IL
A Concise Introduction to Complex Function Theory
COMPLEX ANALYSIS AN INVITATION
Murali Rao Henrik
S
COMPLEX AN...
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IL
A Concise Introduction to Complex Function Theory
COMPLEX ANALYSIS AN INVITATION
Murali Rao Henrik
S
COMPLEX ANALYSIS AN INVITATION
A Concise Introduction to Complex Function Theory
COMPLEX ANALYSIS AN INVITATION
Murali Rao Department of Mathematics University of Florida USA
Henrlk Institute of Mathematics University of Aarhus Denmark
%bworId Scientific
Singapore • New Jersey • London • I-long Kong
Published by World Scientific Publishing Co. Pie. Lid.
P0 Box 128, Farrer Road, Singapore 9128 USA office: 687 Hattwefl Street, Teaneck, NJ 07666 UK office: 73 Lynton Mead, Toucridge, London N20 8DH
Library of Congress Cataloging-in-Publication data is available.
COMPLEX ANALYSIS. An Invitation CopyTig$n e
1991
by World Scientific Publishing Co. Pie. Lid.
All rights reserved. This book, or pilLs thereof. may not be in any form orbyanymeans, electronic or mechanical. includisg photocopying, recording orany storage and retrieval system now known or to be without written pirmi sswnfrcen the Publisher.
ISBN 981-02-03756 981-02-0376-4 (pbk)
Printed in Singapore by JBW Printers & Binders Pte. Ltd.
Preface This textbook is a rigorous introduction to the theory of functions of one complex variable. Much of the material has been used in both graduate and undergraduate courses at Aarhus University. ft is our impression that courses in complex function theory these days at many places axe postponed to leave space for other topics like point set topology and measure theory. Thus we have felt free to assume that the students as background have some point set topology and calculus of several variables, and that they understand
e6 arguments. From Chapter X on we even use Lebesgue's dominated and monotone convergence theorems freely. But the proofs are meant for the students and so they are fairly detailed. We have made an effort to whenever possible to give references to literature that is accessible for the students and/or puts the theory in perspective, both mathematically and otherwise. E.g. to the Chauvenet prize winning paper [Za] and to the fascinating gossip on Bloch's life in [Ca] and its sequel [CFJ. Our goal is not to compete with the existing excellent textbooks for historical notes and remarks or for wealth of material. For that we refer the interested reader to, say the monumental work [Bu] and the classic [SG].
We have tried to reach some of the deeper and more interesting results (Picard's theorems, Riemann's mapping theorem, Runge's approximation theorems) rather early, and nevertheless to give the very basic theory an adequate treatment. Standard notation is enforced throughout. A possible exception is that B[a,r] is the closed ball with center a and radius r in analogy with the notation for a closed interval. An important part of any course is the set of exercises. We have exercises after
each chapter. They are meant to be doable for the students, so we have quite often provided hints about how to proceed. A couple of times we have succumbed to the temptation of making a digression to an interesting topic, that will not be pursued, e.g. Tauber's theorem, Fladamard's gap theorem and the prime number theorem. We hope that the reader will be irresistibly tempted as well. The authors welcome correspondence with criticism and suggestions. In particular about literature on the level of students that axe about to start their graduate studies in mathematics.
V
Contents Preface
facts. ._.4 •
V
.............. 1
Chapter 1. Power Series ...
Section 1 Elementary
1
Section2ThetheoremsofAbelandTauber Section 3 Liouville's theorem Section 4 Important power series Section 5 Exercises
7 8
9
Chapter 2. Holomorphic and Analytic Functions .........
.13
Section 1 Basics of complex calculus Section 2 Line integrals Section 3 Exercises
13 17
22
Chapter 3. The Exponential Function, the Logarithm and the Winding Number 23
Section 1 The exponential function Section 2 Logarithm, argument and power Section 3 Existence of continuous logarithms Section 4 The winding number Section 5 Square roots Section 6 Exercises
24 28 31
35 37
Chapter 4. Basic Theory of Holomorphic Functions ... 43 50 53 58 60
Section 1 The Cauchy-Goursat integral theorem
Section 2 Selected consequences of the Cauchy integral formula Section 3 The open mapping theorem gap theorem Section 4 Section 5 Exercises
Chapter 5. Global Theory .......
