) = 0 we have (g '/0) = O. Hint: Consider f* =fo+ee i8g (e > 0, 0 ~ () ~ 27t). 8.6.8. An analytic fun~tion k of G and defined by the formula
Z E
G depending on a complex parameter
~ E
k(z, 1;) = Ilfoll- 2fo(z, 1;),
where fo is the solution of the extremal problem considered in Exercise 8.6.6 is called Bergman kernel function. Express k(z, 1;) in case G is simply connected, by means of the univalent function w mapping G onto K(O; 1) and such that w(1;) = O. Also verify the uniqueness. Hint: Exercise 8.1.12. ~)
8.6.9. Determine k(z,
for: (i) G = K(O; R); (ii) G
=
{z: imz
> O}.
8.6.10. CAven k(z, 1;) for a simply connected domain G, find the univalent function w mapping G oItto K(O; 1) so that w(1;) = 0, w'(1;) > O. 8.6.11. Verify the reproducing property of k: fm =
~~ f(z)k(z, C)dxdy,
1; E G,/E L2(G).
G
Hint: Consider f(z)/f(1;)-fo(z, 1;); (cf. Ex. 8.6.7).
8.6.12. Show that k(1;,
~)
=
~~ Ik(1;,
z))i 2 dxdy
and
Ik(1;, '])1 2
~ k(1;,
1;)k('], ']).
G
8.6.13. Verify that the reproducing property of Exercise 8.6.11 implies the uniqueness of k. 8.6.14. Suppose {IP.} is a complete, orthonormal set of functions in L 2 (G) (i.e. (lPj, IPk) = ~jk with ~jk = 0 for'j i= k, ~kk = 1 and the linear combinati~ns
• \ ' CklPk
("":"'0
rr f,
g
form a dense set in L 2 (G)). E
L2 (G) and
ab
bk are Fourier coefficients off and g, resp., show that
,,,
00
~ a.b. is convergent and the Parseval identity (f, g) ,.
0
=
L n=O
/lint: Exercise 8.6.11.
L anb,. holds.
n=O 00
8.6.15. (cont.) Prove that k(z, ~)
=
IPn(z) IPn (1;).
8. THE DIRICHLET PROBLEM
120
8.6.16. Show that the functions fPn(z)
=]1:
fPo (z) = (27t log
(l-h 2n
r zn1
1
(n
=
=fl, =f2, ... ),
~) -1/2 Z-l
form a complete, orthonormal system for the annulus A
=
{z: h < Izl < I}.
8.6.17. Find Bergman kernel function for the annulus A
=
{z: h < Izl < I}.
8.6.18. If A
= {z:
h < Izl < I} and J
has the Laurent expansion J(z)
00
=
L
bnzn in A, find the conditions for coefficients bn in order that
J
n=-co
belong to the class L2 (A). 8.6.19. Examine the behavior of k(z, C) under conformal mapping. Show that
yk(" C)ldCl
is a conformal invariant.
CHAPTER 9
Two-Dimensional Vector Fields Potential theory in two dimensions is usually concerned with fields of force in which the vectors of the field are always parallel to a distinguished plane. Moreover, the vectors associated with the points of any straight line perpendicular to the distinguished plane are equal; hence the whole field is characterized by the field in the distinguished plane. If the field does not depend on time, it is called stationary. 9.1. STATIONARY TWO·DIMENSIONAL FLOW OF INCOMPRESSIBLE FLUID
Stationary two-dimensional flow of incompressible fluid is characterized by two harmonic functions: the velocity potential rp and the flow function 1p. If w = u+iv is the vector expressing the velocity of a particle of the fluid past a given point z = x+iy, then u = rpx, V = rpy. On the other hand, the difference 1p(Xl' Yl)-1p(X O ' Yo) yields the volume of fluid passing in one second through a face of height 1 whose basis is any arc joining xo+iyo to Xl +iYl' In absence of singularities and in case the velocity field is a simply connected domain, I - rp+i1p is an analytic function which is called complex potential of flow. The lines rp = const are called equipotential lines and their orthogonal trajectories 'y'(x, y) = const are called the lines of flow. In a stationary velocity field the lines of flow are the paths of the particles of the fluid. 9.1.1. Express the velocity wand its absolute value JwJ in terms of the complex potential of flow. 9.1.2. Prove the following uniqueness theorem for the velocity potential: if the domain D is swept out by regular arcs y and on each y the functions u, U have both a constant value (depending on y), then U = Au+B, where A, B II rc real constants. /lInt: Consider ux-iuy. 121
9. TWO·DIMENSIONAL VECTOR FIELDS
122
9.1.3. The lines of flow are circles x 2+y2-2ax = O. Find the complex potential of flow. Evaluate the ratio of velocity of two particles of fluid passing the points 2a, a(1+i). 9.1.4. The lemniscates r2 = a 2cos2(J (r, (J are polar coordinates) are the lines of flow. Evaluate the ratio of velocity of particles passing the points z = a, z = aC}irt + iJl't) (a is real). 9.1.5. Find the loci of constant velocity for the flow pattern as considered in Exercise 9.1.3. 9.1.6. Discuss the equipotential lines and the lines of flow and evaluate the velocity, if the complex potential of flow is equal: (i) az; (ii) aiz; (iii) Z-l; (iv)
z-b
log--; (v) Z2; (vi) alogz; (vii) ailogz (a z-c
> 0).
9.1.7. If the volume of fluid flowing outside across a cylindrical face of height 1 and basis C(a; r) is equal Q for all r sufficiently small, the point a is called a source of intensity Q. If the flux Q is negative, the point a is called a sink of intensity IQI.
The line integral ~ wsds, where Ws is the tangent component of velocity and y y
is a contour, is called the circulation along y. If the circulation along all contours containing the point a inside and sufficiently close to a is equal 1= 0, then a is called a vortex. If y is a contour containing inside one singular point and f is the complex potential of flow, show that
r
r+iQ
=
~f'(z)dz. y
9.1.8. Given a flow with complex potential log(z2 +Z-2). Evaluate sources and sinks and the corresponding intensities. 9.1.9. Discuss the flow pattern with the complex potential 2ilog(Z2_ a2). Evaluate the circulation along the cir~les Iz=Fal = a. 9.1.10. The complex potential of flow is equal log sinh 1tZ. Evaluate the flux across the circle C(O; t) and the circulation along it. 9.1.11. Evaluate the complex potential of flow of fluid from the lower halfplane into the upper half-plane through an orifice in the real axis between 1 and -1 supposed the velocity at 00 is equal 0 and the flux across the orifice is equal Q. Find the pressure at various points of the orifice. 9.1.12. Evaluate the complex potential of flow past a circular cylinder with cross-section CeO; I) immersed in a parallel stream, assuming that the velocity
9.1. FLOW OF INCOMPRESSIBLE FLUID
123
Woo at infinity is parallel to the real axis. Evaluate the velocity at z = i and z = 2, as well as the difference of pressure at these points.
9.1.13. Evaluate the -complex potential of flow past an elliptical cylinder x2 y2 7 lJ2 = 1 (a > b) immersed in a parallel stream, if the velocity at in-
+
finity is equal
Woo e i «
(0( real, woo> 0).
9.1.14. Consider analogous problems for cylinders with cross-sections:
(i) K(-i; y'2) n K(i; (ii) [-ih, ih].
]/':2);
9.1.15. Evaluate the complex potential of flow with a source of intensity Q at z = a, assuming Woo = O. 9.1.16. Evaluate the complex potential of flow with a source of intensity Q at -1 and a sink of the same intensity at z = 0, assuming Woo = O. 9.1.17. Evaluate the complex potential of flow in the first quadrant with a source of intensity Q at l+i and a sink of the same intensity at z = 0 assuming Woo = O. 9.1.18. Discuss the limiting case as h --+ 0 of flow with a source of intensity p/h at z = h and a sink of the same intensity at z = -h (a dipole of the momentum p). Find the loci of constant velocity. 9.1.19. Evaluate the complex potential of flow with a vortex at z = a involving the circulation Assume Woo = O. .
r.
9.1.20. Evaluate the complex potential of flow in the upper half-plane under the assumption there exists a source at z = ai of intensity Q and the velocity at 00 is parallel to the real axis. Find the velocity at z = O. 9.1.21. Consider analogous problems as in Exercise 9.1.20 assuming that:
(i) z = ai is a dipole of the momentum p and Woo > 0; (ii) z = ai is a source-vortex of intensity Q involving the circulation
r.
9.1.22. Determine the complex potential of flow past a cylinder with the cross-section K(O; 1) with a vortex z = ai (a > 1) involving a circulation Assume Woo = O.
r.
9.1.23. Evaluate the complex potential of flow past a cylinder with the crossNcction K(O; 1) assuming that z = a (a > 1) is a source of intensity Q and II'", . O. 9.1.24. Consider an analogous problem, the source being replaced by a vortex with circulation
r.
124
9. TWO-DIMENSIONAL VECTOR FIELDS
9.1.25. Show that the complex potential function
r logz ( R2) + 27ti
J(Z) = a z+--z
corresponds to an asymmetric flow past a circular cylinder with the cross-section K(O; R). Determine the real constant a so that z = iR is the only point where the stream lines enter (and exit) the cylinder cross-section. Find the velocity at 00 in this case and evaluate the force acting on the cylinder according to the Joukovski lift formula. 9.2. TWO-DIMENSIONAL ELECTROSTATIC FIELD
Two-dimensional electrostatic field is produced by a system of charged, long, parallel wires and cylindrical conductors. The potential is constant throughout each conductor and there are no charges in the interiors of conductors. The electrostatic field can be characterized by an analytic function J = qJ+i'IjJ or the complex potential oj the electrostatic field. The real, single-valued harmonic function 'IjJ is called the (real) electrostatic potential. The field vector w is equal -grad'IjJ = -if'(z). The complex z-plane playing the role of a system of coordinates may be any plane perpendicular to all conductors involved. One single wire with a charge of q units per 1 cm of length which intersects the coordinate plane at z = a gives rise to the electrostatic field with complex potential -2iqlog(z-a) (cf. Ex. 9.2.2). Given the complex potential f= qJ+ i'IjJ , the lines 'IjJ = const correspond to equipotential lines, whereas the lines qJ = const are the lines of force.
9.2.1. An infinite, straight-line shaped wire I is charged with a positive charge of q electrostatic units per 1 cm of length. Show that the Coulomb force w produced by the wire and acting on a unit 'charge at a distance rcm from the wire is perpendicular to the wire and Iwl = 2qr- 1 • 9.2.2. (cont.) Show that w can be derived from the complex potential J(z) = - 2qilogz, where the z-plane is perpendicular to I and cuts it at the origin. 9.2.3. Evaluate the complex potential of the electrostatic field due to two long parallel wires each bearing a positive charge q per unit of length, the distance of wires being 2h. Discuss the equipotential lines. 9.2.4. Evaluate the complex potential of an electrostatic dipole, i.e. the limiting case of electrostatic field due to two long parallel wires each bearing charges of"
9.Z. TWO-DIMENSIONAL ELECTROSTATIC FIELD
125
opposite sign =Fq per unit of length, the distance of wires being 2h, 2qh as h -+ O. Find the field strength and the loci of a constant field strength. 9.2.5. If
(1
-+
M
is the density of charge on the surface of a cylindrical conductor
and "P is the real electrostatic potential, show that
(1
= - :7t .
~~ .
Hint: The flux of the electrostatic vector field across a closed surface S is equal to 47t times the total charge inside S (Gauss theorem). 9.2.6. Given the potential "P outside a charged cylindrical conductor with the cross-section evaluate the total charge on the conductor.
r,
9.2.7. A n-tuply connected domain G whose boundary consists of two disjoint systems r o , r 1 , of closed analytic curves determines an electrostatic condenser. The systems ro, r 1 correspond to the cross-sections with the plane of reference of two systems of conductors (outer and inner coatings), the potential of conductors of either system being kept on the same level. Usually the inner coating is charged and the outer one is grounded so that its potential is equal to zero. Show that the total charges qo, ql on both coatings are equal but have opposite signs.
Hint: Consider the
int~gral ~
ro+r,
~ ~~
hds = -
W
:"P ds and apply the Green's formula: n
~~ (hx"Px+hY"Py}dxdy - ~~ hLJ"Pdxdy G
with
h = 1.
G
9.2.8. (cont.) If "Po, "P1 are potentials of ro, r 1 resp., and q is the charge on 1'1' the ratio c = q /("PI -"Po) does not depend on "Po, "P1 and is called the capacity of a condenser. If w(z} = w(z; r 1 , G) is the harmonic measure of r 1 w.r.t. the domain G, show that
47tC =
~~ (w~+w;}dxdy. G
Hint: Consider
~
iJG
w
~w
un
ds.
9.2.9. (cont.) Show that the capacity of a two-dimensional electrostatic l"Ol1denser is invariant under conformal mapping. 9.2.10. Find the complex potential of the electrostatic· field between two l'o:txial circular conducting cylinders of radii r and R. Show that the capacity or sLich a condenser is equal to t(log(R/r)t 1 cm per unit oflength of gen-
n:ttrix.
9. TWO-DIMENSIONAL VECTOR FIELDS
126
9.2.11. Show that the capacity c of a pair of long conducting parallel wires of radius a separated by a distance b is equal c=
~ (10 2
b+
vbC4a )-1 T
g b-vb 2 -4a 2
9.2.12. Find the complex potential of the electrostatic field between two con-
ducting circular cylinders with cross-sections K(O; 1), K(I; 4) assuming that the potential on the first cylinder is equal 1, whereas the second one is grounded. Evaluate the capacity of the cylinder. 9.2.13. The potential on the cylinder with cross-section 1(5; 4) is equal 1,
whereas the cylinder corresponding to K( - 5; 4) is grounded. Find the extremal densities of charge on both cylinders. 9.2.14. A horizontal wire of negligible radius bearing a charge A per unit length is suspended at a distance h above the surface of a conducting plane. Find the complex potential and the charge per unit area in the plane. Hint: The field will not change, if we replace the plane by a symmetric wIre bearing a charge - A per unit length. 9.2.15. Show that the electrostatic potential outside a system of long, conducting cylinders with cross-sections k (k = 1, 2, ... , n) bearing jointly a positive charge A per unit length and kept on the same potential B, has the form VJ(z) = 2Aloglzl- 1 +o(1) as z --+ 00, if B is suitably chosen. Prove that the ratio BI2A = Y does not depend on A. Express y by means of the Green's function
r
n
g(z) of the exterior of
U
r
k•
The constant
y is
called Robin's constant.
k= 1
9.2.16. Prove that Robin's constant for the exterior of the ellipse x2 y2
(i2+V= 1 is equal to -Iogi-(a+b).
CHAPTER 10
Univalent Functions 10.1. FUNCTIONS OF POSITIVE REAL PART
10.1.1. Iffis a complex-valued function of a real variable t
E
[a, b],J = u+iv, b
and g is a real-valued function of bounded variation, the Stieltjes integral ~ f dg b
a
b
is defined as ~ udg+q vdg. By using the well-known estimate for real-valued/: a
a
h
I~ fdgl ~
V~(g)max If I ,where V~ (g) denotes the total variation of g on the
a
interval [a, b] show that an analogous estimate also holds for complex-valued f. b
I
b
I
Hint: Take real oc such that ~ fdg = e ia ) f dg and transform the last expression. a
a
10.1.2. If the sequence of complex-valued functions {fn} of a real variable t E [a, b] is uniformly convergent in [a, b], and the sequence {gn} of real-valued functions with uniformly bounded total variation on [a, b] converges to g, b
b
show that the limit lim ~ f"dg" exists and is equal to ~ fdg. a
a
10.1.3. Using the Schwarz formula (Ex. 8.2.2) prove the following theorem due to Herglotz: Let f be analytic in K(O; R) with f(O) = 1 and ref(z) > 0 for all z E K(O; R). There exists an increasing function {t of t E [0,2,,], {teO) = 0, {t(2,,) = 1 such that 2" . Ret+z ~ fez) = -R-'t- d{t(t). e -z o
t
flint: Consider the sequence {t,,(t) = (27tr 1 ~ u(R" ei8 ) d() , where u(z) = ref(z),
/(" (I -- .!)
o
R, and write Schwarz's formula in a Stieltjes integral form. 127
to.
128
UNIVALENT FUNCTIONS
10.1.4. Let f!J' be the class of all functions analytic in K(O; 1) and such that = 1, ref(z) > 0 in K(O; 1). Show that the function (1+z)/(1-z) belongs to !?J and cannot be represented by the Schwarz formula in the unit disk. Write a corresponding Herglotz representation.
f(0)
10.1.5 If H is a fixed, continuous, complex-valued function of t E [a, b] and I-' is a variable, increasing function of t E [a, b] subject to the conditions 1-'(0) = 0, 1-'(1) = 1, show that the set of all possible values of the Stieltjes inb
tegral ~ H(t)dl-'(t) is identical with the convex hull of the curve F: w = H(t),
a~ t
•
~
b.
10.1.6. If Cn is the 11th Taylor coefficient at z = 0 off E t!I', express Cn in terms of the function I-' of Exercise 10.1.3. Show that en E K(O; 2). Also show that for any Wo E R(O; 2) and any positive integer n there exists hE!?J such that h(n) (O)/n! = wo. 10.1.7. Using Exercises 10.1.3 and 10.1.5, find the region of variability of the point fez) for fixed z E K(O; 1) and f ranging over !?J. 10.1.8. If f
E
f!J' and argf(zo)
= oc, show that
I
f'(zo) 2cos oc \ f(zo) ~ I-lzol2 . Hint: Consider qJ(z)
= [f( ttz:oz) -f(zo)] [f( t:z:oz) +f(zo)T
1
10.1.9. Let f be analytic in K(O; R) and :1= 0 in K(O; R)",O. Suppose "P is a positive, differentiable function of r E (0, R) such that
I d If'(z) fez) ~ dr log"P(r~ =
"P'(r) "P(r) ,
Izl = r.
Show that for any real, fixed 0 the function If(re i6 )""P(r) decreases in (0, R) and the finite limit lim If(rei 6)""P(r) exists. ' ....R-
10.1.10. If f
E
f!J', show that for any fixed () the finite limit lim (1- r )f(reiO ) ' .... 1-
exists. 10.1.11. If the function I-' of ExerciselO.1.3is continuousatt show that
lim (l-r)f(re i6 ) ' .... 1-
= O.
= ()
(0:1= 0, 21t).
10.2. STARSHAPED AND CONVEX FUNCTIONS
129
10.1.12. If f-t has a jump h at t = () «() #- 0, 27t), I.e. f-t«()+ )-f-t«()-) show that lim (l-r)/(re'6) = 2h.
= h,
Y---=;l-
10.1.13. If () = 0, 27t, show that lim (1-r)/(r) = 2[1-f-t(27t-)+f-t(0+)]. r-+-l-
10.1.14. If IE :!J, show that for any fixed, real () the finite limit lim (l-r )/(rei6 ) = rx«()) exists and is real, nonnegative. Also show that the set r-+I00
{(): rx«())
> O} is at most countable; if ()k are. its elements, show that
L rx«()J :;:;:; 2. k=l
10.2. STARSHAPED AND CONVEX FUNCTIONS
10.2.1. If I is analytic and does not vanish in the annulus {z: r- r5 < Izi < r+ r5}, show that
a
16 zl'(z) aoarg/(re ) = re I(z) .
10.2.2. If I is analytic in K(O; R), does not vanish in K(O; R)""O, 1(0) = 0, f'(O) #- 0, and re{zl'(z)/I(z)} > 0 for any z E K(O; R), show that I is univalent in K(O; R). Hint: The argument principle.
10.2.3. Show that (i) fl(Z) = z(l-z)-3 is univalent in K(O; f); (ii) I(z) = z(1-Z)-1X is univalent in K(O; 1) for 0:;:;:; rx :;:;:; 2. 10.2.4. Let S* be the family of all normalized starlike univalent functions, i.e. IE S* means that I is analytic in K(O; 1),/(0) = 0,1'(0) = 1 and zl'(z)/J(z) has positive real part in K(O; 1). Show that the domain D f =/[K(O; 1)] is starlike (or starshaped) with respect to the origin, i.e. W E Df implies [0, w] c D f . Also prove the converse. 10.2.5. Using Herglotz's formula derive structural formula for IE S*. 10.2.6. Find precise estimates of ill and 11'1 on C(O; r) for IE S*. Hint: Exercise 10.2.5, or Exercise 10.1.8. 10.2.7. Find the region of variability Gz of [z/l(zWJ2 for a fixed z (0 I ranging over S*. /lint: Exercise ]0.2.5 and Exercise 10.1.5.
II nd
< Izl < 1)
10. UNIVALENT FUNCTIONS
130
10.2.8. If IE S*, show that rerJ(z)/z]1/2
t.
>
10.2.9. If I maps the unit disk 1: 1 conformally onto a convex domain B, show that any smaller disk K(O; r), 0 < r < 1, is also mapped onto a convex domain B,. Hint: Consider the function "p = 110 cp, where
cp(z) =
if(;:
assume 1(0)
z)+(1-t)/(Z),
zEK(O; 1) and 0
~.
Hint: Integrate F(z) = (l+z)/(l-z) over the segment [Zl' Z2]. 10.2.13. If lEse, show that re [f(z)/z] > t. Hint: Using Exercises 10.2.11,10.1.3 find a formula expressingf' as a double Stieltjes integral. Integrate under the sign of integral and use Exercise 10.2.12. 10.2.14. If lEse, show that 11-zf(z)I'::;:; Izl, zEK(O; 1). 10.2.15. If IE SC, show that re[zi'(z)/I(z)]
Hint: Consider g(z) = [f'(C)(l-1C1 2)]-1
> t.
[/(C)-/( lC~;Z )l
10.2.16. If IE SC, show that: (i) (l+lzl)-l,::;:; Izf'(z)[f(z) I ,::;:; (l-IZ!)-l;
(ii) larg[zf'(z)/I(z)] I ,::;:; arcsin Izl. 10.2.17. Find sharp estimates of If(z) I and argrJ(z)/z] for IE Sc. 10.3. UNIVALENT FUNCTIONS
10.3.1. Suppose that r is a closed Jordan curve with parametric representation = R«()), (/J = (/J«()), 0'::;:; () ,::;:; 21t, where R, (/J are polar coordinates; R(O). tP«() are supposed to be continuous; moreover, tP«() is piecewise monotonic R
10.3. UNIVALENT FUNCTIONS
131
and the index n(r, 0) = 1. If h is a nonnegative continuous and increasing function of R E (0, +(0), show that 2"
~ h (R«())) dtP«())
> O.
o
10.3.2. Let f be analytic and univalent in the annulus {z: a < Izl < b} and let rr be the image curve of C(O; r) under J, a < r < b. If g is continuously differentiable and If(re i9 ) I = R(r, ()), argf(re i9 ) = tP(r, ()), show that 2"
~
2"
R(r, ())g'(R(r, ())) d 9 tP(r, ())
= r :,
o
~ g(R(r, ()))d(). 0
10.3.3. (cont.) If n(r" 0) = 1, a
0, show that M(r,f) =J(r). Hint: If Wo =J(zo), then g(O, Wo; G) = -loglzol. 10.4.16. Suppose thatJ,f* are analytic and univalent in the unit disk and the image domain G* = J*[K(O; 1)] contains the origin and arises from G = J[K(O; 1)] by circular symmetrization with respect to the positive real axis. If J(O) =1*(0) = 0, show that M(r,f) ~ M(r,f*) for all r E (0, 1). 10.4.17. Suppose that JE S and Df = J[K(O; 1)]. Verify Koebe one quarter theorem (Ex. 10.3.10) by considering r(O; Df ). 10.5. THE METHOD OF INNER RADIUS MAJORIZATION
10.5.1. Let D be a domain possessing the classical Green function g( w, ao ; D) = g(w) and let ff(a o ; D) be a family of functions analytic in the unit disk and such that J(O) = ao , J[K(O; 1)] = Df c D. Prove that 1[,(0)1 ~ r(a o ; D) for any JEff(a o ; D). Hint: Consider h(z) = g(f(z»)+loglzl, z EK(O; 1). 10.5.2. (cont.) Prove an analogous theorem for arbitrary D. Hint: Consider D(P) = J[ K(O; p)] . 10.5.3. Prove following symmetrization principle. Let J be analytic in the unit disk and let J(O) = ao . If the symmetrized domain D* of Df = f[K(O; 1)] with respect to a line (Steiner symmetrization), or a ray through ao (circular symmetrization) is situated in Do, show that 1[,(0)1 ~ r(ao ; Do). 10.5.4. Suppose that 0 < oc < 2 and that the function J(z) = aO+alz+ ... is analytic in K(O; 1) and Df n C(O; r)is for any r > 0 a system of arcs of total length 7tocr at most. Show that lall~~ 20c laol, with equality holding for
( I+Z)" .
f(z) = ao l-z
10.5.5. If J(z) = aO+alz+ ... is analytic in K(O; 1) and R = R f is the least upper bound of all r > 0 such that C(O; r) c Df , show that lall ~ 4(la ol+R) with equality when ao ? 0 and J(z) = ao+4(a o+R)z(l-z)-2. 10.5.6. If J(z) = z+a 2z 2+ ... is analytic (not necessarily univalent) in the unit disk, show that for any 8 > 0 there exists r > {--8 such that C(O; r) c Df . 10.5.7. Deduce Koebe one quarter theorem from Exercise 10.5.6.
10.5. THE INNER RADIUS MAjORIZATION
137
10.5.8. Suppose that/(z) = aO+alz+ ... is analytic in the unit disk and that there exists h > 0 and a real-valued function v(u) , - 00 < u < + 00 such that on any straight line re w = u each point u+i(v(u)+nh) (n = 0, =fl, =F2, ... ) belongs to C,,-Df . Show that lall :;:;:; 2h/1t. Hint: Consider g(z) = exp {27tf(z)/h}. 10.5.9. Suppose that any circle C(O; r), r > 0, contains a point of and I is analytic in K(O; I). Show that (1-lzI2) If'(z) I :;:;:; 4 I/(z) I
C',Pf
for any z E K(O; 1). Hint: Exercise 10.5.5. 10.5.10. If I is analytic, univalent and does not vanish in the unit disk, show that l- lzl )2 ( 1+IZI)2 1/(0)1 ( 1+lzl :;:;:; I/(z) I :;:;:; 1/(0)1 l-lzl . 10.5.11. If cp is analytic in the unit disk and never assumes both values w, -w, show that (1-lzI 2)lcp'(z)1 :;:;:; 2Icp(z)l· 10.5.12. Suppose that I is a Bieberbach-Eilenberg function, i.e. I is analytic in the unit disk, 1(0) = 0 and I(Zl)f(Z2) '" 1 for any Zl' Z2 E K(O; 1). Show that (i) I is bounded in the unit disk; (ii) (1+/)/(1-/) never assumes both values w, -w. 10.5.13. Show that for any Bieberbach-EiIenberg function I we have: (i) (1-lzI2) If'(z) I :;:;:; II-P(z)l; (ii) 11'(0)1 :;:;:; I. 10.5.14. If IE Sand L(r, f) is the angular measure of values omitted by I and situated on C(O; r), then L(r, f) is equal to zero for r :;:;:; -} and any IE S and can be equal to 21t for r;;:;:: 1 (e.g. for I(z) == z). If r E (!f, 1), show that L(r)
=
sup L(r,J)
=
4 arc sin
(2vr -1).
/eS
Show that the extremal function maps K(O; 1) onto
C",({w: Iwl
=
r, largwl :;:;:; 2arcsin(2vr
-In u
[r,
+(0»).
"jnd the extremal function (cf. Ex. 2.9.22). 10.S.15. Solve an analogous problem for starlike univalent functions.
10. UNIVALENT FUNCTIONS
138
Hint: Verify first that the mapping
W=F(t)=(I+Vt)1J I-OV t + t , I-Vt I+OV t + t
0
Ib-ia21.
1.1.24. Circle C(BA-l; IAI-1VIBI2-AC). 1.1.25. The radius: kla-bIII-k 21- 1; the center: (a-k 2b)(I-k2)-1. 1.1.26.
Izl2 IZl12 IZ212 IZ312
z Z Zl Zl Z2 Z2 Z3 Z3
1 1 =0. 1 11
1. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
143
1.1.27. Cf. Exercises 1.1.26, 1.1.24. 1.1.28. (i) If m 1 = 0, then Zo E [Z2' Z3]; if m 1 = 1, then Zo = Zl; if m1 =f. 0, 1, then Zo = m1z1+(1-m1H'1' where C1 = (l-m1)-1(m2z2+m3z3) E [Z2, Z3] and therefore Zo E [Zl' Cd E T. (ii) If the change of mj by hj gives the same point zo, then h1Z1+h2Z2+h3Z3 = 0, or h1 (zl-z3)+h2(z2-z3) = since h1 +h2+h3 = 0. This means that z} are collinear.
°
1.1.29, 30. Induction with respect to n and ExerCise 1.1.28. 1.1.31. If ft =
L" IC- Zkl- 2,
mk = ft- 1!C-Zkl- 2, then mk
k=l
and
L" mk(C-zk) = 0,
k=l
or C =
L"
k=l
>
0,
L" mk =
1
k=l
mkzk'
1.1.32. If P(z) = A(Z-Zl) ... (z-z,,), then P'(z) =_1_+_1_+ + __ 1_. P(z) Z-Zl Z-Z2 ... z-z" '
(cf. now Exercise 1.1.31). 1.1.33. logz" = ;
[2; +0 (:2)] ~
x, or Iz,,1
~ eX;
argz" = narctan(y/(n+x)) ~ y. 1.1.34. The sequence is divergent; tz,,1 2
~ R2 =
Ii (1 + -\-), n
,,=1
however,
. a term 0 f a d'Ivergent series; . . 0 f C(O ; R) arg -z"- = arctan -1 IS every pomt Z,,_l n is an accumulation point for {z,,}.
1.1.35.
I
b"C" is absolutely convergent, since {e,,} is bounded.
1.1.36. The convergence of I Z" implies the convergence of I x,,; since .\'" ~ 0, is also convergent. The convergence of ~>; follows now from the l'OIlVergence of both I
Ix;
Ix;, z; .
1.1.37. Consider first the case
lime"
=
°and show using (i), (ii) that limz"
1.1.38. Put a"k = Pk/(P1+P2+, ... , +p,,) for k,;;;; n, and a"k apply Exercise 1.1.37. 1.1.39. Express
W"
by means of
Z"
and use Exercise 1.1.37.
=
0 for k
= O.
>
n;
144
SOLUTIONS
n
1.1.40. Express lVn in terms of
L
Sn =
Zk
and use Exercise 1.1.37.
k=l
1.1.41. Put p. 1.1.42. If v
=
#;;1 and cf. Exercise 1.1.40.
1 0, consider the sequence lVnlv; note that nv
=1=
1 VI' 0, 0, '" are lines of a matrix of Exercise 1.1.37. If v nv by Vn = g+v n with g =1= O.
•.. , Vn
1.2.1. Use the parametric equation of the straight line NA: X
Z
=
1 nv
V n,
V n_ l , •..
0, replace
=
= Xl t,
Y
=
x 2 t.
I+(x3-I)t.
1.2.2. (coscx; sin'cx; 0), (--j-; -j-; i-), (,;; - 1~;
!;-).
1.2.3 .•C'"K(O; 1); K(O; 1). 1.2.4. Note that all projecting rays are situated in one plane through N and the given straight line. 3
1.2.5. The points of S situated on a circle satisfy:
L
AkXk+B =
0; use now
k=l
Exercises 1.2.1, 1.1.24.
1.2.6. Use the formulas of Exercise 1.2.1: 1.2.7. (i) opposite points on the same parallel; (ii) points on the same parallel symmetric w.r.t. the plane' OXIX3; (iii) opposite poi~ts on the circle Xl = C, x~ +x~ = 1- C 2 • 1.2.8. Put
Xl =
cos()cosrp etc. into the formulas of Exercise 1.2.1.
1.2.9. Cf. Exercise 1.2.1. 1.2.10. Note that the equation of the stereographic projection of the great circle is identically satisfied by -z-I·(Ex. 1.2.9). 1.2.11. The equations Iz-zol2 = I+lzoI2, zz-zoz-zoz-I = 0, are equi. valent (cf. Ex. 1.2.10). 1.2.12. C(zo; R), Zo
=
-14-
2;
i, R
=
-}
VI5I7.
1.2.13. Apply the formula for the distance of two points and express their coordinates by means of formulas of Exercise 1.2.1. da 1.2.14. -d S
=
. a(zl z) lim '2 (cf. Ex. 1.2.13); hence da 2 Zl--+Z
IZ1-zl
this implies preservation· of angles.
=
2
2
A(z)(dx +dy ) alld
f. COMPLEX
~UMBERS.
LINEAR TRANSFORMATIONS
1.2.15. O'(z, a) = 0'([;, a), O'(z, _a-I)
=
145
0'([;, -a-I), hence
+;[; 1= I ;+~z I;
11C
moreover, z and [; are situated on circles througli a, _71- 1 intersecting at an angle cp, hence
C-a C+a- l
arg
arg=---
1.2.16. IZ-alllz-a21-1 with limit points ai' a2' 1.2.17.
t
O'(lal-r,
z-a z+a 1
=
cpo
(1 + laI1 2)1/2(1 + la 212)-1/2 which is Apollonius circle
=
lal+r).
1.2.18. Evaluate e.g. the length of inscribed polygonal line using Exercise 1.2.13 and consider its limit for a normal sequence of partitions. =
1.2.19. The stereo graphic projection of a rhumb line intersects all rays argz const at a constant angle hence it must be a logarithmic spiral (cf. Ex. 1.1.19). 1.2.20. The equation of stereographic projection is r
r I ek6 , k
=
= ~ log ~ , rJ.
.
rl
hence l(r)
=
Jr
22 yl+k- 2 l+r2 dr
--
=
2 j/l+k- 2 (arctanr 2-arctanrl )·
"
The numerical example corresponds to arc tan r1 = 'Ttj6, arctanr2 l(r)
1.2.21. d0'2
=
=
;
[I + ( 61~3
4(dx 2+ dy2) (I + Iz12)2 ' hence IDI
=
rr
,
Tt/3, hence
/2 •
~~ YEG-Pdxdy = ~~ (l+~ZI2)2 LI
=
dxdy.
LI
1.2.22. Without loss of generality we may take the south pole as one of the I'ertices of T. The stereographic projection To of T is bounded by two straight line segments emanating from the origin and an arc of a circle: z = zo+ I ~/ 1-=f-lzoI2e''P, CPI::( cP ::( CP2 (cf. Ex. 1.2.11). Using the formula of Exercise
1.2.21 in polar coordinates we obtain:
~ ~~ (l-~;'iF drdfJ = ~2 1~:i dfJ;
ITI ..
Tn
6,
146
SOLUTIONS
we now introduce a new variable g; putting rei8
,2 =
=
zo+Yl+lzoI2ei'l'. We have:
1+2IzoI2+2 yl+lzoI2(xocosg;+yosing;),
dO d (./ .. dg; = dg; arg zo+ V 1+l z ol2 el'l') =
where xll+iyo
=
r- 2 [1
+ Izol2+yl + IZoI2(xocosg;+Yosing;),]
zo, (cf. Ex. 1.1.20). This gives: '1'2
ITI = ~ dg; = g;2-g;1 = (oc+p+y)-7t, '1'1
which follows from elementary geometric considerations. 1.2.23. The stereographic projection of T is bounded by arcs of C(O; 1), C( -1 + i; y3); hence ITI =
7t
2
[i, i(l+v'Z)] and
.r 1 +2arctan V 2-7t = arctan 2 {i.
1.3.1. The angle of rotation: arga; the ratio of homothety center at the origin: lal· 1.3.3. W = 1ti(z-t). 1.3.4. (i) w = az+b, a> 0, b real; (ii) w = az+b, a < 0, b real; (iii) w = -i(az+b), a> 0, b real.
1.3.5. w =
(b-a)-l[(B-~)z+Ab-aB].
1.3.6. w = (l+i)(l-z). 1.3.7. w =
b:~~l
1.3.8. Zo
b/(I-a); the angle of rotation: arga, the ratio of homothety:
=
(z-ib2)·
lal·
1.3.9. (i) Zo = t(3+i), arga = --t 7t , lal = l/i; (ii) Zo = b2(l-ik)/(b 2-b1-k-i), a = (k+i)/(b 2-b1). 1.3.10. It is sufficient to consider the case W = aZ (cf. Ex. 1.3.8); if Z(O) AlaI 8/"e i8 , -CXJ < 0 < +CXJ, A > 0, arbitrary, and oc = arga ¥: 0, then evidently aZ(O) = Z(O+ oc). Hence the spiral Z = Z(O) remain unchanged after the transformation W = aZ. =
I. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.4.1. If e ¥= 0, then w =
be-ad e
147
a -1 Zl +-, ZI = Zz , Z2 = ez+d. e
1.4.2. If z(O) = r(0)ei6 is the equation of y, then the image curve has the equation w(O) = (r(0)t1e-16. The angle oc. between y and the radius vector satisfies cot oc = r'(O)/r(O); a corresponding value for the image curve after taking the opposite orientation is the same. 1.4.3. Use Exercises 1.4.1 and 1.4.2. 1.4.4. The equality is equivalent with the equality [w-a(laI 2-r 2)-1] [w-aClaI 2-r 2)-I] = r2(laI 2_r2)-2; the image curve of CCa; laD is a straight line. 1.4.5. Use Exercise 1.4.4 with a = bi, r2 = b 2+1. 1.4.6. Use Exercises 1.4.1 and 1.4.4. 1.4.7. Verify that w = az+b, w = Z-l do not change the cross-ratio and cr. Exercise 1.4.1. b-e z-a 1.4.8. w = - - ' - - . b-a z-e 1.4.9. (i) Straight lines rew = (2a)-1; (iii) circles b(U 2+V2)+U+v = 0; (iii) straight lines v = - ku; (iv) circles through w = 0 and w = ZOl; (v) cissoid u2(v+l)+v 3 = O. 1.4.10. Linear transformation defined by the equation (ZI' Z2' Z3' z) c (a, b, e, w), where a, b, c are real and different from each other carries the l'ircle (or possibly the straight line) C determined by Zl' Z2' 'Z3 into the real axis. Since the r.h.s. is real iff w , is peal, so is the l.h.s. iff z E, C. 1.4.11, 12. Verify that identity, inverse transformations and superposition are also transformations of the same type. 1.5.1. The equation satisfied by z* is equivalent to the conditions: Iz-al .: Iz*-al = r2, arg(z-a) = arg(z*-a). 1.5.2. If Z = a+re i6 , z = a+pe''l', then IZ-z*IIZ-zl-l
= /re l6 -
~
el'l'!lrei6_pei'l'l-l
= ;.
X
SOLUTIONS
148
1.5.3.
-t+i.
1.5.4. (i) The straight line re z = t; (ii) lemniscate (.~2+y2)2 = x 2_y2. 1.5.5. If Zo is the center of the circle orthogonal to C(O; 1), its radius is JI Izo12_1. Its reflection satisfies (C- 1-zo)(C- I -zO) = IZoI2-1, which is equivalent to (C-zo)([-zo) = IZo12-1. 1.5.6. This is an immediate consequence of the fact that symmetry is defined in terms of circles and their orthogonality and both properties are preserved under linear transformations.
1.5.7. z*
=
1.5.S. If Lk
(02-o1)[(a2-al)z+a1 a 2-a1a 2]. C(ak; rk), k = 1,2, then
=
Z
=
a2+d(z-al) [(a l -ti 2) (z-al)+rtr1.
Hence (z = a1 ) +-+ (Z = a 2 ). If the same holds after interchanging the order of reflections, then rr+d = lal-a212 which shows that L 1, L2 are orthogonal. The sufficiency is easily verified after an additional linear transformation carrying L 1, L2 into perpendicular straight lines. 1.5.9. An immediate consequence of Exercise 1.5.1. 1.5.10. The roots of the equation. 2 .
a+ _r_ z-a
=
(a2-al)-1[(a2-al)Z+ala2-aI02]'
1.5.11. Circles through -7=f4V3 which are symmetric points with respect to both circles. 1.6.1. (i) The lower half-disk of K(O; 1); (ii) K(O; 1) n (C",K(-i-i; i-)); (iii) {w: imw < O} n {C"-K(t(1-i); +V2)}; (iv) K(t; i-),,-K(i-; -t); (v) {w: rew > i-} n (C,,-K(-t; t)). 1.6.2.
IV
= (z-4)(z-I)-1.
1.6.3. If A (w, b, c, d)
=
(b-a)(d-c)(c-b)-l(d-a)-t, then k
=
2A+I-2 JlX(A-I-1);
= (z, -1, 1, k- 1).
1.6.4. If a, b are real and (z = a) w
+-+ (w = b),
= (a+l)bz[(a-b)z+a(b+l)tl,
then ab(a+l) (b-H)
> O.
f. COMPLEX NUMBERS. LINEAR TRANSFORMATIONS
1.6.5.
C""K(-f; 1); {w: imw < O}
n
149
[C",K(-}; 1)].
1.6.6. z = (1-2w) (1+W)-1. 1.6.7. Z= i(w+1)(w-1)-1; Apollonius circles Iw+11/lw-11
1.6.S. If (z = a)
~
(w = 0), lal
=
r.
< 1, then
w = ei O. This mndition is also sufficient. 1.8.6. zt!(w-z 1 ) = Zl/(Z-Zl)+ih,
with real hand Zl = 1 (consider W
zd(z-zd). e'P(z-a)/(I-az)-l, with real fl and lal < 1. Hence z = e-'P X -- (II' I-ael/i)/(I+awe-//l), thus the inverse transformation is again an h-motion
I.S.7.
IV
=
SOLUTIONS
152
If w = e'Y(Z-b)/(I-bZ), Z = eifJ(z-a)/(I-az), the superposition has the form: w = exp {i({J+ y)} (1 +abe-' fJ) (z-e) [(I+abe ifJ ) (l-cz)]-1,
where e = (a+be-ifJ)/(I+abe- 1fJ ). Note that lei < I and II+abe-ifJl
=
II+abe'fJl.
°
=
[t(Z2-Z1)+Zl(I-Z1Z2)] [l-ZlZ2+tZ1(Z2-Z1W1, ~ t ~ I; the Z)-1 substitution C = (Z-Zl)(1-Z1 gives the Euclidean segment [0, C2] with parametric representation C = tC 2. 1.8.8. z
1.8.9. If w(z) is the given h-motion, (w(z)-b)/(I-bw(z) is a linear transformation with a zero at a and a pole at a-1, hence it has the form A(z-a)/(I-az). Moreover, for Iwl = Izl = I absolute values of both expressions are equal I, hence IAI = 1. 1.8.10.
~8 =
(l-l zI2)/(1-ICI 2) (cf. Ex. 1.8.9); if
u(1.
•
~=
max(tk+1-t,J, then k
with 1.8.11. If the arc-length 8 on y is the parameter and C(o; r) intersects y at two points corresponding to 81' 82, then after removing the open arc 81 < 8 o
luke now k (Ii)
lim
11.-... 0
U 2+V 2 x
x
2. = Uy2+Vy'
= mh which gives
I~wz 12 = u!+v!+2m(I +m2)-1(uXVX+UyVy) , ...
155
or
uxvx+uyVy = O.
156
SOLUTIONS
It follows from (A) and (B) that (u x+iUy)2 = (vx_iVy)2 which means that either J, or I satisfies Cauchy-Riemann equations. 2.1.9. Ux = Vy = 2uuy, Vx = 2uux = -un thus (1+4u 2)ux Hence Vy = Vx = uy = 0.
=
0, i. e. Ux = 0.
III = }/U2+V 2 ; I' = ux+ivx = vy-iuv • Hence Ltl/l = 111-1(u~+v~+u; +v; )-1/1- 3[U2(U~+U;)+V2(V~+V;)- 2uv(ux v x+uyv y)] = 1/1-12If'12-1/1-3(u2+v2)11'12 = 11'1 21/1-1, 2.1.10. If 1= u+iv, then
because uxvx+uyVy = 0. 0 22 I12 2.1.11. ox 1 --
2 2). 0 2 I12 C 2(uuxx+vvxx+ux+vx, oy2 I has an analogous lorm;
note that Ltu = Ltv = 0. 2.1.12. Ur = r- 1 V9 , Vr = -r- 1 U9 ; I'(z) = e- 19 (U,+iVr). 2.1.13. U
r"cosnO, V
=
=
r"sinnO;
I' = e-i 9nr"-1(cosnO+isinnO)
=
nz"-l
(cf. Exercise 2.1.12). 2.2.1. In the ri~t half-plane u has the form rp(y/x) and the condition Ltu =
°
gives ddt [rp'(t)(1+t2)] = 0, where t = y/x. Hence rp(t) = Aarctant+B, or . u(z) = Aargz+B, where A, B are real constants. 2.2.2. u has the form rp(X 2+y2) and Ltu =
d ° gives Yt[trp'(t)] = 0, where
t = X2+y2. Hence rp(t) = Alogt+B, or u(z) = 2Alogz+B, where A, Bare
real constants.
2.2.4. Ux =
~cosy
= v y, uy =
exp(loglzl +iargz)
=
-~siny
= -v x ;
exp(loglzl)[cos(argz)+isin(argz)]
= Izlexp(iargz) Logexpz = 2.2.5. I'(z)
=
=
z;
log~+iArg(cosy+isiny)
= x+iy = z.
ux-iuy, hence
x-iy (Logz)' = - - - = X2+y2
Z-1
and
(expz)'
=
eXcosy+iexsiny =, expz.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
157
2.2.6. If f = u+ iv and Ux = v y, uy = -vx, then
v u
V= Arg-.
Hence Ux
= (uux+oo x) (U 2+V2)-1 = (uvy-VUy) (U 2+V 2)-1 = Vy
and similarly Uy = -Vx ; F' = Ux-iUy =f'1lfr 2 =f'/f. 2.2.7. Uxx = -uy;because L1u = 0; uxy = uyx because the partials of second order are continuous.