.........
1
Section 1 The global Cauchy integral theorem Section 2 Simply connected sets Section 3 Exercises
Chapter 6. Isolated Singularities.....
..........
Section 1 Laurent series
71
75 77 ............
......................
79
79 Vu
Contents
Section 2 The classification of isolated singularities Section 3 The residue theorem Topic 1 The statement Topic 2 Example A Topic 3 Example B Topic 4 Example C Section 4 Exercises
Section 1 Liouville's and Casorati-Weiersuass' theorems
.82 84 84 85 87 89
92
99 100 106 108 112
Section 2 Picard's two theorems Section 3 Exercises Section 4 Alternative treatment
Section5Exercises
Chapter 8. Geometric Aspects and the Riemann Mapping Theorem ...................... 113 113 115 120 122 125 125
Section 1 The Riemann sphere Section 2 The Mobius transformations
Section 3 MonteL's theorem Section 4 The Ricmann mapping theorem Section 5 Primitives Section 6 Exercises
Chapter 9. Meromorphic Functions and Runge's Theorems..... Section 1 The argument principle Section 2 Rouches theorem Section 3 Runge's theorems Section 4 The inhomogeneous Cauchy-Riemann equation Section 5 Exercises
Chapter 10. Representations of Meromorphic Functions ..... Section 1 Infinite products Section 2 The Euler formula for sine factorization theorem Section 3 Section 4 The r-function Section 5 The Mittag-Leffler expansion Section 6 The g- and p-functions of Weierstrass vu'
................. 129 129 131
135 140 144 .......... 147
147 151
153 157 161
163
.165
Section 7 Exercises
Chapter 11. The Prune .169 .173
Section 1 The Riemann zeta function Section 2 Euler's product formula and zeros of ç
Section3Moreaboutthezerosofç
176
Section 4 The prime number theorem Section 5 Exercises
177 181
Chapter 12. Harmonic Functions .............
......
..........
..183
..
Section 1 Holomorphic and harmonic functions Section 2 Poisson's formula Section 3 Jensen's formula Section 4 Exercises
183 187 192
Chapter 13. Subharmonic Functions................ Section 1 Technical results on upper semicontinuous functions Section 2 Introductory properties of subharmonic functions
199 199 201
Section3Onthesetwhereu=_oo
203
Section 4 Approximation by smooth functions Section 5 Constructing subharmonic functions Section 6 Applications Topic 1 Radó's theorem Topic 2 Hardy spaces Topic 3 F. and R. Nevanlinna's theorem
205 208
195
210 210 211
215 216
Sectionl Exercises Chapter 14. Various Applications.............
........
Section 1 The Phragm6n-Lindelöf principle Section 2 The Riesz-Thorin interpolation theorem Section 3 M. Riesz' theorem Section 4 Exercises
......
........
..
219
219 221
223 229
References
231
Index
237 ix
Chapter 1 Power Series SectIon 1 Elementary facts This section contains basic results about convergence of power series. The simplest
and most important example is the geometric series
which converges absolutely for any z in the open unit disc. It emerges in so many other contexts than complex function theory that it may be considered one of the fundamental elements of mathematics (See the thought-provoking paper [Ha]).
Definition I. A power series around
E C is a formal series of the form 00 k
(1) k=O
wherethe coefficients 0k E We
Care fixed and wherez E C. =
shall very often only consider the case (2)
0, i.e.
>.akzk
because it will be obvious how to derive the general case from this more handy special case. We shall in Chapter Vl,* 1 encounter Laurent series, i.e. "power" series in which the summation ranges over Z, not just N. A Laurent series will be a a power series in z and another in the variable l/z.