2.2.8. Cauchy-Riemann equations imply
2.2.9. From Euler's formula for homogeneous functions: u(x, y) = m- 1(xux+Yuy)
it follows by differentiation; Ux = m-l(ux+xuxx+YUyx) = Vy = m-l(ux+YuXY-XUyy),
and similarly, uy = -vx • 2.2.10. (i) u is homogeneous of degree 2, hence by Exercise 2.2.9 fez) = (1-t·i)z 2.
vex, y) = -Hy(2x+y)-x(x-2y)].= t(y2_X2)+2xy;
(ii) v(x,y) = -2X3+3x2y+6xy2_y3, fez) = (1-2i)z3;
(iii) vex, y) = _Y(X2+y2)-1, fez) = Z-l; " 2 4 2 (iv) u = rez z- = rez- , v = i~z-2, f(i) ~ Z-2. 2.2.11. The denominator has the form "(1+Z2)(1+Z2), whereas the numerator is equal t(z+z)(1+zZ) = rez(1+:z2). Hence u v
=
rez(1+z2)-1,
'/= z(1+Z2)-1,
= imz(1+z2)-1 = y(1_x2_y2)[1+2(X2_y2)+(X2+y2)2]-1.
2.2.12. If v exists and F = u+iv J then F' = tix+iv x tive F = P+iQ does exist, then F'
= Px+iQx =
=
ux-iuy = f. If a primi-
Qy+iQ.~ =f= ux-iu,.
lienee Ux = Qy, uy = -Qx which means that Q is a conjugate harmonic function for u. 2.2.13. ux·-iuy = (z+l)e= has a primitive ze", hence v
=
imze" = eX(ycosy+xsiny).
158
SOLUTIONS
[} (
[}F)
2.2.14. rTr rTr 2.2.15. (i) u (ii) u = A
+ [}2F [}e 2
Axy+B, v
=
O.
= =
{-A(y2_ X2)+C;
-Vx+y'X2+y2 +B, v = Ay(x+ y'X2+y2tl/2+C,
i.e. u+iv = AZ 1/2+B+iC (cf. Ex. 1.1.5); (iii) do not exist. 2.2.16. Put cp(t) = ~exp(-bJ(t)dt)dt which means that CP"+CP'''P = 0 in (a, b). Hence L1(cp 0 F) = (F;+F;)cp"
0
F+L1F' cp' 0 F = (F;+F;)(cp" +CP'''P) 0 F = O.
r [} ( [}F) + [}eF]
2.2.17. ( r 2 Fr2 +F92 ) -1 r Tr r ar
[}2
2
=
"Po F.
2.2.18. The equation of the family of parabolas is x+ y' x 2+y2 If F(x,y) = X+J/X2+y2, then (F;+F;)-lL1F= (2F)-1;hence u (cf. Ex. 2.2.15 (ii)) .
= =
P = const. Areyz+B
2.2.19. fez) = exp(Az-1+B) with A real. 2.2.20. fez)
=
exp(AiLogz+B), A is teal.
2.2.21. u(r, e)
=
2.2.22. fez)
exp(-Aiz- 2+c), A is real.
=
A(loglzl-hrgz)+B; A, B are real.
2.3.1.1'(-1) = --t, hence
(X
= 7t, A =-t.
2.3.2. (i) The circle Iz+d/el = lel- 1Iad-ber 1 / 2 ; (ii) the straight line z = -d/e+e- 1y'ad-be t, -
00
< t
dv/ds = a2 and hence w = as+b, a = al+ia2' .
2.3.4. argf'(z) = 0 on the straight line of Exercise 2.3.2 (ii) hence on its image line: argf'(z)z'(t) = argz'(t) = const. 2.3.5. Infinitesimal segments are expanded (i) outside C(O; -t), (ii) outside C(-I;-t), (iii) inside C(O; 1).
2.36. 1f'(zo)1 = 75, argf'(zo) = arg(7+24i)+7t.
2 2 = 1u'" +.IV", 12 2•3•• 7 [}(u, [}(x, v) y) = uxvy -uyvx = u",+v",
=
If'I2
h' h IS . equu I
W lC
to the ratio of area of infinitesimal squares corresponding to each other under f
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
159
2.3.8. The mapping is locally I: I because the Jacobian is =F 0 at Zo; note that the curves u(x, y) = const, vex, y) = const are mapped onto perpendicular I segments. 2.3.9. (i) equilateral hyperbolas x 2_y2 = const, xy = const; (ii) circles ceO; r) and rays Argz = const. b
HIf'(z)1 2dxdy (cf. Ex. 2.3.7).
2.3.10. ~ 1f'(z(t»)llz'(t)ldt;
D
a
2.3.11. e(l; I)
= {z: z = 2cosllei 6}, hence w = 4cos 2 l1e 2 i 6 = 2(I+coslP)ei'l',
IP = 2l1;
n/2
~ 2Iz(lI)llz'(lI)1 dll = 16,
~~ 41z1 2dxdy = 67t.
IQI =
-n/2
K(l; 1)
2.3.12. Under w = Z2 D is mapped opto the rectangle L1: I hence
0
(cf. Ex. 1.1.20) which shows the convexity. 2.6.9. The image domain of C"..jcx,,8] under Z = (z-cx)(z-,8r 1 is C',,(-oo,O] which is mapped under LogZ onto {w: limwl < 'It}; (i) straight lines im w = const; (ii) straight lines re w = const; (iii) w = O. 2.6.10. The mapping is univalent as a superposition of linear transformation and the univalent mapping of Exercise 2.6.9. The mapping W = (,8-cx)w- 1 carries the strip lim wi < 'It onto the outside of two circles with diameters 1,8-cxl'lt- 1 tangent to each other externally at the origin with the common tangent parallel to [cx,,8].
2.6.11. Logf is analytic in D, hence 10g 0, 0< v < tv.
°
2.8.10. The mapping Z = e% carries the strip < y < 7t into the upper halfplane which is mapped by w = ArcsinZ onto the half-strip Irewl < t7t, imw > 0. The vertical boundary rays correspond to the rays: x» 0, y = and x» 0, y = 7t (x+iy = z). Since the mapping function has symmetry property w.r.t. the boundary rays, we can propagate the mapping onto the whole z-plane with removed rays y = k7t, X < (k = 0, =FI, =F2, ... ) and the values will cover the upper half-plane im w > 0.
°
°
2.9.1. The linear transformation Z = (z-a)/(z-b) carries the wedge into an angular domain with vertex Z = and a subsequent mapping w = zrt/lZ gives a half-plane.
°
2.9.2. Mter the transformation Z = zexp[ - ti(IX+P)] we obtain an angle symmetric w.r.t. the real axis; hence
w=
zrt/(P-IZ)
=
z!I/(P-rr.)
exp[ - t7ti(IX+P)/(IX-P)].
2.9.3. The linear transformation Z = (z+ l)/(z-l) carries the semi-disk into lhe quadrant (-; -) and a subsequent mapping w = Z2 gives the upper halfplnne. Hence w = [(z+I)/(z-IW. 2.9.4. The mapping W = Z3 carries the given sector into the upper semi-disk which is mapped under Z = [(W-J-I)/(W-I)]2 onto the upper half-plane. On
SOLUTIONS
166
the other hand, Z imZ > O. Hence
=
i(l+w)j(l-w) maps K(O; 1) onto the upper half-plane
, i(l+w)j(l-w)
=
[(z3+1)j(z3-1)]2.
2.9.5. w = i(z+I)3(z-1)_3 (cf. Ex. 2.9.1). 2.9.6. The mapping Z = (z+i)j(z-i) carries 'the wedge into the angle {-1t < argZ < t1t which is mapped under W = Z2 onto the right half-plane. Hence
w
=
i(W-l)j(W+l)
=
2z(I-Z2)-1.
2.9.7. w = tanht1tz (cf. Ex. 2.8.7); Apollonius circles with limit points -1, 1; circular arcs joining 1 to -1. 2.9.8. The linear transformation Z = (l-Z)-l maps the given domain onto the strip t < reZ < 1 which is carried under similarity W = 2i(Z-1-) into the strip lim WI < we now apply Exercise 2.9.7; finally
t; w
= tanh1ti((l-z)-l_1-) = itan1t(I-z)-l_1-).
2.9.9. The mapping Z = Z-l carries the given domain into the strip limZI
< t which is mapped under w = cotht1tZ onto
C",K(O; 1) (cf. Ex. 2.9.7).
Hence w = coth(1tj2z). 2.9.10. Two cjrcles intersect at an angle -t1t at 0, 2 and the third circle being an Apollonius circle with limit points 0, 2 is orthogonal to former ones. The mapping Z = zj(z-2) yields the circular sector: 0 < IZI < 2, t1t < argZ < f1t which can be mapped onto im w > 0 similarly as the sector of Exercise 2.9.4. Hence . w = [16(z-2)4-z4]2[16(z-2)4+z4t2.
v'
2.9.11. The mapping Z = w+t carries the slit w-plane into the right halfplane reZ> 0 so that (w = 0) (-jo (Z = t); the mapping Z = +(I+z)j(l-z) carries K(O; 1) into the right half-plane, hence w = z(l-z)-2 is the desired mapping. This is the Koebe function which plays an important role in various extremal problems in conformal mapping. 2.9.12. The linear mapping Z = (1+2w)j(I-2w) carries the given domain into ~(-oo, 0] and a subsequent mapping W = liz gives {W: re W > O}. Hence w = z(1 + Z2)-1 .
2.9.13. w = (l+Z)2(l-Z)-2. 2.9.14. The linear transformation Z = (z-ih)j(z+ih) carries the slit halt'plane into K(O;I)",(-I,O] which is mapped under W=Z(l-Zr 2 on(o C",(-oo, 0]; finally w = 2ihVW = 1/z2+h 2.
2. REGULARITY CONDITIONS. ELEMENTARY FUNCTIONS
2.9.15. After the inversion Z thus w = (hZ)-1(Z2+h2)1/2.
=
167
Z-1 we obtain the domain of Exercise 2.9.13,
2.9.16. By Exercise 2.9.11 the given mapping carries K(O; 1) into C",,(-oo, _+p-1(I_p)2]. Note that (-1,0] corresponds to (-tp-1(I-p)2,0].
2.9.17. w = (I+1]z)2(1-1]z)-2, \1]\ I+t 2.9.18. ( l=t
)2
(I_p)2
P
=
1.
Z
(1-Z)2 (cf. Ex. 2.9.17, 2.9.16).
. (1+p)2 2.9.19. The mappmgs Z = 4p
z (I-z)2 ' Z = (I-t)2 carry the slit
disk and the full disk, resp. into the Z-plane slit along (- 00, sired mapping has the following implicit form:
-t] so that
the de-
t(l+p)2p-1 z (I-z)-2 = t(1_t)-2.
2.9.20. The linear transformation
z = -e-ill(Z:-!X)/(I-aC),
where
(b-a)/(I-ab) = eill\b-a\/II-abl,
maps D(a, b) onto K(O; 1)",,(-1, -p] with p = \b-al/ll-abl; moreover (C = a) ~ (z = 0). The mapping function is obtained by substituting the above given values of z, p into the formula obtained in Exercise 2.9.19. 2.9.21. (dC) = (dZ) : (dZ) = _4eill Ib-al\I-ab\(lI-abl-\b-a\) . dt 1=0 dt 1=0 dC C=a (I-\b\2) (ll-abl+\b-al) 2.9.22. If Z1' Z2 are roots of the equation Z2+(p-W)Z-Wp-l.= 0, Z1Z2 = -Wp-1; moreover,
then
Z1(Z1 +p)(I+pz1)-1 = z2(z2+p)(I+pz2)-1 = Wp-1.
Hence Z2 = -(z1+p)(I+pz1)-1 and this means that IZ11 < 1 implies \z21 > 1, i.e. the mapping is univalent in K(O; 1), as well as in K( 00; 1). If Iz\ = 1, then 111'\ = p. The end points of circular slit on' C(O; p) are found by solving the equation dw/dz = O. 2.9.23. Put P = y2 in Exercise 2.9.22; w = z(z+ y2)(1+z }lilt1.
2.9.24. w = Rz(z+p) (1+pZ)-1, p = cosect!X.
CHAPTER 3
Complex Integration In this chapter I denotes the integral to be evaluated
3.1.1. (i) z(t)
=
(l+i)t, rez(t)
t, t
=
[0, 1];
E
1
I
~ t(1+i)dt
=
=
-}(1+i);
u
(ii) z(O) = reiD, rez(t) = rcosO, 0 E [-1t', 1t']; 1
0 implies re t; > 0 for any t; E convr, and this contradicts t; = O.
3.1.15. K(O; 1) is a convex domain; reF'(z) 3.1.16. The half-plane rez for x < 0 (z = x+iy).
=
nre(I+z"-l)
> 0 for z
< 0 is a convex domain; reF'(z)
=
E
K(O; I).
I+excosy
>0
3.1.17. A similar proof as in Exercise 3.1.16.
If dist(zo; y) = 15
3.1.1S. Con tin u it y. we have: IF(z)-F(zo) 1= Iz-zo I
> 0, then for any
z
E
K(zo; -}b)
II ~ (t;-zr1(t;- ZO)-1ep(t;)dt; I< Iz-zo 12MLb-2, 1
where M = sup Iep(t;) I on y and L is th~ length of y. This implies continuity. Differentiability. We have: F(z)-F(zo) Z-Zo
where ep1(t;)
=
= \'
ep(t;)
J (t; -z)(t;-zo) 1
dt; = \' ep1(!;) dt; = F1(z), J t;-z . 1
(t;-zo)-1ep(t;). Since F1 is continuous at any point not on y,
the limit lim F1(z) exists and is equal to ~ (t;-zo)-1ep1(t;)dt;. Z~Zo
y
3.1.19. We have: h'(t) = [z(t)-ar 1z'(t) at continuity points of z'(t). Consequently, u'(t) = 0 except for a finite set of values t, and from the continuity of u(t) it follows that u(t) = const = U(IX) = z(IX)-a = exp( -h({J) [z(fJ)-a] , hence, by Z(IX) = z({J), it follows that h({J) = 2k-ri.
3. COMPLEX INTEGRATION
171
3.1.20. n(y; a) = const as a continuous function of a ED (Ex. 3.1.18) whose all values are integers (Ex. 3.1.19). in
3.1.21. C"'-K is a connected set disjoint with y hence n(y; a) = const = C",-K, since 00 E C"'-K. 3.1.22. n(y; a)
=
n(y; (0)
=
°
0.
3.1.23. If zo = re i8 and r < r(O) , the segment [0, zo] does not contain any points of y, hence 21
(a) = k!fk(a).
3.3.12. lim(z-a)-m/(z) = (ml)-1j'Cm)(a) (cf. Ex. 3.3.11). 3.3.13. lim(z-a)"'/(zr 1 = m![J<m)(a)r1 #: O. Z-+Q
3.3.14. By Exercise 3.3.10 I(z) = Pm(z)/Pn(z), where Pm, P n are polynomials of order m and n, resp.; if e.g. m > n, then I has q = m zeros, n finite poles and z = 00 is a pole of order m-n. 3.3.15. If 0
< r < R and
M = suplf(rel 9l) I then rp
J,.(z)
=
(27ti)-1
~
(C-a)-n(C-z)-1/(C) dC,
Iz-al
loga, there is only one simple pole inside Rn with residue a- 1Ioga.
or
1= 27ta- 1Iog[(1+a)/a], If 0
< "
~
-n
a
< 1, then
l/a
a>
> 1 and this gives
xsinx 1+ a 2 - 2a 1 cosx -dx
=
27talog(1+a),
i.e.
1.
3. COMPLEX INTEGRATION
181
3.6.7. We have 2",
~
e% z-n- I dz = i ~ exp(cosO) exp[i(sinO-nO)] dO
C(O;I)
0
2",
2",
=
~ exp(cosO) sin(nO-sinO) dO+i ~ exp(cosO) cos(nO-sinO) dO
=
27tijn!
o
0
since the integrand has a pole of order n+ 1 at the origin, the residue being equal to 1 jn!. Compare now the real and imaginary parts of both sides. 3.7.2. The inequality sinO >
~0 7t
is a consequence of convexitY'of sine in
(0, +7t); ",
",/2
~ exp(-AsinO)dO
",/2
2 ~ exp(-AsinO)dO
=
o
< 2 ~ exp(-2AOj7t)dO
a, then ' Rl'
~ e imx (a 2+x2)-2dx+ ~ e im%(a 2+z 2)-2dz -R ",
I ~ I
1. k=O
After describing the point exp(2k7ti/n) the values on the opposite side of u corresponding slit are multiplied by exp(27ti/n); moreover, the integ~als over
3. COMPLEX INTEGRATION
187
both edges of any slit are equal to [1-exp(27ti/n)]I. Hence n[l-exp(27ti/n)]I = 27tires(00;f) = -27tiexp(i7t/n).
3.9.1. We have: I-zl = I > IF(z)l; hence by Rouche's theorem both equations: -z = 0, -z+F(z) = 0 have the same number of roots in K(O; I), i.e. exactly one. 3.9.2. Consider the variation of argP(z) as z describes in the positive sense the boundary y of K(O; R) n (+, +): Ll[O.R]argP(z) = 0; Ll[IR.O]argP(z) = -arg[l+i(y8+5)-1(-3y 2+7y)] = O(R- S ); on the circular arc LlargP(z) = Ll arg[z8(1 + o(R- S ))] = 47t+o(I). Hence LlyargP(z) = 47t+o(l) = 47t for all R sufficiently large. Now, n(r,O) = (27t)-1Ll yargP(z) = 2.
3.9.3. P has no roots on coordinate axes; it is real on the real axis and attains a positive minimum at x = -!-; on the other hand, reP(iy) = y4+10 ~ 10. Suppose now that z describes the boundary y of K(O; R)n (+; +). We have: Ll[O.R]argP(z) = 0 because P is real and positive on [0, R]; on the arc of C(O; R) LlargP(z) = Llargz4+Llarg[l+z-4(2z 3 -2z+10)] = 4.-!-7t+o(l) = 27t+o(1) as R-+ +00, moreover, on [iR,O] the initial value of argP(z) is arg[I-2i(R4+ + 10)","1 (R+R 3 )] = 0(1) as R -+ + 00, and all the time reP(iy) ~ 10, the final value argP(z) being 0 which means that Ll[iR.O]argP(z) = 0(1). Finally, Llyargp(z) = 27t+o(1) for large R and being a multiple of 27t it is equal to 27t for all R sufficiently large. A similar reasoning can be made for remaining quadrants. 3.9.4. P(iy) = (_I)n[y2n_ia 2y2n-1]+b 2, hence w
=
P(iy)
=
y2n+b 2_ia2y 2n-1
for even n.
Suppose that z ,describes y = a[K(O; R)n {z: rez > O}] and R is large. For z = iy moving on [iR, -iR] the point P(iy) describes an arc with end points W1 = R2n+b 2_ia2R2n-1, W2 = R 2n+b2+ia2R 2n-1 situated in the half-plane {w: rew > b2}, hence Ll[iR._IR]argP(z)
as R
-+
= argw2-argw1 = 2 arc cot a- 2(R+b 2R-2n+1) = 0(1)
+ 00. If z is moving· on the arc of C(O; R), then Ll argp(z) = Ll argz2n +Ll arg(1 +a2z-1+b2z-2n) = 27tn+o(I).
Consequently, Llyargp(z) = 27tn for all R sufficiently large. If n is odd, then AargP(z) = 27tn+o(l) on the arc of C(O; R). If z = iy moves on [iR, -iR], then w = P(iy) = b2_R2n+ia2R2n-1 describes an arc 1 with end points W1 b 2- R2n+ la 2R 2n-1, W2 = b2- R2n_ ia 2R 2n-1, the equations of 1 being: u b2_y2n, V = a ZyZII-l. We have: dargw/dy = a2Iwl-2(b 2+y4n-Z) > 0 which
r
r
SOLUTIONS
188
means that argP(iy) decreases as z moves on [iR, -iR]; if y
+ iv E ( - ; +); WE
if 0 (+; -); if y
l-z+161 on C(O; 2), hence the polynomials zS, zS-z+16 have exactly 5 roots in K(O; 2). Moreover, l-z+161 > 14 > IzsJ on C(O; 1), hence the polynomials -z+16, zS-z+16 have no roots in K(O; 1), and consequently, all the roots are situated in {z: 1 < IzJ < 2}. If z moves on [iR, -iR], then P(iy) = 16+iy(y4-1), i.e. LI[iR,_iR]argP(z) = -7t+o(I), whereas on the right-hand half of C(O; R) LlargP(z)
=
5L1argz+o(1) = 57t+o(1).
Hence LI argP(z) = 47t+o(l) = 4n for all R sufficiently large which means that there are two roots of positive real part.
u
FIG. 3
3.9.7. If z = iy moves on [iR, -iR], then W = u+iv = P(iy) = 1-2y2+ +iy(y2-1)(y2_4) describes an arc F1 which can be outlined by means of the following table indicating the signs of u, v as depending on y (Fig. 3): y
u v
R
2
-
+
o
0
o +
0 _2- 1 / 2 +.i\ 0 0
-1 0
-2
+
-R
0
We have: argw1 = -t7t+o(I), argw2 = --t7t+o(1) and Llr,argw = -7t-J-o(l) as R -+ +00; moreover, Llargw = 57t+o(1) on the right-hand half of C(O; R), hence n(r,O) = 2--1-0(1) = 2 for R sufficiently large.
3. COMPLEX INTEGRATION
189
3.9.8. We find the variation of argP(z) as z moves on the contour y consisting of [- R, R], two arcs of C(O; R) contained in D and the segment IR = hi R2 1+ +i, _yR2 I+i]. We have: P(x) > 0 on the real axis and LlargP(z) = 0(1) on both circular arcs, henc"e LI argP(z) = 0(1) on all three arcs. If z = x+i, then P(z) = x 4-6x 2 +3x+4+i(4x 3 -4x+3). The. polynomial 4x 3 -4x+3 has a unique root Xo E (-2, -1) and x~-6x5+3xo+4 < 0 which follows from the inequality: (x 2_2)2 < x(2x-3), x E [-2, -1]. Hence, if R is large and z moves on fR' then at the beginning argP(z) = 0(1) and imP(z) > 0, afterwards P(z) meets the negative real axis for z = xo+i and then for x decreasing we have imP(z) < 0 and at the end point again argp(z) = 0(1). This means ,that LlI~argP(z) = 27t+o(I) and also Llyargp(z) = 27t+o(I) = 27t for all sufficiently large R. 3.9.9. [iz 3 [ = 27/8 < 4 < 81/16-1 ~ [z4+1[ on C(O; f) and by Rouche's theorem both polynomials Z4 + 1, Z4 + iz 3 + 1 have four roots in K(O ; f). The number of roots in (+; +) can be found similarly as in Exercise 3.9.2.
3.9.10. (i) [-5z+II > 3 > IZ41 on C(O; 1), hence q = 1; (ii) 1-4z5 -II > [Z8+ Z 2[ on C(O; 1), hence q = 5. 3.9.11. laz"[ = a > e > eX = [-e"[, z = x+iy E C(O; 1); hence both equations: azn = 0, azn-e'Z = 0 have n solutions in K(O; 1) by Rouche's theorem.
3.9.12. If rez > -f, then Iz+21I1z+I[ > 1, whereas I-e-'Z[ = e-~ ~ 1 for z = x+iy in the right half"plane. Hence on the boundary of K(O; R) n {z: rez > O} we have Iz+2I/iz+I[ > II-e-'ZI. Since (z+2)(z+1)-1 has no roots in the right half"plane, so does (z+2)(Z+1)-1_ e-'Z. 3.9.13. Compare Il-zl, I-e-'Z[ on the boundary of K(O; R)n {z: rez > O} for R> l+1. If z = iy E [-iR, iR], then [l-iYI;;?; l > [-e-:.'[ = 1; if [zl = R, rez;;?; 0, then [l-zl;;?; [zl-l = R-l > 1 ;;?; [-e-'Z[. Now apply Rouch6's theorem. 3.9.14. The case a = 0 is trivial. If [a[-l ~ 2n and Zl, ... , Zn are roots of the ro1ynomial, then [ZlZ2 ... zn[ = la[l ~ 2n, hence IZkl ~ 2 for some k. If la[-l > 2n, then [al- 1 > (2+ [a[-l > (2+ Ibl on C(O; 1); by Rouch6's theorem withf= F,g= -bthe equationsF = O,F-b = 0 have n roots in K(O; 1). 3.9.18. Apply Rouche's theorem with f= F, g = -F(O). 3.9.19. Suppose that r < R1 < R2 < R. By the argument principle n('YR 2 , a) -n('YR1' a) = Na-P, whereNa is the number of roots of the equationf(z)-a = 0 and P is the number of poles off in the annulus {z: R1 < Izl < R 2}. 'We have Na = P = 0 by our assumptions. 3.9.20. Suppose that D~ = {z: loglf(z) I ~ I, and n is sufficiently large, then lun(z)1 ::(; 2/R"+1. This proves a.u. convergence. Now, un(z) = (l_zn)-1_(I_zn+1)-1+ n
+zun(z) and hence putting sn(z)
L
=
Uk(Z) we
obtain sn(z) = (l-Z)-1-
k=l
< 1 and n ~ +00, then s(z) = limsn(z) =
-(I- z n+2)-1+ zsn(z).1f Izl
(I-Z)-1-
n
-I+zs(z); if Izl
> I, then
s(z) = (l-Z)-1+ ZS(Z) and s(z) can be evaluated in
each case. 4.1.3. Isinz[2 = sin2x+sinh 2y (z = x+iy), hence [yl::(; Olog3, 0::(; 0 implies: 3-n [sinnzl < 3-n(l+exp(nlyl) ::(; 3-n+3-(1-9)n
which proves the uniform convergence in the strip limzl
r. n3n=1 00
1'(0) = 4.1.4. Suppose that fez)
= -~ 2m
rJ
Z E
t·
= f(z)-f(O) = _1_. 2m
C(O; '1)
rJ
zf(i;,) di;, i;,(i;,-z)
C(O; '1)
hence [f(z) I "-( M(r)lzl, where M(r) L.I~zn) has a majorant
=
< Olog3;
= -}(l+r) and 0 < r < 1. We have
K(O; r), r1
f(i;,) di;, i;,-z
n
< 1,
L
=
2(1-r)-1suplf(rleil1)l. This implies that o M(r}r" in K(O; r). 191
SOLUTIONS
192
4.1.5. im(z+n)-1 = -y[(x+n)2+ y 2rl,
I
hence the series
(_I)n+ lim(z+n)-1
is absolutely and uniformly convergent in K(zo; r), since it has a convergent majorant IAn- 2 ; re(z+n)-1 = (x+n)[(x+n)2+ y 2]-1 and from the fact that t(t 2+y2)-1 strictly decreases for t ;> Iyl it follows by Leibniz test of convergence that -1)n+lre (z+n)-1 converges. Now, the rest in an alternating series is bounded in absolute value by the first term omitted, i.e. by Ix+n+ 11 [(x+n+ 1)2+ +y2r1 ~ Ix+n+ll- 1 ~ (n+l-lx ol-r)-1 for large n which proves the uniform
I(
convergence in K(zo; r). 4,1.6. The points -n (n =
=
1,2, ... ) are simple poles of
f and res(-n;f)
(-I)n+\ hence
~f(z)dz
= 27ti
2: (-I)k+ In(y,-k) ,
y
the sum being finite. 4.1.7. We have T(n)
~
n and this implies that the power series on the righthand side is convergent (and also a.u. convergent in K(O; 1). If 0 < r < 1 and z E K(O; r) then
Izn(l-zn)-11
~
r n(1-r)-1
and this proves the a.u. convergence of the series on the left-hand side. Moreover, both sums represent analytic functions in K(O; 1). The identity of both functions OC!
follows from the fact that the double series
OC!
LL
•
zmn is absolutely conver-
n=lm=l gent and its sum does not depend on order of summation.
4.1.8. Differentiate k times the identity (1-Z)-1 = l+z+z 2+ ... 4.1.9. (i) K(O; 1), K(oo; 1);
(ii) the real axis; (iii) C""N 1, where N1 is the set of negative integers; (iv) the annulus {z: q < Izl < q-1}; no domain of analyticity does exist in
(ii).
n [1 + (R/n)2] exists and m
4.1.10. The finite limit H(R)
=
lim m
consequently
n=1
the given series has a convergent majorant
I
R2 H(R) (n+ It2 in K(O; R).
4.1.11. The series is a.u. convergent in C""N where N is the set of all integers, hence it. can be differentiated term by term; the sum to be found is equal to 7t 3 COS 7tz(sin 7tZ)-3. .
4. SEQUENCES AND SERIES
193
4.1.12. Suppose that Go is a subdomain of G such that Go c G and aGo is a cycle consisting of a finite number of contours. If w does not lie on I(aG o), the same is true for I" (aGo) for all n ;;:: N, hence
~ (J" (z)- wtlI~ (z)dz since
I"
~ (f(z)-wtlI(z)dz = 0, 1
-+
oGo
oGo
are univalent. This proves the univalence of f, unless it is a constant.
4.2.1. If ql
=
l/al, q2 = at/a2, ... , q" =;= an_liaR, then la"I- 1/" = (lqlllq21···lqnl)1/"-+ Iql·
4.2.2. (i) {-; (ii) e; (iii) 1; (iv) min(l, lal- 1). 4.2.3. (i) For any e
>
° there exists an integer k such that
la"1 1/n < R 1 1+e,
Ib"11/" < Rzl+e
for all n;;:: k;
hence
la"b" 111" < (RIR2)-1+s(Rl1+Rzl)+e2 for all n;;:: k and consequently (ii) a" = b"(a"/b"), cf.(i); (iii) both series L a"z", L b"z" are absolutely convergent in K(O; R o), hence after multiplication and rearrangement according to increasing powers of z we obtain a convergent series for any z, Izl < Ro. 4.2.4. Take real increments of z and verify by induction that imI(k)(z) for all z E( - 15, b). Note that ak = j t. Now, the shortest distance from the origin to the points where p ceases to be analytic (z =.1) is equal 1, hence
< 1;
R=1. 4.3.13. By equating Taylor's coefficients of both sides we obtain a1 = 1, 3a 2 = 2a1, ... , (2n+l)a n+1 = 2na n, ... , hence
2n a n+1 = 2n+l an· 4.3.14. Putting z
u2 in the formula of Exercise 4.3.13 we obtain
=
(l-u2)-1f2Arctanu(l-u2)-1f2
=
(l-u2r1/2Arcsinu
_ 2 3 2·4 5 - u+ 3 u +3":5 u
+ ... ,•
integrate now both sides.
4.3.15. Rn(z) = J(z)-sn(z) = zn+1 pn(z), where Sn is a polynomial of degree at most nand Pn is analytic in K(O; R). We have: Pn(z)
z-n-1 [J(z)-sn(z)]
=
=
~
(27ti)-1
(C -zr 1Pn(C)dC
C(O; ,)
(27t0- 1
=
~
c n- 1(C-zr 1J(C) dC
C(O; r)
by Exercise 3.1.28. 4.3.16. We have
sn(z) = J(z)- Rn(z) =
~
(27ti)-1
(C-z)-lJ(C)dC-(27ti)-1
C(O; r)
=
~
(27ti)-1
~
(zg)n+ 1(C-Zr1J(C) dC
C(O; r)
c-n-1(C-Zr 1(ClI+1_ Zn+1)J(C)dC.
C(O; r)
4.3.17. If Iz-al < R, then J is analytic in K = KC-}(a+z); R--}Iz-al) and the Taylor series with center -}(z+a) has the following form:
Jm =J(-}(a+z))+(C--}(a+z))J'(-}(a+z))+-}(C--}(a+z)Y!,'(-}(a+z))+ ...
Put first C= a and then C= z and subtract both sides. 4.3.18. If an = lin+i(Jn, then
.L (lin+i(Jn) (cosnO+isinnO)r 00
n=l
n
= P(O)-HQ(O)
4. SEQUENCES AND SERIES
199
and consequently co
P(O)
=
L:(cxncosnO-PnsinnO)rn. n=O
Thus by Euler-Fourier formulas 2n
2n
7tCX nr n = ~ P(O)cosnOdO,
-7t{3nrn
=
o
\
o
P(O)sinnOdO,
or 2",
7tan r n = 7t(cx n+i{3n)
=
~ P(O)e- in6 d8 o
and similarly 2n
7tan r n = ~ iQ(O)e- in6 dO. o 2n
4.3.19. lanl ~ (7tr n)-1 ~ P(O)dO by Exercise 4.3.18 and the equation P(O) o 2n
=
(27t)-1 ~ P(O)dO (cf. Ex. 3.2.5) yields lanl ~ 2r-n. Since r E (0, 1) can be aro
bitrary, we have lanl
~ '2.
4.3.20. IJ(z) I ~ I+la11· Izl+la21·lzI2+ ... ~ I+2Izl/(1-lzj) = (1+lzl)/(1-lzl). In order to obtain a lower estimate, note that I/J is analytic in K(O; 1) and satisfies the assumptions of Exercise 4.3.19. Equality holds for real z and J(z) = (1+z)/(I-z). 4.4.1. (i) divergent everywhere on C(O; 1); (ii) divergent only at z = 1; (iii) absolutely convergent on C(O; 1); (iv) conditionally convergent everywhere on C(O; 1) except at z = -1 and
z = +(I=fi}l3). 00
4.4.2. an = (-I)n; lim
L
r-----+1- n=O
anrn = lim (1+r)-1 = r~l-
i.
4.4.3. Zn = ei6 (1-r n expicx n), where Icxnl ~ +7t-b, b
IZnl = (I-2rncoscxn+r,;)1/2 = I-rncoscxn+O(r,;),
> 0; we have: rn = le"6_ zn l.
Note that a necessary and sufficient condition for Zn to be situated inside a Stolz angle is: (cOSCXn)-l = rnlrncoscxn = 0(1). 4.4.4. We may assume e l6 = 1, hence the radius of convergence of l:ak Zk is ~ 1.
SOLUTIONS
200
By Exercise 4.2.3 (iii) the esries L8kZk is convergent in K(O; 1), moreover
00
Thus
L
akz! can be considered as a transform of a convergent sequence {8n } by
k=O
means of a Toeplitz matrix.whose nth row has the following form: l-zn' zn(1-zn), z;(1-zn), ... Now, cf. Exercise 1.1.37 and Exercise 4.4.3. 4.4.5. (i), (ii) The series Z_-}Z2+-tZ3- '" whose sum in K(O; 1) is equal to Log(l+z), is also convergent for all Z E C(O; 1) except for z = -1, which follows from Abel's test of convergence. If z = cosO+isinO and 101 < 'It, then by Abel's limit theorem:
L (-I)n+ln-l(cosnO+isinnO). 00
Log (1 +e1fl) = log (2cos 0/2)+;0/2 =
n=l
Separate now real and imaginary parts. 4.4.6. The seriesz+-tz3+tzs+ .,. is convergent on C(O; 1) except at z = =fl and its sum is equal to tLog[(I+z)/(1-z)] in K(O; 1). By Abel's limit theorem:
tLog[(1+ei fl)/(1-e1fl)] = tlogl cot (0/2) I+i-i'lt = eifl+-te3ifl++eSifl+ ...
= cosO+-tcos30+tcos50+ .. , +i(sinO+-tsin30+ ++sin50+ ... ),
0
lim Ia_n 11/n = r and divergent
K(O; 1) and the principal part is convergent in
4.5.2.
4. SEQUENCES AND SERIES
201
for Izl < r, whereas the regular part is convergent for Izl < (limlanI1/n)-1 = R and divergent .for Izl > R. Hence the annulus of convergence is not empty iff r max(lal, IbJ), with
n= 1
+00
4.5.14. The sum of the series
L
n=
(z-n)-2 is an analyticfunctionh in C",N.
-00
Suppose F is a compact set such that F (IN = 0. If bn
=
inf 11- z /nl then bn
-+
1
ZEF
as n -+ =j=oo and consequently Iz=j=nl- 2 < +n- 2 for all n sufficiently large which implies a.u. convergence and analyticity of h. Obviously z = n is a pole of second order with principal part (z-n)-2. If t = z-n, then 7t 2 cosec 2 7tz- (z-n)_2 = 7t 2 cosec 2 7tt- t- 2 = [7t 2t 2_(7tt-(7tt)3/3!+ ... )][t 2(7tt-(7tt)3/3!+ .. Yr 1 = 0(1) as t -+ 0 which means that 7t 2 /sin 2 7tz- h(z) has a removable singularity at z = n .. 4.5.15. Evidently both h(z) and 7t 2 /sin 2 7tz are periodic with period 1. Since g(z) = 7t 2/sin 27tz-h(z) has removable singularites at zEN, and has period 1, it is bounded in the strip limzl < 1. Suppose now that limzl ;;: 1 and 0 < rez < 1. We have: 00 00 1 00 00 2 2 In-zl- = (x-n)2+y2 .< 2 (n +y2)-1 < 2 (n 2 +I)-1.
nboo
nboo
b
k
Hence h is bounded in {z: limzl:;;:' I}. The same is tr~e for 7t 2 /sin 2 7tz and consequently g is bounded in C, i.e. g = const = 0 which follows from Exercise 4.5.14. 4.5.16. The series is a.u. convergent, hence it can be differentiated term by term and this gives the equality of Exercise 4.5.15. Note that 7tcot7tZ-Z- 1 -+ 0 asz-+O. 4.5.17. (i) 2z/(z2-n 2) = (-2z/n 2) (1 + (z/n) 2 +(z/n)4+ ... ) and consequently 7tcot7tZ = z-1-2s1Z-2s2Z3_2s3ZS- ... Hence (27t)2kBk= 2(2 k)!Sk; (ii) 3z- 1 +2z- 3+2z- s + ... -2(sl-I)z-2(s2-1)z3-2(s3-I)zs- ...
4. SEQUENCES AND SERIES
203
4.6.1. We have Icot7tzl ~ M for each Z E frQN and each N, where M does not depend on' N, (cf. Ex. 2.7.3, 2.7.7). Hence
I ~ J(z)cot7tZdZI ~ M oQ N
~ IJ(z)lldzl ~ 8MICNIIJ(CN)I, oQ N
where sup IJ(z) 1= IJ(CN) I· Hence ~ J(z) cot 7tzdz -+ zefrQN
oQN
°asN
-+
+
00.
If a1, ... , am
are inside QN, then
~ J(z)7tcot7tzdz oQN
= 27ti {
N
m
n=-N
k=1
L res[n; 7tJ(z) cot 1tz]+ L res [ak; 7tJ(z) cot 7tz]} ° -+
as N
-+
+00.
Note that res[n; 7tJ(z)cot7tz] = limJ(z)coS7tz(7tz-7tn)/(sin7tz-sin7tn) =J(n). z->-n
(z2+ z+1r 1 satisfies the assumptions of Exercise 4.6.1, hence s = -7t(b1cot7tal+b2cot7ta2) where a1 = -(I+q/3)/2, a2 = a1 are poles of J and b1 = res[al;!] = i/]l3,b 2 = res[a2;J] = -i/]l3. Hence 4.6.2. The function J(z)
s
=
3- 1 / 27ti[cot7t(I+i]l3)/2-cot7t(I-i]l3)/2] = -3- 1 / 227titani7t ]13/2 = 27t3-..1 / 2tanh 7t ]13/2,
=
because cot ( ; +cx) = -tancx. 4.6.3. (i) If a =F 0, =j=i, =f2i, ... thenJ(z) = (Z2+ a2)-1 satisfies the assumptions of Exercise 4.6.1 and therefore
L 00
(n 2 +a2 )-1
= -(2ai)-1 7t [cot 7tai-cot(-7tai)] = a- l 7tcoth7ta;
11==-00
(ii) if a4 =F 0, -14, _24, ... thenJ(z) = (Z4+ a4)-1 satisfies the assumptions of Exercise 4.6.1; it has 4 simple poles a. = 2- 1/ 2 a(=fl =fi) with residues b. = -{-a- 3 a. and therefore 4
~
L n~-~
(n 4 +a4 )-1
=
-}7ta- 4
L a. cot 7ta. '=1
= 2- 3 /2a-3 7t [(1 +i)cot 7ta(I+i)/~/2+ (l-i)cot 7ta(I-i)/t!2]
= 7ta-- 3 2- 1 / 2 (sin 7ta]l2+sinh 7ta~2)/(cosh 7taJl2-cos7taV2) hy Exercise 2.7.16 and the identity zcotz ,,- -zcot(-z);
SOLUTIONS
204
4
L 00
2s =
-i- 7t
n 2J(n 4+a4) =
n==-oo
a;lcot 7tay
v=1
where a y are the same as in (ii); hence by a1a2 2s
L
_a 2 we obtain:
=
(2a2rl7t(a2cot7ta1 +a1 cot7ta 2) = r 3 / 2a- 17t[(1+i)cot( -1 +i)a7tjV2+( -1+i)cot(1 +i)a7tjV2] =
and the result follows similarly as in (ii); (iv) if a i= ni Y (v = 1, ... ,4; n = 0,1,2, ... ), then
L 00
(n 4_a4)-1 =
-i- a- 47t
4
L
a y cot7ta y ,
\1=1
n=-oo
where a y = ai Y , v = 1, ... 4; --}a-47t(a1 cot 7ta1 +a 2cot 7ta 2)
=
--}a- 3 7t(cot7ta+icot7tai)
=
-ta-37t(cot 7ta+coth 7ta).
4.6.4. The function fez) = (z-a)-2 has a double pole· at z = a; moreover, res[a; 7t(z-ar 2cot7tz] = --7t 2 jsin2 7ta, hence by Exercise 4.6.1
I
00
(n-a)-2 = 7t 2 jsin2 7ta.
n=-oo
4.6.5. fez) = (z-a)-l(z-b)-l satisfies the assumptions of Exercise 4.6.1. We have: res[a; 7tf(z)cot7tz] = 7t(a-b)-lcot7ta and consequently s = -7t(a-b)-l X X
(cot7ta-cot7tb).
4.6.6. Putting a = i, b 2s+1
=
-i in Exercise 4.6.5 we obtain:
=
7ticot7ti
7tcoth7t
=
or
s
=
-}7tcoth7t--}.
4.6.7. If Z E frQN' then Isin 7tzl- 1 ~ 1 for each N, hence
I~
(sin 7tz)-lf(z)dz/
~ ~
oQ N
If(z)lldzl
~ 81 CNf(CN) I =
0(1)
oQ N
as N ~ +00 (CN being the point yielding the maximum of If I on frQN). If all a, are inside QN, then \ (sin 7tz)-l7tf(z)dz = 27ti{
". oQ N as N
~
~ res [n; ~(Z)] + ~ res [ak; ~f(Z)]} L.J SIll 7tZ L.J SIll 7tZ k=l
n=-N
+00. Note that res[n;
~f(Z)l
Sill 7tZ
=
fen)
cosn7t
=
(-I)nf(n).
=
0(1)
4. SEQUENCES AND SERIES
205
4.6.8. E.g. (iv). The function J(z) = Z2 /(Z4 +a4 ) satisfies the assumptions of Exercise 4.6.7. It has 4 simple poles ak = 2- 1/2 a(=j=I=fi), k = I, ... ,4 with residues (4ak)-1. Hence 4
1 "'., ( . )-1 = - 21 7t [alslll ( . 7ta1 )-1 + ( . )-1] , S = - 4 7t L...J ak Slll7ta k a2S1117ta2 k=l where a1, a2 are situated in the upper half-plane. Now, a l a 2 = -a 2 and therefore s
(2a2r17t(a2 sin 7ta2 +a1 sin 7ta1)J(sin 7tal sin 7ta2)
=
and the result follows. 4.6.9. We have: Isin az/sin 7tZ12
=
(sin 2ax+sinh 2ay)(sin 27tx+sinh 27ty)-1,
x+iy; therefore Isinaz/sin 7tzl ~ 1 on vertical sides since sin 27tx = 1 sin 27ta, sinh27tY ~ sinh 27ta. Moreover, Isinaz/sin 7tzl ~ 0 uniformly as Iyl ~ +00. Hence ~ J(z)dz ~ 0 as N ~ +00 (f(z) = 7tsinaz(z3sin 7tZ)-1). Now,
z
=
~
iJQN
~
=
27ti
iJQN
f
res[n;f],
res[O;j] = a(7t 2-a 2)J6,
res[n;j] = (-I)nn- 3sinna,
n=-N
n = =fl, =f2, ... and consequently N
a(7t2-a2)/6+2L(-I)nn-3sinna~ 0 n=1 (ii) put a = 7t/2 in (i).
as
N~ +00;
4.6.10. Similarly as in Exercise 4.6.9. we verify that Icosaz/sin 7tzl ~ 1 on v~r tical sides of QN, whereas Icosaz/sin 7tzl ~ 0 uniformly as limzl ~ + 00. Hence IN ~ 0 as N ~ +00. IfJ(z) = 7tcosazcosec7tz/(x 2_z 2), then res(n;f) = (-I)n X x cosna/(x 2-n 2) , res(x;j) = -7tcosax/(2xsin7tx) = res(-x;j), and consequently for all N sufficiently large R
IN
=
27ti{-7tCosax/(xsin7tx) +
N
L
(-I)ncosna/(x2-n2)} ~ O.
n=-N
4.6.11. If J(z) = z(sinh 7tazsin 7tZ)-l, then for real a "" 0 and z on the vertical side of RN we have: IzJ(z) I ~ (N+i-)2(I+a-2)/sinh7tlal(N+i-) --+ 0
as
N ~ +00;
for z on the horizontal sides of RN we have: IzJ(z) I < (N+t)2(l+a-2)/sinh7tlal-1(N+i-) ~ 0
I'his implies IN
~
O.
as
N ~ +00.
SOLUTIONS
206
(i) There are following singularities of f inside RN: simple poles -N, -N+l, ... , -1,0,1, ... , N, with residues (-I)"n/(7tsinh 7tan) , n i= 0, as well as simple poles -Ni/a, ... , -i/a, i/a, ... , Ni/a with residues (-I)mm X X [7ta 2sinh(7tm/a)tl; res(O;f) = 1/7t 2a. Hence IV
IN
=
27ti [(7t 2 atl+2
I
(-I)mm(7tsinh 7tamt 1
m-l N
+22.".:(-I)mm(7ta2sinh(7tm/a))-1]~0
as
N~+oo.
m=l
Note that each sum has a finite limit. (ii) Put a = 1 in (i).