Proposition 2. Considerthe power series (2), and let us assume that the set {a&(klk = 0,1,2,...) is bounded/or some C. the power series (2) converges absolutely and umformly in the Then/or any p < closed disc B[0,]] = {z E CIIzI p}.
M for some M and all k. Thus if
Proof: Because of boundedness IakII(kI Izi
p
jf(z)I then
m. 7
b
Chapter).
Power
Of course Liouville's theorem does not hold for smooth functions of one or more real variables. For example, the function f(x) = sin x is a bounded, smooth function of one real variable, but it is certainly not constant.
Liouville's theorem can be generalized to harmonic functions (See Chapter XII. Exercise 10). Pmof. It suffices to prove (ii). The coefficient found by the formula
J
=
may for any n =
for any R >
0,
1,2,..- be
o
which crops up when we introduce the power series expansion of f on the right hand side. In particular we get the estimate
lani
b Letting R .—' 00 we see that
thepolynomialf(z)=
N
n0
=
0
whenever n > m, so the power series reduces to
,whereNistheintegerpartofm.
0
For proof without calculus (!) see [Le].
SectIon 4 Important power series In this paragraph we collect the most important power series expansions. Some of them will be derived later. In particular we will discuss the exponential function in the next chapter.
Exetcises
Section 5. 00
1
IzI— n! n=O
e+e_ix
cosz=
2
n=o 00
—
e
S1flZ
fl
00
=
2i
— —
for z E C
(2n) (—1)
n0
n
(2n+l)!r
2n-i-1
z
C
for z
C
for
00
cosh
z=
2
n0 00
—
sinhz=
1
—
2
n0
(2n+l)!Z
2ti+1
00
z" for IzI
00
arCSiflZ=>
n
for IzI
0 ,
then
0 for all IzI > 1
10
Section 5.
Exercises
(Hint: Abel's partial summation formula). (B) Deduce from (A), that
> an_I >
> ai >
> 0 implies that EOL.Zk
hasall its roots in the open disc IzI 0. Show for each
=
that
1
and such that
1 for all n
as not a root of unity [i.e.
such
11 that
=0 uniformly in z on compact subsets of B(0, p). Deduce the Cauchy inequaluy
sup lf(z)t foreach rE]0,p(
zlr
9.
(A generalization of Abel's theorem)
Let us assume that the power series f(z) := >2akzL converges for z = hence for each Izi
0. Show that Eakzk
(.8) Let B E
1
that Eokzk converges to f(z) uniformly on the set
!)
{zEB(0,1)II1—zIcos9 , 10. theorem
Show by dint of an example that the assumption kak cannot be deleted. 11
—.
L—oo
0
in Tauber's
Chapter 2 Holomorphic and Analytic Functions SectIon 1 BasIcs of complex calculus
f
Definition 1. Let f be a complex valued function defined on an open set is said ta be complex differentiable at E C if the limit (1)
lim
in the complex plane.
f(zo+h)—f(zo) h
h—O
exLct.s in C.
The limit is called the complex derivative off at zo and is denoted is complex differentiable a: each point of Il. The Lv holomorphic on derivative of I or just the derivative. If C is then called the complex : —. function f there exists a holomorphic function F defined on such that F' = f. we say that F is a primitive off. III is holomorphic in all of C then f is said to be entire.
f
Like in real variable theory we find that I is continuous on an open set Q if it is holomorphic on ft By routine calculations the usual rules for differentiation also hold in the complex case:
Theorem 2. (a) 1ff andgare holomorphic on Il then so are f+gandfg, and(f + g)' =
(fg)'
f'g + fg'.
(b) Let f and g be holomorphic on ft If g is not identically 0 on c then f/g is holomorphic on the open set {z E O}, and on this set
(fV_ f'g—fg' g2 (c) The chain rule: 1ff is holomorphic on g on then the composition g o f is holomorphic on fl, and
(go f)'(z) = g'(f(z))f'(z) for all
and f(
is contained in IZ1,
E fl
As is easy to see from the theorem, the set of functions which are holomorphic on fl forms an algebra over the complex numbers. We denote this algebra
Examples. (a) Constants are holomorphic and their derivatives are 0. 13
Chapter 2.