°
4.6.12. Suppose that h,O are arbitrary real numbers satisfying: h > 1, < 0 < 7t/2 and D(h, 0) is the bounded, closed domain whose boundary consists of: 1 the segment I on the straight line rez = h such that -0 < argz < 0, 0
20 the circular arc y arising from I by inversion, 3° two segments joining the end points of I and y.
FIG. 4
Obviously a E D(h, 0) implies a- 1 each z E D(h, 0) and therefore
E
D(h, 0). If b = h-1coS 20, then rez ~ b for
1m/sinh 7tmzl ~ 1m/sinh 7tmbl ~ Am- 2 for all sufficiently large m and all Z E D(h, 0). This implies that both series or Exercise 4.6.11 (i) are uniformly convergent in D(h,O) and represent analytic functions of a in the right half-plane. By Exercise 4.6.11 (i) the difference or both sides vanishes identically for real a E [h-t, h], hence being analytic in the right half-plane, it vanishes there identically. An analogolls identity for I he
4. SEQUENCES AND SERIES
207
left half-plane is obtained by a change of sign. On the imaginary axis both series are divergent. 4.6.13. If N+iexceed
> lxi, the sum of integrals over the horizontal sides does not
"4(N+i-) [I +sinhlal(N+i-)] V2(N+i-) [sinh 7t(N+-iW t [(N+t)2_lxI 2]-t which tends to 0 as N -+ + 00. The parametrized integral over the vertical right-~
d
hand side (0
=
argz
E
[-i-7t, i-7tJ) has the form ~ + -/j
rr/4
~ + ~ , where the first -rr/4
/j
integral can be made arbitrarily small since the integrand is bounde~ independently of N and in the remaining intervals Isinazl/lsin7tzl-+ 0 uniformly as N -+ +00. This'implies that the integral over the vertical right-hand side (and also over the left-hand side) tends to zero. Hence N
res(x;f)+
I
res(n;f) -+
o.
n=1
Note that •
res(x;f) = -7tsinax/(2sin 7tX),
_
res(n,J) - (-I)
n
nsinan X
2
-n
2·
4.6.14. (i) Put a = 7t/2, x = 2z in the formula of Exercise 4.6.13; (ii) replace Z by iz in (i). 4.6.15. cot7tZ and coth7tz are uniformly bounded on frQN' where QN is the square of Exercise 4.6.1. Hence
~ f(z)dz -+ 0
as
N -+ +00.
iJQN
The integrand has following singularities inside QN: 2N simple poles on the real axis, 2Nsimple poles on the imaginary axis with res(n;f) = res(ni;f) = n- 7 coth 7tn and a pole of order 9 at the origin. In order to evaluate res (0 ; f), we find the Laurent series expansion of the integrand near z = 0 by using the formulas: 7tzcot7tZ = ,1-Bt (27t)2z 2/2!-B2(27t)4z4/4!+ ... ,
7tzcoth7tz = 1+Bt (27t)2z 2/2!-B2(27t)4z4/4!+ ... . Hence res(O;f) = 7t-t(27t)8[(B2/4!)2_BtB3/6!-2B4/8!] = 287t7(-19)/(6!7!). ()hserve that N
4~n-7coth7tn+reS(O;f)-+O
as
N-++SO.
SOLUTIONS
208
2>n
4.7.1. Suppose F is a compact subset of G and an arbitrary convergent series with decreasing positive terms. Suppose {An} is an increasing sequence such that b
I ~ W(z, t)dtl :;:;:; Un for any b > An and any ZE F . . An OC> An+l +00 The series 2..: ~ W(z, t)dt is uniformly convergent on F, hence ~ n=1 OC>
+ 2..:
An
a
A, ~
+
a
An+l ~
n=1 An
is analytic in G.
4.7.2. Suppose F is a compact subset of the right half-plane. We have: rez ;;?: b
> 0 for all z E F; if s > 0, arbitrary, and 2d- 1 e- M < s, then
B
I ~ e-ztdtl
< s
b
for any z E F and any b, B half-plane. Evidently
>
A, hence the integral is a.u. convergent in the right
A
hence
~ e-ztdt = _Z-l e- Az +z - l ~ Z-l o
A ~ +00,
as
4.7.3. The integral is a.u. convergent in C and can be differentiated under the sign of integral. +OC>
4.7.4. Put g(z)
=
~
(x-z)-2q;(x)dx and verify that
-OC>
Ig(z)- [f(z+h)-f(z)]jhl
=
O(h)
as
h
~
0.
4.7.5. Suppose that Zo is a pole of F and Zo is not an integer. Since F(zo+n) = (zo+n-l) ... (zo+l)zoF(zo), also zo+n would be a pole of F which is impossible for F is analytic in the right half-plane. 4.7.6. Consider G(z)jF(z). 4.7.7. Put x"
=
t.
4.7.8. If z = Re iO and s = a+ir, we have:
le-zzs-11 ::s;; Ra-lexp( - RcosO); hence by Exercise 3.7.2:
I. ~ I::s;; y(R)
re/2
Ra
~
0
exp(-RcosO)dO::S;;
_}7tR
0
209
4. SEQUENCES AND SERIES
as R
~
+00
< 1). Similarly
(for a
rr./ 2
I ~ I ~ (j" ~ exp(-(jcosO)dO ~ -~:,(j" ~ 0 0
y(6)
0
0
4.7.10. Putting t = sin 2 0 we obtain the integral 1
-i ~ t
P/
2- 1 (l-t)Q/2- 1 dt =
fB(P/2, q/2)
o
which is equal to +r(p/2)r(q/2)/r(f(p+q)). 4.7.11. (i) Put p = a+l, q (ii) put p = 3/2, q = 1.
=
I-a in Exercise 4.7.10;
4.7.12. (i) +7t- 1 / 2 r 2 (+); (ii) 27t 3 / 2 [r(+W 2 • 4.7.13. If e hence
=
exp(27ti/n), then (z-I)(z-e)(Z-e 2 )
(l-e)(I-e 2 )
...
(l-en -
1) =
...
(Z-e n- 1)
lim(zn-l)/(z-l)
=
Z----7'-l
=
z"-1 and
J
n. .
n sin(k7t/n)
n-l
Now, [1-e k [
=
2sin(k7t/n) which gives the identity
=
n2 1 -". Use
k=1
now the formular(z)r(l-z) = 7t/sin7tz with z = 7t/n, 27t/n, ... , (n-l)7t/n. 4.7.14. If f is nonnegative and decreasing in (0, 1] then 1
~ f(x)dx ,,10
1-10
n- 1 fJ( I /nHI(2/n)+ ...
+f(n/n)]'(
~ 0
f(x)dx
SOLUTIONS
210 1
which implies that the improper integral ~ f(x)dx exists, iff the finite limit o
lim n- 1 [f(1ln)+f(2In)+ ... +f(n/n)] exists and both are equal. Note that logr(x) n-+<Xl
is nonnegative and decreasing in (0, 1] and use the equality of Exercise 4.7.13. 4.7.15.
We
have:
I'(a) = logr(a+1)-logr(a) = loga, = +log27t.
therefore
l(a)
= aloga-a+b; b = 1(0+)
4.8.1. If I is an arc in the z-plane through zo', III its length and 11 is the spherical image of f(l) , then p (zo,f) = lim III I / III
4.8.2. p(O,f)
=
as 1 shrinks to Zo. 2 2Ial/(1+lbI ) is unbounded.
4.8.3. Note that any normal family in which the case of a.u. dive'rgence to is excluded, is necessarily compact.
00
4.8.4. Suppose that K(zo; 2r) c: D and M is the common bound of If I in this disk. Then for z E K(zo; r) by Cauchy's formula:
If'(z) 1= i (27ti)-1
~
(C-z)-2f(C)dCi ~ 2Mr- 1 ;
C(zo;2r)
hence f' are locally uniformly bounded and form a compact family: Marty's condition shows that n(z2- n2) form a normal family, whereas their derivatives do not. 4.8.5. If H is a compact subset of D, then If(z) I ~ R for all z E H, f E§". Hence IFof(z) I ~ sup IF(w)1 for all z E H. Hence {Fo!} is a compact family. Iwl";;R
4.8.6. {az} form in Izi > 1 a normal family by Marty's condition. Consider the sequence f.(z) = F[(2n+ )z]; we have:
+
fn(7t) = exp[(2n+-})7t] ~
f.(27t) =
+00,
°
which shows that {Fo!} is not a normal family. 4.8.7. By Marty's criterion {llf} is a normal family. Hence llf. is either a.lI. convergent in D and the limit function 1/g is analytic which makes g(z) = 0 impossible, or 11fn tends a.u. to 00 which gives g = 0.
°
4.8.8. By Ex. 4.8.7, g(z)-g(a) is either never in D"",a (which means that is univalent since a is arbitrary), or g(z)-g(a) == 0, i.e. g = const.
°
4.8.9. The sequence n- 1 (e nz -1) is equal to at z hence the family considered cannot be normal.
=
°and tends to
00
~
at z ' 1/2.
4. SEQUENCES AND SERIES
211
4.8.10. Let g = logf(z)/z with f E To be this branch which is equal to 0 at the origin. Since g does not take the values =j=27ti, so {g} is normal and even compact (cf. Ex. 4.8.3), since g(O) = O. By Exercise 4.8.5 To is compact. 4.8.11. The functions gn form a normal family (gn i= =j=27ti). Since gn(O) = 0, contains a subsequence gnk a.u. convergent in K(O; 1) whose limit g is analytic and g~/O) -+ g'(O) = (3 with finite (3. Now, g~k(O) = -1/IY. nk -+ 00 which is a contradiction.
gn
4.8.12. Obviously we can assume IY. = O. Since a family is normal, iff it is locally normal, we can also assume that D = K(O; 1). Now, 10g[f(z)/f(0)] = g(z) are analytic in K(O; 1) and omit the values =j=27ti; hence {g} is normal, and also compact (Ex. 4.8.3; g(O) = 0). By Exercise 4.8.5 {f(z)/f(O)} form a compact family, as well as {f(O)/f(z)}. Hence there exists a finite mer) such that If(z) I ~ If(O)lm(r) and also If(O)1 (m(r)tl ~ If(z) I in K(O; r). From this the normality readily follows. 2~
4.8.13. (J(zo)Y
=
(27t)-1 ~ f2(zo+re'8)dO, by Cauchy's formula: hence o 2~
If(zo)1 2 ~ (27t)-t ~ If(zo+re iO )1 2dO. o
After multiplying by r and integrating W.r.t. 'r over [0, R] we obtain 7tR2If(zoW
~ .. ~~ K(zo; R)
which shows that
If(z)1 2dxdy
~ ~~ If(z) I2dx dy ~ M D
CHAPTER 5
Meromorphic and Entire Functions 5.1.1. In what follows H denotes an arbitrary entire function. <Xl
(i) F(z)
=
L z2[n(z-n)rl;
H(z)+
"=1 00
(ii) F(z)
=
L
H(z)+
z/(z-a");
"=1 00
L Z2 /[n(z- yn)];
(iii) F(z) = H(Z)+Z-1+
"=1
(iv) we have n(z-n)-2 = n- 1[1+2(z/n)+3(z/n)2+ ... ], Izl < n, and smce In(z-n)-2- n- 11~ (n- )/71)-2 in K(O; yn), therefore
L z(2n-z)/[n(z-n)2]; 00
F(z)
H(z)+
=
"=1 00
(v) F(z)
=
H(z)+
L
[n 2(z-n)-2-1-2z/n+(z-n)-1+1/n];
"= 1 00
(vi) F(z)
=
H(Z)+Z2
L
00
[n(z-n)]-1 +Z2
"=1
L
[(z+n)r 1
"=1
00
L
(vii) F(z)
=
[n(z2-n 2)r1; "=1 H(z)+7tcot7tZ (cf. Ex. 4.5.16);
(viii) F(z)
=
H(Z)+Z-1_ Z
=
H(z)+2z 3
00
L [n(z+n)]-1. "=1
00
5.1.2. F(z)
H(z)+
L
(-l)"/(z+n). "=1 5.1.3. (Z-W)-1 is the principal part at a pole wand =
I(Z-W)-1+ W-1+ZW-21 ~ 21z1 2 1wl- 3
for
Izl (r)
=
2nlogr-2nlog1t'-210g[t·
f .. · (n-t)],
(n- t)1t' ~ r ~ (n+ t) 1t'; now subtract both expressions for if> for cosine and
sine. 5.3.5. Obviously logx = log+x-Iog+(llx). Since the zeros of of 1If, we have only to show that r
10g[r Ilb l b2 O
...
bol]
=
~ t-ln(t.!)dt. o
r
~ t-lnk(t)dt o 00
Observe now that
I
k=l
nk(t) = n (t,f).
= log[rllbkl].
f are poles
5. MEROMORPHIC AND ENTIRE FUNCTIONS
217
5.3.6. f is analytic and 1= 0 at the origin and moreover n(r,!) = nCr, llf). 5.3.7. n(r,f)
= 2[r/7t+1/2]; nCr, 1If) = 2[r/7t].
5.3.8. Since n(r,f) == 0, we have 1 O. Therefore
Alunl2+un ~ Log(1+un) ~ Blunl2+un =
for all
E
K(O; r) and suffi-
n ~ no.
5.4.2. The series L Un is convergent as an alternating series, whereas L I/n is divergent (cf. Ex. 5.4.1).
L u~
5.4.3. (i) P n = PIP2 ... Pn = (n+I)/2n; (ii) P n = -;-[1 I/n(n+ 1)];
+
(iii) P 2k = 1,
P 2k + 1 = (2k+2)/(2k+ 1).
m
544 • ..
D. (1 +z2n) = n=O
1+Z+Z 2 +
'" +Z 2 m + 1 - l ~ '
5.4.5. (i) The finite plane; (ii) K(O; 1/e); (iii) the finite plane; (iv) E"'--.N1 , Nl is the set of all negative integers. 5.4.6. Put un(z) = (z-a n)/(z-a;l)-1. If 0 < r < 1, and z E K(O; r), we have lun(z)1 ~ (1-la nI2)/(1-r) = A.n. Since. LAn < + 00, the product (1 + Un (z)) represents a function F analytic in K(O; 1). If z 1= an, then all the factors are different from 0 and consequently 'F(z) 1= 0 which follows from the definition of convergence for infinite products. Moreover,
n
l(z-an)/(z-a;l)1
=
la nIICz-a n)/(1-anz)1 < lanl < 1
K(O; 1) and this implies W(z) I < 1 in K(O; I). 5.4.7. We have: 1(1-z)e -11 = IzI 21(1-1/2!)+(1/2!-1/3!)z+(1/3!-]/4!)z2+ ... 1 ~ IzI 2[1-1/21+(1/2!-1/3!)+ ... ] = Izl2
for
Z E
Z
for z E K(O; I). Suppose now that z and this implies
E
K(O; R); if n > R, then Izl/n < R/n < 1
Iun(z) I = 1(1 +z/n)e-z/n-II ~ Iz 2 [/n 2 < R2/n 2. 5.4.8. We have: n
hn(z)
=
zexp[z(1+1/2+ ... +I/n-Iogn)]
Il (/+z/k)e- z/
k•
k=1
5. MEROMORPHIC AND ENTIRE FUNCTIONS
219
Note that Yn = 1+1/2+ ... +l(n-logn -+ Y = 0.5772 ... (y is the so-called Euler's constant) and use the result of Exercise 5.4.7. 5.4.9. (i) With the notation of Exercise 5.4.8 we have: hn(z+l) = z- l hn(z)(z+n+I)/n and therefore F(z+I) = zF(z); (ii) hn(l) = I+l/n-+ I, hence F(I) = I; now apply (i). 5.5.1. (i) q = 2; (ii) no finite genus exists; (iii) q = 2; (iv) q = 3; (v) q = 4; (vi) q = 3.
n co
5.5.2. (i) zexpg(z)
(I-z/n)e zfn ;
n E(z/log(l+n), n), n=l (vi) zexpg(z) n [E(z/n, 2)]n. "=1 co
(ii) zexpg(z)
mn
=
n;
co
5.5.3. If F is an entire function with zeros a1 , a 2 , ••• , an, ... (0 < Iall la21 ~ ... ) written as many times as their order shows, then F'!Fis a meromorphic function which has at an simple poles with principal parts (z-a n)-1 (for a zero of order k, we have k terms of this kind). Now, according to MittagLeffler formula, ~
{I 1[ z-a an
(an
F'( ) -~= ~ - - + - I+~+ ... + ~)mn-1]} +h'(z) <Xl
F(z)
L...J
n=1
n
an
where h is an entire function. If G is a simply connected domain such that 0, z and all an belong to C,"", G and y c G joins 0 to z, then
f F'(t;)
J
E
G
F(z)
F(Cr dt;
=
log F(O)
o
=
fIOg{(l-
n=1
~)exp[~ + ... + ~n (:nf n]}+lo gexP(h(Z)/h(O»,
where log denotes a single-valued branch of logarithm in G. The term by term integration is admissible by uniform convergence on y. 5.5.4. If bo = 0, b 1 , b 2 , ••• , bn , ••• are poles of/and h is the Weierstrass product for {b n } , then H = /h is an entire function. 5.5.5. Suppose that 0 < la 1 1~ la21
Kn ,., K(O;
~ lanl)
and
~
... ,
Mn
=
~~f..I-W'(;n)~;~a") I·
220
SOLUTIONS
Choose qn so that 2- qn M n :s;; 2- n. If N > n, then the Nth term of (A) is dominated by 2-N in Kn and this implies a.u. convergence of (A). If Z ~ an, then the only term with a removable singularity at an tends to 'Y}n' while all the remaining ternis tend to zero because they contain the factor w(z). 5.5.6. We can take sinz instead of w in Exercise 5.5.5. We have Isin 7tZI K(O; n/2) hence e.g.
< exp(n7t/2) in
F(z) =
~ (_I)n ~ n=1
(n-I)!sin7tz 7t(z-n)
(~)n2 n
5.6.1. The genus of the sequence of zeros is equal to 2, hence
IT (l-z/n)e 00
sin7tz = 7tzexpg(z)
00
zfn
= 7tzexpg(z)
"=-00
I1 (l-Z2/n2). n=1
Taking the logarithmic derivative of both sides and using Exercise 4.5.16 we see thatg=O. 5.6.2. (i) Replace z by iz in Exercise 5.6.1; (ii) use Exercise 5.6.1 and the identity: cos iz-cosz = 2sin[(1 +i)z/2]sin[(I-i)z/2]; (iii) a particular case ,)" (iv); (iv) replace z by (a-b, "''(,, in (i). 5.6.3. Cf. Exercise 5.6.1. 5.6.4. (i) (coshz7t V2-cosz7t V2) [27t 2z 2(I+z 4)t1 (cf. Ex. 5.6.2 (ii)); (ii) (sin 7tzsinh 7tz)[7t 2z 2(I-Z 4)tl; take the limits of both sides as z ~ 1. 5.6.5. (i) If 'Y} =
1- (1 + i y'3),
then (l-z2'Y}2/n 2) (l-z 2i?/n 2)
=
I+(z/n)2+(z/n)4-
and using Exercise 5.6.1. we obtain 00
I1 [1 + (z/n) 2+ (z/n)4] =
(7tz)-2sin7tz'Y}sin7tzrj;
n=1
Oi) if
T
= +(V3+i), then
I+(z/n)6 = [1+(z/n)2] [I-(Tz/n)2] [I-(Tz/n)2] and using Exercises 5.6.1, 5.6.2 (i), we obtain that the value of the product is equal to (7tzF3sinh 7tZ sin 7tTZ sin 7tTZ; (iii) (7tz)-3sin 7tZ sin 7t'Y}Z sin 7trjz.
5. MEROMORPHIC AND ENTIRE FUNCTIONS
5.6.6. (i), (ii). Put z
=
221
1 in Exercise 5.6.5 (i), (ii).
5.6.7. Integrate both sides of the formula of Exercise 5.2.8 similarly as in Exercise 5.5.3. 5.6.8. If z
C+t, then by Exercise 5.6.3 we have:
=
sin7':z
---:-:----:- 7':z(l-z) -
4cos7':C 4n°o n 2+n ( Z-Z2) I+ 2 7':(1-4C ) - -;- n=1 (n+f)2 n+n2
n (n 2+n) (n+i-)-2 which gives the desired 00
making z
--+
0 we obtain I
=
(4/7':)
n
1
result. 5.6.9. Taking
~he
logarithmic derivative of the right-hand side we obtain:
~,[( 3 -~)-2(---~~ n7':+3z n7': n7':+z
n=-oo
__ I n7':
)+(~I +_1)] z-n7': n7':
=
3cot3z-cotz
which is the logarithmic derivative of the left-hand side. The two expressions differ from each other only by a constant factor which shows to be equal 1 (put z = 7':(2). 5.6.10. Integrate both sides of the equation of Exercise 5.2.1: 00
F'(z) F(zf
=
F'(O) F(O)
~
+~ n=1
(1 1) z-a + a::
(An
=
1).
n
5.6.11. The function F(z) = sin 7':(z+a)/sin 7':a vanishes at z = -a-n, n being an integer, moreover its logarithmic derivative F'(z)/F(z) = 7': cot 7':(z+a) is uniformly bounded on squares with corners (n+t)(=f1 =fi)-a. Hence by Exercise 5.6.10: F(z)
J~l (1+ a~n)exp[-z/(a+n)]
=
exp(z7':cot7':a)
=
(1+~)expz(7':cot7':a-a-1) a
=
oo
(1+ :)expz{ 7':cot7':a-a- 1 -
X
Jl(l+
IT '(I+_z_)exp(_~+~ __ z_) a+n n n a+n
n=-oo
nt~ [(a+n)-l- n-ll} X
a_tn)exp(-z/n).
Nole that the expression {... } vanishes identically.
SOLUTIONS
222
5.6.12. F(z) = cos(7tz/4)-sin(7tz/4) = y2sin[7t(I-z)/4]; F'(z)/F(z) = - t7tcot[7t(I-z)/4] is uniformly bounded on the boundary of squares with vertices (4n+2)(=t=I=t=i)+l, moreover F(z) = 0 for z = 4n+l,
where n is an integer. Hence F(z)
=
exp(-7tz/4) (l-z)ez(l+z/3)e-z/3(I-z/5)ez/5 ...
=
(l-z) (l+z/3) (I-z/5) ...
because 1-1/3+1/5- ...
=
7t/4.
5.6.13. We have
[r(z)r(l-z)t1
=
(_Z)-1 ze YZ
=
[-zr(z)r(-z)t1 00
00
"=1
"=1
IT (I +z/n)e-z/"( -z)e- Yz IT (l-z/n)e z/"
n (I-1/2n)e 1/
=
sin7tz/7t.
00
(i)
5.6.14.
[r(-t)t1
=
-te-y/2
,
=
now,
2 ";
r(-t) (-t)
n=1
r
8
only) since loglf(re i9 ) I < 10gM(r)
for r sufficiently large. Moreover, ~ t- 1m(t)dt
°
2r
2r
m(r)log2
=
mer) ~ t- 1dt
< ~
since m(t)' is increasing. Hence mer)
1
6.1.12. Iq;(z) I < expJ.IY[lexps(x 2-y2+2ixY)1 ~ exp(sa 2+J.IYI-ey2)
if Iy[ is large enough. Making e
~
< C,
0 for a fixed z, the result folldws.
6.2.1. z-lf(z) is analytic in K(O; 1) since it has a removable singularity at the origin, hence the maximum of Iz-Y(z) I in K(O; r) does not exceed l/r and making r ~ 1 we obtain Iz-Y(z) [ ~ 1 in K(O; 1). If Iz-lf(z) [ = 1 for some z e K(O; 1), then z-Y(z) == eiCt:.
6.2.2. The function [f(z)-f(O)][I-f(O)f(z)]-l satisfies the assumptions of Exercise 6.2.1. 6.2.3. From Exercise 6.2.2 it follows that 11'(0)[ ~ l-lf(0)[2, hence 11'(0)[ ~ 1. Moreover, if 11'(0)1 = 1, thenf(O) = 0, thus f satisfies the assumptions of Exercise 6.2.1 and If(z)/z[ attains a maximum at z = 0, hence it is a constant. 6.2.4. By Schwarz's lemma for f-l(W) we have [1'(0)[-1 then fez) = eirxz by Exercise 6.2.3.
< 1;
if 11'(0)1
=
1,
6.2.5. F(z) = M[f(z)-a o][M 2-aof(z)r l satisfies the assumptions of Exer1. cise 6.2.1, hence lim IF(z)/zl z-+o
6.2.6. Put Ihe function
OJ =
0
CHAPTER 7
Analytic Continuation. Elliptic Functions 7.1.1. The former series represents log2+Log[1-t(1-z)]
Log(l+z) in
=
K(I; 2), whereas the latter series represents Log(l+z) in K(O; I)
7.1.2. Both series represent (1-zr1 in K(O; I) and K(i; K(O; l)nK(i;
y'2),
c
K(I; 2).
resp.;
y'2) =I 0.
7.1.3. The former series represents -Log(1-z) in the disk K(O; 1), while the latter one represents 'lti-Log(z-l) in K(2; 1), both disks being disjoint. The function element (-Log(l-z), {z: imz > O}) is a direct analytic continuation of both series which is easily verified for z approaching 0 through the upper half-plane. 7.1.4. Suppose that D1
11(Z)
=
=
C",(-; +), D2
=
loglzl+iargz,
--'--'It
= loglzl+iargz,
0
C",(+; -), and
< argz < t'lt,
while 12(Z)
< argz < t'lt.
12 in (+, +), while 11 =I 12 in (-, -). 7.1.5./(z) = -z-1Log(1-z), z eK(O; 1); if this element is continued along C(1; 1), then after encircling the point 1 we obtain the element
Obviously 11
=
11(Z)
= -[=f2'1ti+Log(1-z)]Z-1,
z e K(O; 1)",0
(the sign depends on the sense of encircling). 7.1.6. No radial limits at points z of C(O; 1) do exist.
=
exp(2'1tim2-n) which form a dense subset 00
7.1.7. Suppose that the point z
=
1 is a regular point of I(z)
=
) .......'
n=O
the Taylor series of I with center h:
00
2: n-O
129
(z-h)nJ 1. Obvicusly Ipn>(he iO ) I ~ pn>(h) and this implies that also the series
L (z-heioypn>(heiO)/n! 00
n=O
is convergent in K(he iO ; r) for any real O. This implies that f is analytic in K(O; h+r) which is a contradiction. 7.1.8. The functionfis analyticin H+and in H_ which is easily proved by using the M-test. Suppose that there exists a disk K(xo; r), xo real and a function F analytic in K(zo; r) and such thatf = Fin H+n K(zo; r). There exists a rational number Wk E K(zo; r) and by our assumptions lim f(Wk+iy) = F(Wk). However, Y--"O+
as
y
-+
0+
which is a contradiction. 00
7.1.9. E.g.
L 2-nz2n ; the derivative cannot be continued beyond the unit disk n=l
and the same necessarily holds for the function itself. 7.2.1. After reflections w.r.t. the real axis we obtain fez) = fez) and after reflections w.r.t. the imaginary axis we obtain f(Z) = -f(-z)· which gives f(z)+f(-z)
=
o.
7.2.2. After reflections we obtain a function meromorphic in the extended plane, i.e. a rational function. 00
7.2.3. If fez) =
L
An (z-ay, then
n:::::-oo 00
feZ)
=
L
An (z-a)n.
n:::::-oo
7.2.4. We have: z
=
a+R 2 /(Z*-a), b = a+R2 /[b*-a);
z-b = R 2 (b*-z*) [(Z*-a) (b*-a)rl
and consequently
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
231
where g is analytic in some neighborhood of b *. Observe that res[b*; ck(z*-a)\a-b*)k R- 2k (z*-b*r k] = (-l)kkck (b*_a)k+1 R- 2k . 7.2.5. Take f1
1
=
2
[f(z)+f(z)] , f2 =
zi1 [f(z)-/(z)].
7.2.6. f is a rational function with poles fez)
=
Zl'
l/z1, hence
(AZ2+Bz+C) [(Z-Z1) (I_Z1Z)r1
for Z1 i= 0 (note that f is analytic at
or fez) = Az+B+Clz for Z1 = 0;
00),
moreover, fez) =f(llz) which implies: C
=
A, B is real.
7.2.7. After reflections we obtain a function meromorphic in the extended plane which has two poles 0, 00 and maps C onto a two-sheeted w-plane; hence fez) = az+blz+c; 1'(=[=1) = 0 because the angles with vertices at these points are doubled. Moreover f(=[=I) = =[=2 which gives a = b = 1, c = O.
7.2.8. f is a rational function which has a double pole z = 1 (the angles with vertices at z = 1 are doubled), moreover, fez) == f(llz) which implies f(O) =/(00) = 0 and finally fez) = Az(1-z)-2. f(-I) = gives A = 1.
-+
7.2.9. Ai'(Z-Z1) (Z-Z2) ... (z-ze) [(I-z\z)(I-z 2z) ... (l-z/z)rl, k+1 = n, IAI = 1. 7.2.10, 11. After reflections we obtaina function f analytic and univalent in C such that 00 ~ 00, which means that fez) = az+b. 7.2.12. After suitable rotations of D and K(O; 1) round the origin we can achieve that the real axis in the w-plane is the axis of symmetry of D and/, (0) > O. Suppose now that cp =f- 1 and consider cp(!(z» = '!j!(z); thus '!j! is analytic and univalent in K(O; 1). Moreover '!j![K(O; 1)] = K(O; 1), '!j!(0) = 0, '!j!'(O) > 0, hence '!j! must be identity which implies fez) == fez), i.e. f is real on (-1, 1). 7.2.13. If the branch f2 arises from f1 by an analytic continuation along an arc I, then the arc I (with both end-points in the upper half-plane H+) intersects the real axis an even number of times. To each intersecting of the real axis there corresponds a reflection with respect to some boundary arc of D, hence /2 arises from f1 by an even number of reflections. Observe that two reflections can be replaced by one linear transformation. 7.2.14. We have: F' = (ad-bc)f'(c!+d)-2,
'·fence
and also
F"IF'
=
f" ff'-2cf'(cf+dr 1.
SOLUTIONS
232
and moreover, (F" JF') 2 =(f" J/)2 +4c2f' 2 (Cf+d)-2_4cj"(cf+ d)-1 which implies {F, z} == {t, z}. 7.3.1. Suppose (j, D) is a function element of the given global analytic function and K is a disk contained in {w: rew < O} such that z = expw is univalent in K and exp(K) cD. The function element (f0 exp, K) can be continued along any arc situated in the left half-plane and hence it determines a single-valued function F(w). Note that F(w) = fo expw, WE K, or fez) = F(logw) with suitably chosen branch of log. 7.3.2. Suppose g(zo) = Wo, Zo being arbitrary. Since g'(zo) i= 0, there exists a branch of g-1 in a disk K(wo; r) = K, say (j, K), such that f(w o) = zoo By our assumptions the element (j, K) can be continued arbitrarily in C, thus it defines a single-valued inverse function g_1 in C. "Hence g as an analytic, univalent function mapping C onto itself must be a similarity transformation.
7.3.3. Sincef(z) i= 0, we havef= exph, where h is an entire function .. Hence f' = h' exph and f'(z) i= 0 implies h'(z) 0 for all z E C. Suppose h does not take a finite value a. This means that fez) i= ea , 0 which is impossible for nonconstantfby Picard's theorem. Thus h assumes all finite values, whereas h'(z) i= 0 for all z, and consequently, h(z) = az+b (cf. Ex. 7.3.2).
*"
7.3.4. Let z = z(t), 0( ~ t ~ f3, be the equation of Y and ~ = dist(y, C ",D). Let 0( = to < t1 < ... < tn = f3 be a partition of [0(, ,8] such that for m = 1,2, ... ... , n-l botp. arcs Ym-1, Ym of Y corresponding to [t m_1, tm], [t m, tm+1] are situated inside Km = K(zm; ~), where Zm = z(tm). Put Fm(z) = ~ f(C)dC+C m , [Zm. Z ]
where Cm are such that Fm_ 1 = Fm in Km- 1 ("\ Km and Co = O. Evidently Fo admits analytic continuation along y, Fm being functions of the chain. Moreover,
~
=
Fm(zm)-F,n_1(Zm_1)'
Ym
Hence n
~
=
L ~ = FnCzn) = F(Z). m=lY m
7.3.5. Take K(zo; r)
(
~
=
KeD
and
continue
mdC, K) along Yi which gives (Fi' K(Z;
the
function
~), j = 1,2. Now Fl
element
=
F2 =F
[zo ... ]
by theorem of monodromy since Yl' Y2 are homotopic w.r.t. D. However,
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
thus both integrals have the
s~e
233
value.
7.3.6. Let y be an arc situated in D and r = /(y). If argw is a continuous branch of argument of r, then 10glf(z)l+i argf(z) is analytic and single-valued in any sufficiently small disk K(z; rz ), z E y. This implies that any initial element of logf(zo), Zo ED can be continued "along any arc y cD starting at Zo and defines a single-valued branch of log / in D.
= g(y). Since
7.3.7. Take any finite Zo and any arc y. starting at Zo and let r
r omits 0, 1, with each CE y, we can associate a disk K, such that r
0 g is analytic in K" r being a branch of ./l-1, and moreover the corresponding elements (rog, K,) are identical for C sufficiently close to each other. This procedure defines a single-valued entire function G(z) with reG(z) > 0, which evidently implies G = const. 7.3.S. If fez) = a, b, then g = (f-a)/(f-b) is an entire function which omits the values 0, 1 and therefore g = const. 7.3.9, 10. It follows from the definition of ./l and from the reflection principle that ).(r+2) = ./l(r); moreover, im(Iogw/ni) > 0 in Iwl < 1. Hence Q is singlevalued and analytic in K(O; 1). An open segment (-!5, (5), 0 < () < 1, is mapped under Q onto another segment (-n, n) of the real axis, hence Q(w) = A 1 w+ +A2w2+ ... and all Ak must be real. Ai > 0 since the angle of local rotation at w = 0 is equal to O. An element of the inverse function w = Q-l(W), 0 = @-1(0), can be continued on IWI < 1, and its values cover a part of K(O; 1), hence Id~/dWlw=o < 1, which implies Ai > 1.
7.3.11. The function element (Q-l 0/, K(O; (5»), where !5 is sufficiently small, can be continued along any arc situated inside K(O; 1) and determines a singlevalued analytic function ()) with 1())~z)1 < 1. Hence / = Q 0 ro, i.e. / -< Q. 7.3.12. A consequence of Exercises 7.3.11, 6.3.2 and the remark given in Exer-
cise 7.3.10. t
7.4.1. If !(t)
=
A~
n (r-b,,)fJkn
1
dr+B maps {t: imt
> O} onto the inside
Ok=l
of D and z
=
(b n -t)-1, then
t-bk = (bn-bk)z-l(Z-XIc), k = 1,2, ... , n-l, n-l
dw/dz
=
dw/dt· dt/dz
=
t-bn
CIT (z-X")',,.-l. k_1
=
-z-1,
SOLUTIONS
234
7.4.2. w X3
=
= co; IXI
z \
o
C- 2/3 (1-0- 2/3 dC;
a = (27t)-13 1/ 2 [r(+W;
Xl
= 0,
X2 = 1,
+.
= 1X2 = 1X3 = z
7.4.3. w = ~ C,,-l(1-C 2r"dC; a
=
i-[r(IX)COS(IX7t/2)r1 r2(-tlX).
o
7.4.4. We have dw = z-l[z/(I-z 2 )]"dz, hence argdw = const on the circle z = ei8 , as weII as on the line z = iy, Y > 0; z = i corresponds to the center of the rhombus. 7.4.5. The image domain is bounded by the polygonal line with interior angles ~7t, f7t, f7t, -}7t, hence z
W=
~ [(C 2- r')2)/(C 2-a2)]1/2dC, a a
h
~ [(t 2- r')2)/(a 2_ t 2)]1/2dt,
=
0 in K(O; 1).
7.4.25. The integrand has in K(O; 1)",-0 the Laurent expansion:
Z-2_
n
_Z-l
L
k=l
ct kCk+ ... , hence it does not depend on the line of integration joining n
in K(O; 1)"'-0 the point 1 to z, iff
L ctkCk =
k=l
n
0, i.e. iff
L
k=l
ctkCk =0. Then W(z)
is meromorphic in K(O; 1) and has a simple pole at the origin. Moreover, W(z) is continuous in K(O; 1)",-0 since all ctk > -1. Hence W(e i9 ) is a closed curve. If ()k = arg Ck> 0 ~ ()1 < ()2 < ... < ()k < 27t and Z = e;9, then n
dW!d()
= iz- 1
IT (l-ze- i9k ) 0 onto a rectangle with
00
sides a, b. After reflections we obtain a function w = w(u) meromorphic and doubly periodic with periods 2a, 2ib which can be identified as fp because it has double poles at 2ma, 2nib and the principal part at the origin is u- 2. 7.6.10. If one factor has a pole (which is double) then another one has a double zero, hence the l.h.s. is analytic in C and must be constant. Now put u = w'. 7.6.11. 2K(e1 -e 2 )-1/2 ,
since fp is homogeneous of degree - 2. Hence both sides have periods 2K, 2iK'. Moreover, principal parts of both sides at u = 0 are equal to (e1-e2)u-2, hence their difference reduces to a constant which can be found by putting u = iK' which gives e2 • 7.6.12. If z is replaced by z+wk> t;(z) being an odd function increases by 2t;(-}w k ); after integration we obtain 2'1}lW2-2r12w1 = =[=27ti because res(O; t) = 1; the sign depends on the orientation of parallelogram of integration.
7. ANALYTIC CONTINUATION. ELLIPTIC FUNCTIONS
243
7.6.13. (i) this is obvious; (ii) if 'l}1 = C(j-) then by Exercise 7.6.4: 1f(z+l)
(iii) if 'I}
=
=
exp (-'I}lZ2-2'1}lZ-'I}1)a(z+ 1)
=
-exp(-'I}lZ2-2'1}lZ-'I}1)a(z)exp(2z+1)'1}1
-1f(z);
C( r /2), then by Exercise 7.6.4:
1f(z+r)
since r'l}l-'I}
=
=
1ti
=
exp(-'I}lz2-2'1}lrZ-'I}lr2)a(z+r)
=
-exp (-7h z2-2'1}1 rZ-'I}l r2) exp(2z+r)'I}a(z)
=
-1f(z) exp [-m'(2z+r)]
(cf. Ex. 7.6.12).
7.7.1. The angles (on the sphere) at u = 0 are doubled, hence u = 0 is a double pole, the only singularity in: Ireul < 2a, limul < b; fis an even function, hence after a suitable choice of A, f-Ap is analytic and hence it reduces to a constant B. Putting u = 2a, ib, 2a+ib we obtain for f(u) the values 0, -h, _h- 1 and this enables us to find A, B, h. 7.7.2. Put z
=
10gC/Q, a
=
7.7.3. Put w = ei6 , 0 < () < # O.
10gQ, b 1t,
=
1t
in Exercise 7.7.1.
and verify that du/d()
=
ie i9du/dw is real and
7.7.4. Use Exercise 7.7.3; note that the image of [K, K+-}iK'] is the segment [Vk, 1]; horizontal sides of the rectangle are mapped onto the upper and lower half of C(O; 1), whereas vertical sides correspond to the slits [-1, -y'k], [y'k, 1]. 7.7.5. Under the mapping w = Vksn(u, k) the rectangle R: Ireul < K, limul < -}K' is mapped onto K(O; 1)",,[(-1, - y'k] u hlk, 1)]. On the other hand; z = csn 1tu/2K maps R conformally onto an ellipse with foci =fc slit along [-a,-c] and [c,a]; if (u=-}iK')--(z=ib), i.e. b=y'a 2 -b 2 sinh1tK'/4K, or 1tK'/2K = log a + bb -, we obtain the given ellipse H. The slits can be removed
a-
by reflection principle.
7.7.6. w = y'ksn(u, k) maps 1: 1 conformally the rectangle Q: Ireul < K, 0< imu < -}K', onto K(O; 1)"'-{[-yk, y'k] u [0, i)}; u = 2Ki1t- 1 Logz maps A"'-( - R, -1] onto Q; the slits can be removed by reflection principle. 7.7.7. Consider W cise 7.7.6.
= [w(Y~)r2,
where w(z) is the mapping obtained in Exer-
SOLUTIONS
244
7.7.S. If K'(k)/K(k)
=
7t/logQ, then w = sn(Klogz/logQ, k).
7.7.9. Find k from the equality of cross-ratios: (al, a2 , bl , b2) = (-k-l, -1,1, k- l ) and put (W, a2 , bl , b2 ) = (w, -1, 1, k- l ), where w is the mapping of Exercise 7.7.8 continued by reflections. 7.7.10. Use the reflection principle and show that fez) has the same periods
and singularities as cn(z, 1/v':2). 7.7.11. After reflections the mapping function can be continued over the whole plane, has simple zeros at ~+ifJ+2ma+2inb, -~-ifJ+2ma+2inb and simple poles at ~-ifJ+2ma+2inb, -~+ifJ+2ma+2inb; it is moreover, doubly-periodic with periods 2a, 2ib. The function .
a (z-~-ifJ) a (z+~+ifJ) [a(z-~+ifJ) a (z+~- ifJ)tl has analogous properties and is an elliptic function of order 2 (cf. Ex. 7.6.4). Both functions are identical up to a constant factor whose absolute value is equal to 1 (put Z= 0 and use Exericise 7.6.2).
CHAPTER 8
The Dirichlet Problem 8.1.1. The inverse mapping would be a bounded, entire function, thus a constant by Liouville's theorem. 8.1.2. The sets of values taken by g on (- 00, -1), (-1,0) coincide. Moreover, if u = -vcotv and vcosecv = exp(vcotv) which holds for. some vE(21tn+1tt4, 21tn+1t/2) and all n large enough, then g(ll+iv)=g(-I). Thus we need to verify that g takes every finite value in C. Suppose that weW - a 1= 0 for Some a and all w. Then also (w +21ti)ew - a 1= 0 and the quotient I+21tie w /(we W -a) which is an entire function 1= 0, 1 must be a constant. This is an obvious contradiction. If h(z) = 4z(I-z)-2, then h[K(O; 1)] = G and 1= {- g 0 h has the desired properties. 8.1.3. Suppose h is a (many-valued) conjugate of g and the increment of h over a small circle with center at Zo and positive orientation is equal to A; then 1= exp o 27tA-1 (g+ih) has all th~ desired properties. Univalence follows from the argument principle; since A is real and negative, I(zo) = O. 8.1.4. The mapping It oli 1 carries the unit disk onto itself; the mapping is a homeomorphism in K(O; 1) and after reflections it becomes a 1: 1 conformal mapping of the extended plane onto itself and consequently it is a linear mapping with 3 fixed points, i.e. an identity. 8.1.5. Both mappings I(z), (f(z*)*, where the stars indicate corresponding reflections carry Zk into Wk (k = 1,2,3), hence they are identical. This means that I(Yl) remains unchanged after a reflection w.r.t. Y2' i.e. Y2 = I(Yl)' 8.1.6. Arcs of: C(I; 1), reA =
t,
C(O; 1) situated in the upper half-plane;
A((I+iV3)/2) = (I+iV3)/2, A(i) = 1/2, A(1+i) = 2.
8.1.7. Suppose h is a positive, continuous and nowhere differentiable function with the period 27t and let D be the domain whose boundary D has the equation r = h«()), 0 ":( () ~ 27t, in polar coordinates. If I maps 1: 1 conformally K(O; 1) onto D, then C(O; 1) is obviously a natural boundary of f
246
SOLUTIONS
8.1.8. If the function f/J mapping D onto K(O; 1) is such that 0 = f/J(a), then all the functions with this property have the form eilXf/J(w); hence q?(w)
= eilXf/J(w)/f/J'(a)
and
r(a; D)
=
1/1f/J'(a)l.
8.1.9. In view of Exercise 8..I.8. r(a; G)
=
Idw/dCJ~=o,
w = J[(C+zo)/(l+zoC)].
where
8.1.10. (i) R- 1(R 2_laI 2); (ii) 2h; (iii) 4d. 8.1.11. If Do = J[K(O; 1)], ao = J(zo) , then r(ao; Do)
= (l-l zoI2)1f'(zo)1
and
r(a; D)
= (l-lzoI2)1f'(zo)IIq?'(ao)l.
8.1.12. Suppose that z = f/J(w) maps Do onto K(O; r). Then q? conformally K(O; r) onto D and IDI
~~ I(q?
=
0
0
f/J-l maps
f/J-l),1 2dxdy
K(O;r)
and since q?
0
f/J-l (z)
= z+ A2Z2 + ... , we have IDI
=
1tr2(l+2IA212r2+ •.. ),
(cf. Ex. 4.2.12). Hence IDI is a minimum if q? i.e. f/J = q?
0
f/J-l is an identity mapping,
8.1.13. We may assume that 00 Ern' We apply Riemann mapping theorem to the domain C"", rn and the mapping function carries G into th~ unit disk minus n-l continua. After a suitable linear transformation one of the boundary continua, say Yn-l will contain 00. We again apply the Riemann mapping theorem to C"",Yn_l and obtain an image domain whose boundary already contains two analytic Jordan curves, C(O; 1) and the image curve of a circle under the latter mapping function. After n analogous steps we finally obtain as an image domain of G the unit disk with removed n-l interior domains of closed analytic Jordan curves.
2n:
8.2.2. J(z)
=
(21t)-1 ~ U«()) (Rei6 +z)/(Rei6 -z)d()+iv(O) , where v(O) is an aro
bitrary real number. 8.2.3. This is a consequence of (8.2A), the mean-value property: 2n:
u(O)
=
(21t)-1 ~ u(Rei6 )d() o
8. THE DIRICHLET PROBLEM
247
and the inequality R-izi R2_lz12 R+lzi :::;; IRe'8_ z I2
8.2.4. u(z)
=
R+lzi
~ R-izi .
-}re(1+z2) (cf. Ex. 3.6.3 (i)).
8.2.5. 21t'u(z) = ~ (1-lzI2) IC-zl- 2dO = ll(Cl-Z)/(C-z)ldO because IC1 -zliC -zl ~
=
~
l-lzI2; moreover, by considering the triangles z, C, C+L1Candz, C1 , C1 +L1C1
FIG.