Holomo$ic
and
Analytic Functions
is holomorphic and its derivative is 1. is holomorphic on C for n = 0,1,2,... , and on C\{0} (C) The function z for n = -.1, —2,.... In both cases with derivative (z")' = (b) The function z
z
(d) A polynomial p(z) = Eaazk is entire and p'(z) = Theorem 3.
Let 1: Q —' C
, where is an open subset of the complex plane, and write u and v are real valued functions. f= (a) 1ff it holomorphic on ci then all the first order partial derivatives of u and v exist in ci and the Cauchy-Riemann equations are satisfied there, i.e.
u
+ iv .
where
OuOv ,40u
Ov
or in a more compact notation
Of
=0, where 0 := 110
.0
+z
(b) 1ff E C'(f 1) satisfies the Cauchy-Rie,nann equations in ci then f is holomorphic
Proof (a) This is seen from (I) by letting h tend to 0 along the real and the imaginary axes respectively. (b) Left to the reader as Exercise 4.
0
Remarks on the Cauchy..Rlemann equations: The differentiability conditions in point (b) of the theorem can be relaxed considerably. One generalization is the LoomanMenchoff theorem:
Let f
C(ci). Then f
HoI(f), if Of/Ox and Of/Oy exist in all of ci and
satisfy the Cauchy-Riemann equations there (See Theorem 1.6.1 p.48 of [Na]). Another generalization: It suffices that the Cauchy-Riemann equations are satisfied in the sense of distributions (See [Tr,Theorem p.36]). For more information see [GM].
The Cauchy-Riemann equations say roughly speaking that I does not depend on
1, and so it is a function of z only. If f is holomorphic on an open set then (use the Cauchy-Riemann equations) = Igradul2 = Igradvl2 , and lgradul2 = (Ou/Ox)2 + (Ou/Oy)2. In particular, 1' = 0 throughout ci implies that f is constant on each connected component of ci. Not all functions are holomorphic: E.g. z is not. As we saw in an example above, the functions fo = 1 and fi(z) = z are entire. Hence so is each polynomial in z. And each rational function, i.e. each function of 14
Secdcn 1.
Basics at conipici calculus
form / = P/Q where P and Q are polynomials, is holomorphic off the zeros of the denominator Q. It turns out (Propositions ilLS and 11L20 of the next chapter) that continuous roots and logarithms are holomorphic, too. So are power series; we give a direct proof here. A shorter, but more sophisticated proof is given below following Lemma 12. the
ProposItion 4. The swn of a power series is holomoiphic inside its circle of convergence, and its derivative can be go: by term-by-term differentiation: If
1(z) =
—
has radius of convergence p > 0 , then I is holomorphic on B(zo, p) and
f(z) = >aRn(z
z
—
B(zo,p)
In particular all the complex derivatives f', f",... exist in B(zo, p). Proof: Observe that the formula for f makes sense by Comflaiy 1.4. Let us for convenience of writing assume that zo = 0. The technical key to the proof is the following inequality: + h)R — ZR —
+
validfornEN,h,zEC suchihat O<JhIo. f(z+h)—f(z) =
—
—
as h—eO The inequality follows from the binomial formula:
(z +
h)R
—
Zn —
— z'1 —
=
=
k=2
=
{t2 15
and Analytic Funcdons
Chapter 2.
so
I(z +
—
—
k=2
=
k=2
k=2
= (Izi +
Definition 5. Let be an open subset of C. Afuncnon f : —. C is said tobe analytic on cZ if there exists open for each point zo C disc B(zo,p) in Q in which can be written an as the swn of a power series centered at zo, i.e. I can be written
f
1(z) =
— zn)"
for z E B(:o,p)
We have already observed that the coefficients are uniquely determined by f (Proposition 1.6). By say what the coefficients are
= So,
for n
} in the power series around zn help of Proposition 4 we can even
= 0,1,2,...
restricting to the real line, we see that the power series expansion of an analytic
with
function coincides
complex numbers the always exists a C°° in fact
(Sec
its
power
function
Taylor
series
expansion. may or
series E on the
Given any
sequence (an) of
may not converge around
real line with E an
as
its
Taylor
0. There
series. Many
e.g. [Me]). But such functions will in general not be restrictions of power
series to the real line.