6
(cf. Fig. 6) we have dO/IC-zl = dOdIC1 -zl, hence 21t'u(z) = ~dO = length 1
of y. 8.2.6. If KeD, where D is the domain where the functions Un are defined, then the formula (8.2A) may be applied, hence the limit function is harmonic, in K and also in D. 8.2.7. In view of Exercise 8.2.6 the function u(z)
= u(re i"') = -}a o +
ao
L (r/R)n(ancosrtq;+bnsinnq;) n= 1
is harmonic and tends to U(q;) as r formulas we have:
-+
R- (cf. Ex. 4.4.4). By the Euler-Fourier
2",
an
=
(21t')-1 ~ U(O)cosnOdO, o
2",
bn
=
(21t'r 1 ~ U(O)sinnOdO, o
248
SOLUTIONS
and hence 2,...
u(re''If) = rc- 1 ~
co
[t + L (r/R)nCos n(p-O)] U(O)dO
OJ
n=1
2,...
=
co
(2rc)-1 ~ re[I+2L(zm n]U(0)dO o
n=1
2,...
=
(2rc)-1 ~ re[(C+z)/(C-z)] U(O)dO, o
.
2,...
8.2.8. u(z) = u(rei'P) = (2rc)-1 ~ re[(z+O/(z-C)]U(O)dO, where C = Re w, o
8.2.9. u(rei'P) = C(rc-1+2~-1
co
L n- (r/R)nsinnr1..cosnp. 1
n=1
8.3.1. A function harmonic in 0 < Izl < 1 and continuous in K(O;I) is harmonicinextensiontoK(O; 1) and hence it is determined by its values on ceO; 1). 2,...
If U(O) =1= (2rc)-1 ~ U(e i8 ) dO , the Dirichlet problem has no solution. o
8.3.2. (i) u(z) = A(1og(lzl/R2»:(log(R1/R2»+B(log(lzl/R1»: (log(R2/R1»; (ii) the coefficients an, bn, Cn, dn may be evaluated from Euler-Fourier formulas: 2,...
ao+bologR k = rc- 1 ~ Uk (0) dO , o 2,...
anR~+bnR"kn
=
rc- 1 ~ Uk (0) cosnOdO , o • 2,...
cnR~+dnRkn = rc- 1 ~ Uk (0) sinnOdO , o
n
=
1, 2, "', k
=
1, 2.
8.3.3. The mapping function 1). The solution of the latter problem is: 10g(JwIIR)+A. Hence u(z) = loglz+ yz2- c2 1-log(a+b)+A; if A = log [(a+b)/2] , then u(z)-loglzl follows from the maximum principle.
=
0(1) as z --.
+00.
The uniqueness
8.3.12. The mapping w = 2(Z2+-}) carries the given domain into that considered in Exercise 8.3.10. Hence u(z) = 1t'-limLog(2z2+1+2zyz2+1) = 1t'- l imLogF(z). We have
Ux
=0
on the real axis, hence u, = aulan = 1t'- l IF'(z)1 = 21t'-1(x 2+1r 1/2.
8.4.1. re[(l+z)/(l-z)] = (l-lzI 2 11-zl-2 • 8.4.2. (i) Circular arcs with end points a, b situated in the upper half-plane (cf. Ex. 8.3.5); (ii) circular arcs with end points eilJ1 , e11J2 , situated inside K(O; 1).
8.4.3. £O(z; /2' G) = oc-1Argz, £o(z; f2' G) = l-£O(z; fl' G). 8.4.4. If O(z) = imLog[(z-l)/(z+l)], then £O(z; T, G) = 21t'- l O(z)-1; £O(z; 'Y, G) = 2 [1-1t'- 10 (z)]; circular arcs with end points 1, -1 situated inside G. 8.4.5. If z = x+iy, then £0 (z;
'Y, G) = 21t'-1 [1t'-0 (z)] = 21t'-1[arctan(yl(r-x»)+arctan(yl(r+x»)]
= 4Y(1t'r)-1(1+0(1»). 8.4.6. The mapping z = w2 carries Go. 'Yo into G, 'Y and due to the conformal invariance of £0 we have: £O(z; 'Y, G)
=
21t'- l Arg[(z-1)/(z+1)]
=
21t'- l Arg[(w2-1)/(w2+1)]
8.4.7. We have £O(r,.O)
= A(l-logrflogR) +
L n=l 00
[(anrn+bnr-n)cosnO+(cnrn+dnr-n)sinnO] ,
where
=
[1t'n(1-R- 2n)rlsin21t'nA,
an
= [1t'n(l-R 2n)r 1sin21t'nA,
bn
Cn
=
dn = [1t'n(1-K2n)rl(I_cos21t'nl).
[7tn(1-R 2n)r 1(1-cos27tnA),
8. THE DIRICHLET PROBLEM
8.4.8. If O(z)
251
Arg[(z-b)/(z-a)] and CPk
=
w(z; Yb G)
=
(k, 1= 1,2, k i= f).
[O(Z)-CP1]/(CPk-CP1)
=
O(C), CE Yb then
8.4.9. (i) w(z; Y2' G)-w(z; Yl' G) is a bounded harmonic functions with nonnegative boundary values, hence it is nonnegative in G; (ii) consider w(z; Y, G2)-w(z; y, G1), z E G1; again the difference has nonnegative boundary values. 8.4.10. The function
h(z)
=
10g/f(z)l-w(z;
0(,
G)logm-w(z;
p, G)logM
is harmonic except possibly at zeros Zk off and bounded from above. If z ~ Zk, then h ~ - 00; if z ~ CE frG, limh(z) ,;;;; 0 except possibly for end points of 0(. Hence h(z) ,;;;; 0 in G by the maximum principle.
fJ
8.4.11. If Z = x+iy is fixed, then by Exercise 8.4.10 (with m = 1,0(= (-r, r), r) (") H+ and G = K(O; r) (") H+) and by Exercise 8.4.5 we have:
= C(O;
log/f(z)/ ,;;;; w(z; p, G)logM(r) = 4Y(7tr)-110g M(r) (1+0(1)). Suppose that lim r- 1IogM(r) = (],;;;; 0; then there exists a sequence {rn}, r~+co
rn ~ +00, such that r;llogM(rn) ~ (],;;;; 0, and consequently loglf(z)/ ,;;;; 4Y7t- 1 (]
,;;;;
0,
i.e.
If(z) I ,;;;; 1.
8.5.1. If h(z, zo) is the solution of the Dirichlet problem with boundary values 10gIC-zol, then
g(z, Zo; G)
=
h(z, zo)-loglz-zol.
8.5.2. Suppose t =f(z) maps 1:1 conformally G onto K(O; 1). Then
g(z, zo; G) 8.5.3. If dist(w, frG w) if w ~ wo, then
~
=
10g[ll-f(zo)f(z)lIlf(z)-f(zo)l].
0, then dist(z(w), Gz )
~
0, hence g(w, wo; G)
~
0;
g(z(w), z(wo); G2 ) = -loglz(w)-z(wo)I+O(I) =
-loglw-wo/-log[jz(w)-z(wo)l//w-wol]+O(I)
= -loglw-wol+O(1). 8.5.4. u+iv = 10g
_(27t)-1L1 rk h(C, zo),
=
g+ih, hence for w describing C(O; 1):
L1log€l> = iL1h = -27tiw(c; C(O; 1),A) = -27ti(1=
:::~)
L1logwlo.gc/logh-L1logF(w),
i.e. L1logF(w) = 27ti, similarly for w describing C(O; h) we have L1logF(w) = O.
8.5.16. D is a convex domain, hence there exists a half plane H containig D and such that c E C",,-H. In view of Exercise 8.5.13 we may assume that D = H. Therefore we may take c = 0, D = {z: rez > O}. By Exercise 8.5.6 g(a, b; H) =
log
::~!\
.
If the points a, b are rotated round the origin, la-bl remains
unchanged and la+bl :::;;; lal + Ibl with the sign of equality in case a, 0, -b are collinear. Hence supg(a, b; D) = 10g[(lal+lbJ)/la-bl] in case c = 0; in the general case: D
s~pg(a, b; D) = log
la-cl + Ib-cl la-bl .
SOLUTIONS
254
8.5.17. If g(z) = g(z, 00; G), then Ag(Z)+B q;(z) = { B
for for
z EG, ZEH.
The uniqueness follows from the maximum principle in the usual manner. 8.6.1. If f, g
E
L 2(G) , so does al+bg for any complex a, b because lal+bgl 2 :0::;; laI2IfI2+lbI2IgI2+labl(lfI2+lgI2).
8.6.2. Ilgl :0::;; fClfI 2+lgI 2), hence ~Ug is finite. 8.6.3. Suppose q;:
c"'-.r--* K(O;R)
R = r(C; C"'-.T)
is such that q;'(C)
=
1; then
~~ 1q;')1 2dxdy = 'rCR 2
and
C,,",r
(cf. Ex. 8.1.12) is finite. Now, G c
c"'-.r and
hence q;'
E
L2(G).
8.6.4. If dist(z, C"'-.G) = p, then II(z) I :0::;; (M/7tp2)1/2. 8.6.5. I/(a) I :0::;; (M/7t)1/2(1-lal)-1. 8.6.6. The family of functions IE L2 (G) such that I( C) = 1 and 11I1 t :0::;; y~R, where R is defined as in the solution of Exercise 6.8.3, is compact and nonempty. 8.6.7. If e > 0, 0:0::;; () :0::;; 27t, then f* We have Ilfoll :0::;; Ilf*ll, hence
=
fo+ee i9 g
E
L 2(G) for any g
E
L2(G).
(fo,Jo) :0::;; (fo,Jo)+2ere[e i9 (g,Jo)]+e 2(g, g), i.e. 0:0::;; 2re[ei9 (g,!0)]+e(g, g) for any e > 0 and any () (g,!o) = O.
E
[0,27t] which implies
8.6.8. Any function analytic in G has a primitive, hence we need to find q; analytic in G and such that q;(C) = 0, q;'(C) = 1 and ~ ~ 1q;'12 has a minimum. In view .
G
of Exercise 8.1.12 the extremal function fo = q;' is equal to w'(z)/w'(C); if R = r(C; G), then IlfoW = ~ ~ 1q;'l2 = 7tR2 = 7tlw'(nl-2 by Exercise 8.1.8 'and hence G
k(z, C) = 7t- 1Iw'(C)1 2w'(z)/w'(C) = 7t- 1w'(z)w'(C). 8.6.9. (i) 7t- 1R 2(R 2-zC)-2; (ii) 7t- 1 (Z-C)-2. 8.6.10. w'(z) = V7t/k(C, C)k(z, C). 8.6.11. If g(C) = 0, then (g,!o) = 0 (Ex. 8.6.7) and also (g, k) = 0, I.e. g(C) = ~ ~ f(z)k(z, C)dxdy = 0; G
255
I. THE DIRICHLET PROBLEM'
if I(C) =F 0, then g(z) = l(z)IfW-/o(z, C) vanishes for z = C, hence ~ ~ [/(z)If(C)-/o(z, C)]/o(z., C)dxdy
=
°
G
and consequently
~ ~/(z)/o(z, C)dxdy = I(C) 11/0 112. G
8.6.12. Put 1= k(z, C) in Exercise 8.6.11 and use Schwarz's inequality. 8.6.13. Suppose that I(C) = (f, k 1 ) = (f, k 2); then (J, k 1 -k 2) = 0 for any leL2(G) and also for 1= k 1 -k 2. This implies IIk1-k211 = 0, i.e. kl = k 2 • n
8.6.14. If Sn =
L akf/Jk>
k=O
n
Gn =
L bkf/Jk,
k=O
then from the minimal property of n
Fourier coefficients and from the density property of II/-snil
0, IIg-Gnil
-+
-+
o.
L Ckf/Jk
k=O
it follows that
Hence n
(/-sn,g-Gn) = I(ig)- ~akbkl,::;;; II/-snll·llg-Gnll-+.O.
8.6.15. From Exercise 8.6.11 it follows that k(C, rJ) = (kq, k,), where k, = k(z, C), kq = k(z, rJ); moreover, an = (k" f/Jn) and hence by Exercise 8.6.11: an = (f/Jn, k,) = f/JiC); similarly bn = (kq, f/Jn) = f/Jn(rJ). Using Exercise 8.6.14
we obtain: 00
(kq, k,) =
2:
f/Jn(rJ) f/JnW = k (C, rJ)· n=O 8.6.16. Using polar coordinates we easily verify that {f/Jn} form an orthonormal system; the completeness can be proved by verifying Parseval's identity, the integral /112 being expressed by Laurent's coefficients.
H
+00
8.6.17. k(z, C) = (27tlog(1/h)t 1 (zC)-1+7t- 1
L
(l-h 2n )-ln(zC)n-l.
"=-00 00
8.6.18. Both series
00
L (n+ lr1lbnl2, L
n=O gent, cf. Exercise 4.2.5.
n=2
(n-lrllb_nI2h-2n should be conver-
8.6.19. If z = z(t), C = z('r), then after the change of variables in the formula of Exercise 8.6.11 we obtain: I(Z(T)) = ~ V(z (t))k (z (t), Z(T)) 1z'(t)1 2dudv Gl
and due to uniqueness of kwe obtain: kl (t, T) = v'k(z, z) Idzl.
=
k(z, C) Idz/dt/ 2. Hence ykl (t, t) Idtl
CHAPTER 9
Two-Dimensional Vector Fields 9.1.1.
W
=f'(z);
Iwl
= 1f'(z)l.
9.1.2. g = ux-iuy, G = Ux-iUy are analytic in D and we may assume that g and G are not identically 0; then g, G do not vanish on D""-.H, H ED being an isolated set. If z E H ny, then the normal vector of y has components ux , uy, or Ux , Uy, and consequently arg((Ux-iUy)/(ux-iuy)) = 0, hence the quotient
G/g must reduce to a constant (which is real). 9.1.3. The lines of flow are circles im(ilz) = const, hencef(z) = kilz; hence IW11/1w21 = 1/2. 9.1.4. The lines of flow can be represented in the form im(iz-2) fez) = kiz- 2 (k is a real constant); IW11/1w21 = 2- 3 / 2.
=
= f'(z) ,
W
const, hence
9.1.5. Circles ceO; r). 9.1.6. (i) Equipotential lines : straight lines x = const; lines of flow: y velocity: w = a; (ii) y = C1 , X = C 2 , W = -ai; (iii) X2+y2 = C1x, X 2+y2 = C 2y, W = Z-2; (iv) Iz-bl/lz-ci = C 1 , circles through b, c,
IV
const;
(b-c)[(z-b)(z-C)r1;
=
(v) x 2_y2 = C1, xy = C 2 , (vi) X2+y2 = C 1 , Y = C 2i, (vii) y = C 1x, X2+y2 = C 2 ,
IV
=
W
= 2z.;
W =
W
=
alz; -ailz.
9.1.7. If W is the velocity and f = u+ iv is the complex potential of flow, then and this means that u x , U y are components of w. Hence
= f' = ux-iuy
~f'(z)dz = ~(ux-iUy)(dx+idy) y
=
y
=
~uxdx+uydy+i~ uxdy-uydy y
~ wsds+i ~ wnds = r+iQ. y
y
(wn denotes the normal component of w). 256
y
9. TWO·DIMENSIONAL VECTOR FIELDS
257
9.1.8. The points -=t=(1 =Fi)/J/2: are sources of intensity 21t, the points 0, sinks of intensity 41t.
00
are
9.1.9. Equipotential lines: equilateral hyperbolas through =Fa with center 0; lines of flow: lemniscates with foci =Fa, r = -41t. 9.1.10. Q = 61t, I'_= O. 9.1.11. J(z) = Qi1t- 1Arcsinz; according to Bernoulli law the pressure
P = C-tlw'12 = C_Q2/[21t 2(I_x)2],
C
= const =
Poo'
9.1.12. J(z)
= woo (Z+Z-l), w(i) = 2woo , w(2)
=
fw
oo ,
p(2)-p(i) = ~w!.
9.1.13. The complex potential of flow Ji outside the circular cylinder lei > R has the fotmJl(C) = w(e-icrC+R2ei t(2-rx) in K(O; 1). 10.2.4. From the solution of Exercise 10.2.2 it follows that the conditions are sufficient for Df to be starlike. We now prove the converse. Suppose Df is starlike, f being univalent with/CO) = 0,/'(0) = 1. We first prove that also f[K(O; r)] is starshaped for any r E (0, 1). Obviously w(z) = /-1 (if(z» satisfies the conditions
SOLUTIONS
of Schwarz's lemma for any 0 < t < 1. Hence lro(z) I ~ Izl. If Wl Ef[K(O; r)] and Wl =f(Zl)' IZll < r, then lro (Zl) I = Irl(twl)1 ~ IZll < r. Therefore tWl = f(Z2) willi Z2 E K(O; r) which means that f[K(O; r)] is starshaped. Hence f[c(O; r)] is a curve starlike w.r.t. the origin and so argf(reiIJ ) increases with () which yields re[zf'(z)lf(z)] ~ 0 as in Exercise 10.2.1. The case re[zf'(z)lf(z)] = 0 in K(O; 1) is excluded by the maximum principle. 2"
10.2.5. !,(z)lf(Z)-Z-l = Z-l [P(z)-I] = 2 ~ (e!IJ-zrldft«()
and the integra-
o
tion gives 2"
fez)
= zexp{-2 ~ Log (I-ze-i9)dft«()}. o
10.2.6. If(z)/zl ~ exp{210g(I-lzJ)-l} = (I-lzJ)-2, hence If(z) I ~ Izl(I-lzJ)-2 and similarly (1+ IzJ)- 2Izl ~ If(z)l. Hence using the estimates
(I-lzJ)(I + IZJ)-l ~ Izf' (z)lt(z) I ~ (1 + IzJ)(1-lzJ)-l, cf. Exercise 10.1.7, the estimates of!' easily f(jIlow: (1-lzJ)(I+lzJ)-3 ~ 1f'(z)1 ~ (1+lzJ)(I-lzJ)-3. 10.2.7. Log[zlf(z)]l/2
=
2"
~
Log(I-ze-ilJ)dft«() by Exercise 10.2.5. Using Exer-
o
cise 10.1.5 and Exercise 2.6.8 we see that Log[zlf(z)]l/2 ranges over a convex domain being the image of K(1; r). under Log. Hence [Zlf(Z)]l/2 ranges over K(I; r) for varying f E S* and fixed z, Izl = r. . 10.2.S. If f, z range over S* and K(O; 1) resp., then [z/f(zW/ 2 ranges over
U
K(I; r) = K(1; 1), i.e. ff(z)/Z]l/2 ranges over {w: rew
> t}.
rE(O.l)
10.2.9. Suppose that IZll ~ IZ21 < rand Wk = f(Zk) , k = 1,2; by Schwarz's lemma I1J'(z) I ~ Izl and for z = Z2 we have
Ir l [twl +(1-t)W2] I ~ IZ21
< r;
if Zo =f- l [tw+(1-t)W2], then IZol < r andf(zo) = tWl+(I-t)w2 which means tlaat Br is convex since t E (0, 1) can be arbitrary. 10.2.10. If z!' E S*, then the tangent vector of f[c(O; r)] turns monotonically and its argument increases by 2" as z descril}es C(O; r). Hencef[C(O; r)] is a convex curve whose interior domain is convex. A converse can be ,proved in an analogous manner. 10.2.11. !' = F/z with FE S* (cf. Ex. 10.2.8).
10. UNIVALENT FUNCTIONS
265
10.2.12. We have reF(z) > 0 for z EK(O; 1) and also for z
o
R~k) > ... > R~r:.>k+l and if LIe]) is the measure of the corresponding angle, then the integral sum is equal to L Lle])k[h(R~k»-h(R~k»+ .,. +h(R~~+l)] > 0 because the k
SOLUTIONS
266
orientation is positive. Take now the limit for a normal sequence of partitions. 10.3.2. We have:
r-;'
2"
2"
o
0
~ g(R)dO = r ~
2"
:r
g(R)~O = r ~ g'(R)R
2"
=
~
2"
g'(R)R
~~ dO = ~
o
10.3.3. If g(R)
=
~
Rg'(R)d6W(r, 0).
0
R2, then by Exercise 10.3.2 we obtain
2"
r ;,-
:r 10gRdO
0
2"
If(re"6)1 2d8
o
=
r ;,
~
2"
R 2dO
=
0
which is ~ 0 in view of Exercise 10.3.1 (h(R)
~
2R2(r, O)d6W(r, 0)
0
=
2R2 ).
o FIG. 7
10.3.4. The values of F do not cover'C completely (Liouville's theorem); if F(z) i= Wo for any z E K( 00; 1), then by the argument principle the images of C(O; r) by F(z)-wo contain w = 0 inside, hence 2"
I(r)
=
~ IF(reiB)-woI2dO = 27t[r2+lbo-woI2+lblI2r-2+lb212r-4+ ... ] o
is an increasing function of r and consequently I' (r) ?- O. N
10.3.5.
2..: n IbnI2r-2n-l :S;; r; .=1
then make N
-+
+ 00 .
N
make first r -+ 1 which gives
2..: n Ibnl2 :S;; 1 .=1
and
10. UNIVALENT FUNCTIONS
267
10.3.6. If IE S and Izl > I, then F(z) = IIf(z-1) belongs to 1:0 • If FE Eo and ICI < I, then/(C) = I/F(C- 1) belongs to S. The function F(z) = z(1+z-3)Z/3 corresponds in this way to a function IE S*. 10.3.7. G(z) = [F(zZWIZ = z+tboZ-1+ ... E Eo, hence by Exercise 10.3.5 we have: -} Ibol ~ I. The case of equality occurs only for G(z) = z+e i "z- 1, or F(z) = z+2e'''+e Zi ..z-1, with oc real. 10.3.8. F-Wk
E
Eo, hence IWk-bol ~ 2 (k = 1,2) and
IW1-wzl ~ IW1-bol+lbo-wzl ~ 4. 10.3.9. If IE S, then F(C) = 1/I(C-1) = C-a2+ ... E Eo so that lazi ~ 2 (cf. Ex. 10.3.7). Equality holds only for F(C) = C+2e i "+e Z:"C- 1, i.e. for I(z) = I .. (z) = z(1+ei"z)-2. 10.3.10. cp is univalent and analytic in K(O; I) being a superposition of a linear transformation and IE S; moreover, cp(z) = z+(a z +h-1)zZ+ ... , hence laz+h- 11 ~ 2, i.e. Ih- 11 ~ 2+la21 ~ 4. Equality holds only, if la21 = 2 which means that I = I ...
10.3.11. rp is univalent in K(O; I), moreover, rp(O) = 0, cp'(O) = I. Hence cp E S. Now, z(t) =f:. b and this implies that rp(t) =f:. rJ(b)-/(a)]t'(a)If'(a) = h for any t E K(O; I); t(z) denotes here the inverse of z(t). The result follows immediately from Exercise 2.9.21 and Exercise 10.3.10. 10.3.12. (i) Take b = 0 which gives laf'(a)lf(a) I ~ (I + lal)(I-lal)-1 ; (ii) on integrating both sides in (i) we obtain: I/(a) I ~ lal (I-Ial)-z; taking a = 0 in Exercise 10.3.11 we obtain Ibl(I+lbl)-z ~ I/(b)l; (iii) "P E S, hence 1"P(t)1 ~ Itl(I-ltl)-z by (ii) and putting t = -z we obtain Izf'(z)lf(z) I ~ (I-Izl) (1 + Izl)-1 .
Equality can be attained in all cases for I = I ... 10.3.13. Multiply both estimates for I/(z) I and Iif' (z)/I(z) I side by side. 10.3.14. We have I'(b) = rp'(O) (1-lbI Z)-l,
f'(a) = rp'(C1)(I-lbl z ) (l-a6)-2
with C1 = (a-b) (l-ab)-1. Hence f'(a)If'(b) = (1-lbI Z)Z(I-ab)-zcp'(C1)/rp'(0).
Now, [rp(C)-rp(O)]/rp'(O) = "P(C)
E
S, thus
1"P'(C1)1 ~ (1+IC1 1) (1-IC1 0- 3
268
SOLUTIONS
and finally
I ~ (1-lbI ) (ll-abl+ la-b l )2 If'(a) f'(b) (1-laI 2) 11-abl-la-bl 2
(cf. Ex. 1.1.8 (iii»). The lower estimate is obtained by interchanging a, b. 10.3.16. (i) (Jzl-l)2Izl- 1 ~ IF(z) I ~ (Izl + 1)2Izl-1; (ii) (Izl-l) (lzl+1)-1 ~ IzF'(z)/F(z) I ~ (Jzl+l) (lzl-IJ)-l, which is easily obtained by using Exercise 10.3.12 and Exercise 10.3.6. 10.3.17. (i) Take a sequence {zn} such that IZnl -.. 1, F(zn) is convergent and
limIF(zn)-bol = lim IF(z)-bol. JzJ-l+
Then Wo = limF(zn) is not a value of F, hence F-w o E Eo. Now, by Exercise 10.3.7, Ibo-wol ~ 2. • (ii) lim !z(F(z)-z-b o)!2 ~ lim JF(z)-b ol+l ~ 3, moreover, !z(F(z)-z-b o)! IzJ-+l+ JzJ-l+ -.. Ib11 ~ 1 as z -.. +00, hence !z(F(z)-z-b o)! ~ 3 by the maximum principle.
10.3.18. (i) IF'(z)-11 ~ Iznl·lb11+V2Izl-1y'2Ib21+ ... ) ~ Izl- 2 y(l-lzl- 2)-2 Vlb112+2Ib212+ '" ~ (lzI2-1)-1
by Schwarz's inequality and the area theorem. (ii) IF' (z)1 ~ IF'(z)-11+1 ~ IzI2(lzI2-11-1. 10.3.19. F(z) is a superposition of a(C +C- 1)+b and C = i(h2-1)-1/2(hz-l) hence it must be univalent. The latter transformation carries C(O; 1) into a circle through =F1 and by Exercise 2.5.2: C'"F[K( 00; 1)] is a circular arc. 10.3.20. (ii) g
E
S hence in view of Exercise lO.3.12 (ii) we have
If(z) I 11 +ei":{(z)/MI ~ Izl (1-lzJ)-2 for an arbitrary real ce. Thus
If(z)I(I-lf(z)IIMt2 ~ Izl(I-lzJ)-2 = Iwl(I-lwl/M)-2, where w = fM(Z). Now, u(l-uIM)-2 is an increasing function of u e:'(0, M). hence If(z) I ~ Iwl = IfM(Z)I. A similar proof for the lower estimate. (i) M(lzl,f)
= Izl+la2i1zI2+O(JZJ)3 =
~
M(jzl,fM)
Izl+2(I-M- 1)lzI 2+O(JzI3), hence la21 ~ 2(1-M-1).
10.3.21. g(z) = z+C-i-(I-ICI 2)f"(C)lf'(C)-i-(I-ICI 2)2{f. C}Z-1+ ... Now, by the area theorem (Ex. 10.3.5) the coefficient of Z-1 is at most 1 in absolute
10. UNIVALENT FUNCTIONS
269
value, hence Itf, 01 ~ 6(1-ICI2)-2. If h is analytic and univalent in K(O; 1), then f = ah+b E S for suitably chosen constants a, band {h, z} = {J, z} by Exercise 7.2.14. If h is meromorphic and univalent in K(O; 1) and h(z) ¥= Wo for any z E K(O; 1), then H = (h--wo)-l is analytic and univalent and again
{H, z}
{h, z}.
=
10.4.1. The proofs for Steine~ and P6lya symmetrization are similar. For sake of simplicity we give here the proof for Steiner symmetrization Suppose that a(xl), a(x z ) meet G* and also G. Then a(x) meets G for any x E [Xl' x 2] since in the opposite case G would be disconnected. This implies that [Xl, X 2 ] C G* and obviously two arbitrary points Zl' Z2 E G* with rezl = Xl' rez 2 = X2 can be joined by the polygonal line [Zl' Xl' X2' z 2] in G*. Hence G* is arc-wise connected, if G is connected Evidently C",-G* is also simply connected in this case. Suppose now G is an open set and Zo = xo+iyo E G*. It is easily verified that the linear measure lex) of a(x) n G is a lower semicontinuous function of X for open G (note that a(x) n G is an at most countable system of open segments: replace the countable system by a finite system of closed sub-segments with a slightly less total measure and use Heine-Borel theorem for the latter system). Now, in our case l(x o) > 21Yol. From lower semicontinuity it follows that for any Yl > 0 such that l(xo) > 2Yl > 21Yol we can choose lJ such that flex) > Yl for any X E (xo-lJ, xo+lJ). Then the open rectangle {x+iy: Ix-xol < lJ, lyl < yd c G* which proves that G* is open. In case of Steiner symmetrization G* is always simply connected if G is a domain. In case of P6lya symmetrization we should make an additional hypothesis that G is a simply connected domain. 10.4.2. If G is a domain starshaped w.r.t. the origin 0 and WO EC",-G, then the whole ray emanating from Wo whose prolongation contains 0 belongs to C",,- G. This implies that the function l(r) introduced in the definition is decreasing which again implies that G* is starshaped w.r.t. the origin. 10.4.3. Suppose G is a convex domain. It is sufficient to consider bounded, convex domains since each convex domain is a sum of an increasing sequence of bounded convex domains e.g. G n K(O; n). Now the boundary of a bounded convex domain consists of the graphs of 2 functions rpl' rp2' a ~ X ~ b, and possibly two segments rez = a, rez = b, the functions rpl' -rp2 being convex. This implies that t(rpl -rp2) = lex) is convex and non-positive, hence G* being bounded by two convex arcs and possibly two segments must necessarily be convex. D = [S n K(O; 2)] u [S n K(O; 4) n {z: rez > O}], where S
=
{z: IJim zl < I}, is convex, however n* is not convex.
270
SOLUTIONS
00
~~
1f'(rei6 )1 2 rdrd() =
7t
K(O; 1)
Suppose that
< +00.
L nlanl2 n=1
8> 0 is arbitrary and choose m ~ 2 such
00
that
2: nlanl
2
r5,
dist(z"; frG m)
~
r5,
m and also
Ilogr(z'; Gn)-logr(z"; Gn)1 Z 2Iz'-z"l/r5
for all n ~ m. Hence either both sequences r(z'; G.) tend to to a finite limit. In the limiting case we have:
+ 00, or both tend
Ilogr(z'; G)-logr(z"; G)I Z 2Iz'-z"I/r5
which proves continuity of r(z; G). We have proved even more: r(z; G) is Lipschitzian on compact subsets of G. 10.4.8. r(z; H) attains a finite maximum at some point CE H in view of Exercise 10.4.5. Consider now Steiner symmetrization w.r.t. a line through Cparallel to the major axis. This gives r(C l ; H) ~ r(C; H), Cl being the projection of C on the major axis. Another Steiner symmetrization w.r.t. a line through Cl parallel to the minor axis shows that reO; H) ~ r(C; H), 0 being the center of H. 10.4.9. If G E and the real axis is the line of Steiner symmetrization then G* = C which is not of hyperbolic type.
c"""
( - 00,
+}
10.4.13. From the conformal invariance of Idwl/r(w; G) (cf. Ex. 8.1.11) it follows that Idwl/r(w; G) = Idzl/r(z; K(O; 1)) = (l-lzI2)-lldzl, hence
)[r(w; G)]-lldwl
P(WI' W2; G),
=
r
rbeing the h-segmentjoining WI to w2 • We now associate with r a curve y c G* joining WI to W2 in the following manner. Suppose that w = Rei E Then the point CE Y associated with w is the point C= R. From Idwl 2 = dR2+R 2d¢2 it follows that IdCI < Idwl. We consider a ray Ow, w E r and the symmetrized domain G! of G w.r.t. ray Ow. We have: r(w; G!) ~ r(w; G) due to P6lyaSzego theorem. However, G!, G* are congruent and a rotation round the origin by the angle q;, carries G* into G!. Since the inner radius is invariant under motion, r(w; G!) = r(C; G*). Thus IdCl/r(C; G*) < Idwl/r(w; G) and after integration we obtain
r.
~ [r(C; G*)tlldCI 1
L(r,/o) for some / E S. We symmetrize/[K(O; 1)] w.r.t. the negative real axis and obtain a domain D* such that /o[K(O; 1)] c D* and D*""-/o[K(O; 1)] :1= 0. Thus we have reO; D*) > r(O'/o[K(O; 1)]) = 1 which contradicts the inequality r(O;/[K(O; 1)]) = 1 :::.;; reO; D*). 10.5.15. Obviously fez) = F[(t-toz)!(I-z)], where y~ = to+iYl-f(P. Similarly as in Exercise 10.5.14 we can show that the extremal domain is the image domain of K(O; I) under the mapping g(z) = /(z)//'(O) and arises
276
SOLUTIONS
from H«()) by a similarity with the ratio r = 1/'(0)1-1 , The set of values not taken by g and situated' on C(O; r) is a circular arc subtending the angle (2-())1t = mp(r), where mp(r) is the maximal value to be evaluated. We have: 11'(0)1 = t[(2+())2+B(2_())2-0]1/2, hence rp(r) = 2-() satisfies the equation:
Bibliography 1. BASIC TEXTBOOKS
1. L. V. Ahlfors, Complex Analysis. McGraw-Hill, New York, 1966. 2. H. Behnke, und F. Sommer, Theorie der analytischen Funktionen I!!iner komplexen Veriinderlichen. Springer, BerIin-Gottingen-Heidelberg, 1955. 3. C. Caratheodory, Funktionentheorie, Vols. I and II. Birkhliuser, Basel, 1950. 4. B. A. Fuchs, V. I. Levin, and B. V. Shabat, Functions of a Complex Variable and Some of Their Applications, Vols. I and II. Pergamon, Oxford, 1961. 5. M. Heins, Complex Function Theory. Academic Press, New York, 1968. 6. E. Hille, Analytic Function Theory, Vols. I and II. Ginn, Boston, 1962. 7. A. Hurwitz, und R. Courant, Vorlesungen iiber allgemeine Funktionentheorie und elliptische Funktionen. Springer, Berlin-Gottingen-Heidelberg, 1964. 8. G. W. Mackey, Lectures on the Theory of Functions of a Complex Variable. Van Nostrand, Princeton, 1967. 9. R. Nevanlinna, und V. Paatero, Einfiihrung in die Funktionentheorie. Birkhliuser, Basel, 1965. 10. S. Saks and A. Zygmund, Analytic Functions. PWN, Warszawa, 1965. 11. E. C. Titchmarsh, The Theory of Functions. Oxford University Press, 1947. 2. COLLATERAL READING
12. S. Bergman, The Kernel Function and Conformal Mapping. Amer. Math. Soc., New York 1950. 13. R. P. Boas, Entire Functions. Academic Press, New York, 1954. 14. R. Courant, Dirichlet's Principle, Conformal Mapping and Minimal Surfaces. Interscience Pub!., New York, 1950. 15. W. H. J. Fuchs, Topics in the Theory of Functions of One Complex Variable. Van Nostrand, Princeton, 1967. 16. G. M. Golusin, Geometrische Funktionentheorie. VEB Deutscher Verlag der Wissenschaften, Berlin, 1957. 17. W. K. Hayman, Multivalent Functions. Cambridge University Press, 1958. 18. W. K. Hayman, Meromorphic Functions. Oxford University Press, 1964. 19. M. Heins, Selected Topics in the Classical Theory of Functions of a Complex Variable. Holt, Rinehart and Winston, New York, 1962. 20. J. A. Jenkins, Univalent Functions and Conformal Mapping. Springer, Berlin~Gottingen Heidelberg, 1958. 21. G. Julia, Exercices d'analyse, Vol. II. Gauthier-Villars, Paris, 1969. 22. J. E. Littlewood, LecflIres Ofl 'he Theory of Functions. Oxford University Press, 1944.
278
BIBLIOGRAPHY
23. l. P. Natanson, Theorie der Funktionen einer reel/en Veriinderlichen. VEB Deutscher Verlag der Wissenschaften, Berlin, 1954. 24. Z. Nehari, Conformal Mapping. McGraw-Hill, New York, 1952. 25. R. Nevanlinna, Eindeutige analytische Funktionen. Springer, Berlin-Gottingen-Heidelberg, 1953. 26. I. I. Priwalow, Randeigenschaften analytischer Funktionen. VEB Deutscher Verlag der Wissenschaften, Berlin, 1956. 27. M. Tsuji, Potential Theory in Modern Function Theory. Maruzen, Tokyo, 1959. 3. PROBLEMS AND EXERCISES
28. M. A. Evgrafov, Yu. V. Sidorov, M. V. Fedoryuk, M. I. Shabunin :and K. A. Bezhanov, Collection of Problems on the Theory of Analytic Functions (in Russian). Moscow, 1969. 29. K. Knopp, Problem Book in the Theory of Functions. Dover, New York, 1948. 30. G. P61ya und G. Szego, Aufgaben und Lehrsiitze aus der Analysis. Vois. I and II. Springer, Berlin, 1925. 31. M, R. Spiegel, Theory and Problems of Complex Variables. Schaum, New York, 1964. 32. L. I. Volkovyskii, G. L. Lunts, and I. G. Aramanovich, A Collection of Problems on Complex Analysis. Pergamon, Oxford, 1965.
Index Abel's limit theorem, 65 Almost uniformly convergent sequence of functions, 57 Almost uniformly convergent series of functions, 57 Analytic continuation of a function, 95 Analytic element, 95 Analytic function, 19 univalent in a domain, 23 Area of a spherical triangle, 10 Area theorem, 131 Argument principle, 5 4
Bergman kernel function, 119 Bernoulli numbers, 63 Bessel function, 68 Biberbach-Eilenberg function, 137
Cauchy's coefficient formula, 61 Cauchy's integral formula, 38 Cauchy's theorem for a rectangle, 38 for simply connected domains, 38 homological version of, 38 Cauchy-Hadamard formula, 59 Cauchy-Riemann equations, 19 Center of Taylor's series, 60 Chain, 33 Circular symmetrization, 134 Circulation along a contour, 122 Close-to-convex function, 104 Compact family of analytic functions, 75 Complete analytic function, 95 Complete Legendre elliptic integral, 105 Complex function, 19
Complex potential of an electrostatic field, 124 of a flow, 121 Conformal mapping, 23 Conjugate harmonic functions, 21 Continuation of a function analytic, 95 direct analytic, 95, 97 Contour, 34 Convergence of an infinite product, 82 disk of, 59 Convergent infinite product, 82 Convex hull of a set, 6 conv {Zlo z" ... , z.}, 6 Criterion Marty's, for normality, 75 Montel's, 75 Curve Jordan, 34 regular, 17, 18, 33 starshaped image, 94 Cycle 33 homologous to zero, 38 Decomposition of an entire function, 84 .Derivative of a complex domain, 19 Schwarzian, 98 Differentiable complex functIon, 19 Direct analytic continuation of a function, 95, 97 Disk of convergence, 59 Domain Jordan, 110 of hyperbolic type, 135 regular with respect to the Dirichlet problem, 113
INDEX
280 Eilenberg-Biberbach function, 137 Elliptic linear transformation, 15 Elliptic modular function, 99 Electrostatic potential, 124 Entire function, 76 of finite order, 88 Equipotential lines, 121 Essential isolated singularity, 41
Fibonacci sequence, 62 Field of force, 121 stationary, 121 two-dimensional electrostatic, 124 Flow complex potential of, 121 function, 121 stationary two-dimensional, 121 Formula Cauchy's coefficients, 61 Cauchy's integral, 38 Cauchy-Hadamard, 59 for nth derivative, 39 Green's, 125 Jensen's, 80, 81 Legendre's, 107 Poisson'S, 111 Pringsheim's interpolation, 85 Taylor's -42 Formulas, Schwarz.-Christoffel, 100 Fourier series representation, 112 Function analytic, 19 analytic, univalent in a domain, 23 Bergman kernel, 119 Bessel, 68 Biberbach-Eilenberg, 137 cIose-to-convex, 104 en, 105 complete analytic, 95 • complex, 19 differentiable complex, 19 dn, 105 elliptic modular, 99
Function entire, 76 entire, of 'finite order, 88 flow, 121 gamma, 73 Green's, 114 harmonic, 21 holomorphic, 19 integrable complex-valued, of real variable, 33 Koebe's, 131, 166 Pick's 133 regular, 19 subordinate, 92 Weierstrass, t, 106 Weierstrass, f.J 106 Weierstrass, (1 106 Function element, 95 Functions, conjugate harmonic, 21 Gamma function, 73 Gauss and Lucas theorem, 6 Gauss theorem, 125 Genus of a sequence, 84 Goursat lemma, 38 Green's formula, 125 Green's function, 114 Hadamard's three circles theorem, 90 h-area of a regular domain, 18 Harmonic function, 21 Harmonic measure of a system of arcs, 114 h-boundary rotation, 17 h-distance, 18 h-Iength of a regular curve, 18 h-line at infinity, 17 h-parallels, 17 h-rotations, 17 h-segment, 17 h-translation, 17 h-triangle, 18 HeIIy's selection principle, 260 Herglotz theorem, 127 Holomorphic function, 19 Homotopic paths, 98
INDEX
de I'Hospital's rule, 43 Hurwitz theorem, 76 Hyperbolic distance, 17, 18 Hyperbolic length of regular curve, 18 Hyperbolic linear transformation, 15 Hyperbolic metric, 111 Hyperbolic motions, 16 Hyperbolic straight lines, 16
Index of a point, 36 Infinite product, 82 Inner radi us of a domain at a point, 111 of a simply connected domain, 133 Isolated singularity, 41 Intensity, 1~2 Invariant point of a linear transformation, 14 Involution, 14 Integrable complex-valued flmction of a real variable, 33 Integral complete Legendre elliptic, 105 line, of a complex-valued function, 33 Schwarz-Christoffel, 105 StieItjes, 127 unoriented line, 34
Jensen's formula, 80, 81 Jordan curve, 34 Jordan domain, 110 Jordan's theorem for curves starlike w.r.t. an origin, 36 Koebe function, 131, 166 Koebe one quarter theorem, 136
Luarent coefficients, 67 Laurent expansion, 45 Laurent series, 66 principal part of, 66, 67 Legendre formula, 107 Line integral of a complex valued function, 33
2
Lines equipotential, 121 of flow, 121 Liouville's theorem, 40 Lucas and Gauss theorem, 6
Marty's criterion, 75 Mercator projection, 163 Mittag-Leffler representation 77 Montel's compactness condition, 75 Montel's criterion, 75 M -test of Weierstrass, 57
Nevanlinna's characteristic, 81 Normal family of analytic functions, 75 . n-sheeted unit disk, 28
Order of an entire function, 88 ofa pole, 41 Osgood-Taylor-Caratheodory theorem, 110
Parabolic linear transformation, 15 Parallel axiom, 16 Parseval's identity, 59, 119 Pentagram, 103 Phragmen-LindeIOftheorem,115 Pick flllction, 133 Point regular, 61 singular, 61 Points of a hyperbolic plane, 16 Poisson's formula, 111 Poisson's kernel, 112 Pole, 41 Polynomial represented as a product, 84 Potential, 124 electrostatic, 124 complex, of an electrostatic field, 124 velocity, 121 Power series, 58
282 P61ya symmetrization, 134 Principal part of the Laurent series, 66, 67 Pringsheim's interpolation formula, 85 Product convergent infinite, 82 infinite, 82 Weierstrass, 84
Radius inner, 133 of a domain, 111 of convergence of a power series, 59 Reflection of a point with respect to a circle, 12 Regular curve, 17, 18, 33 Regular function, 19 Regular part of the Laurent series, 66 Regular point, 61 Removable singularity, 41 Representation Fourier series, 112 Mittag-Leffler, 77 Residue of an analytic function, 43 Residue theorem, 45 Riemann ~here, 8 Riemann theorem, 11 0 Robin's constant, 126 Rouche's theorem, 54
Schwarzian derivative, 98 Schwarz-Christoffel formulas 100 Schwarz-Christoffel integral 105 Series power 58 Taylor's 60 Sequence Fibonacci 62 of functions almost uniformly convergent 57 Single-valued branch of a function 95 Singular part of the Laurent series 66, 67 Singular point 61
INDEX
Singularity essential isolated 41 isolated 41 removable 41 Sink of intensity 122 Solomon's seal 103 Source of intensity 122 Sphere of Riemann, 8 Spherical derivative of an analytic function, 75 Spherical distance between two points, 9 Spherical image of a complex number, 8 Starshaped image curve, 94 Stationary field, 121 Stationary two-dimensiomil flow, 121 Steiner symmetrization, 134 Stereographic projection, 8 Stieltjes integral, 127 Stieltjes-Osgood theorem, 75 Stolz angle, 65 Subordinate function, 92 Symmetric points w.r.t. a circle, 12 Symmetrization circular, 134 P6lya's, 134 Steiner's, 134 Symmetrization principle, 136 Szego, 134 Taylor coefficients, 61 Taylor's formula, 42 Taylor's series, 60 Theorem . Abel's limit, 65 area, 131 Cauchy's, for a rectangle, 38 Cauchy's, for simply connected domains. 38 Cauchy's, homological version, 38 Hadamard's three circles, 90 Herglotz's, 127 Gauss', 125 Gauss-Lucas, 6 Hurwitz's, 76 Jordan's, 36 Koebe's one quarter, 136
INDEX
Theorem Liouville's, 40 on decomposition of an entire function Weierstrass, 84 Osgoo d-Taylor-Caratheodory, 110 Phragmen-Lindelof, 115 residue, 45 Riemann's, 110 RoucM's, 54 Stieitjes-Osgood, 75 Toeplitz's, 7 two constants, 115 'uniqueness, for the velocity potential, 121 Weierstrass' mean value, 35 Toeplitz's theorem, 7 Toeplitz's transform, 65 Two constants theorem, 115 Two-dimensional electrostatic field, 124
28 Uniqueness theorem for the velocity potential, 121 Unoriented line integral, 34 Velocity potential, 121 Vortex, 122 Weierstrass C-function, 106 Weierstrass SO-function, 106 Weierstrass mean value theorem, 35 Weierstrass M-test, 57 Weierstrass primary factors, 84 Weierstrass product, 84 Weierstrass a-function, 106 Weierstrass theorem on decomposition of an entire function, 84 Winding number, 36