A consequence
of Proposition
4 is that an analytic
function
is holomorphic, a result
which is nice, but certainly not surprising. What is surprising and remarkable is that
the converse
is
true (as will be proved
assumption that I
is
in
Chapter IV). Under the (apparently) humble
complex differentiable, we shall infer that I
is
infinitely often
even that f around each point in its domain of definition can be expanded in a power series. The situation is vastly different in the case of functions on the real line. There the differentiable and
derivative of a differentiable function need not even be continuous, let alone expandable
in a power series.
complex plane the word "holomorphic" is synonymous with the word "analytic". However, until we have proved that result, we must distinguish between the two concepts. But for
functions in
the
16
Sccuon 2.
Line integrals
SectIon 2 LIne Integrals In this paragraph we introduce the line integrals which axe important tools in complex function theory. We shall until further notice not require any finer theory of integration and convergence theorems. All we need is to consider piecewise continuous functions on the real line and uniform convergence.
Definition 6. [a, bJ —. C where —oo < a < b < oo, is said A curve (i.e. a continuous map) to be a path it is piecewise thfferentiable. This means that there are points •
such that the derivative exists in the open subi,uerval Jt,, t,+i[ and extends to a for each j = 0,1,2,.... continuous function on the closed subinten'al [t1, The length 1(7) of the path is the number 1(7) :=
I
17'(t)Idt E [0, oo[
To be precise, we distinguish between the curve which is a map, and its image -(([a, bJ) which Is a compact subset of C. We will say that a complex number z belongs (or does not belong) to the curve We express the same thing by saying that 7 passes through z z E 7' (resp. z (resp. does not pass through z). The curveis saidto be closed -,'(a) = Is contained in a subset (1 of C then we will say that is a curve in (1.
If'
Examples 7. (a) The positively oriented circle
around
(t) = zo + Re" for I
C with radius R> 0, i.e. the curve (0,2x1
is a closed path with length 2,rR. We will often write Iz — zoI = R instead of . (b) The line segment [A, BJ, where A and B are complex numbers, i.e. the curve
[A,BJ:=(1—t)A+tB for IE[O,1J isapath of Length l([A,B1) = lB—Al
(c)LetA,B,CE C. and C (the order is important!) we understand the closed curve defined by
([A,BJ(t) fcrt€[0,1J [B,C](t—1) foriE[1,2] { L[C,A](t—2) 17
fortE[2,3]
chapter 2. Holomoqthic and Analytic Functions
his apath of length
= IB—AI+IC—BI+IA—Cf.
The length of a path does not depend on its parametrization. More formally:
Proposition 8. Let-y: (a,b) C bea path. monotone map of [c,dJ onto [a,b].
Ic,dj Then
[a,b] bea
1(7) = 1(70
Proof:
0
The change of variables theorem.
Definition 9. Let [a,b] —. C be apath, and letf : = -,((a,bJ) function. The line integral of f along 7 is the complex nwnber :
C be a continuous
Ji = J f(z)dz :=J To emphasize what the variable is, we sometimes write f f(w)dw or the like instead
of ff(z)dz. The next proposition collects properties of the line integral which will be useful for us in the sequel.
Proposition 10. Let (a, 6) —. C be
a
path, and let
f be
a continuous, complex-valued function
ony'. Then (a)
J f(z)dz
sup zEir
7
(b) (A primitive version of the fact that the line integral is invariant under orientation-
preserving changes of parameter). Let —oo < c < d < 00. If 4: [c, d]
(a, bJ has the form
c6(s)=ks+1,sE(c,d),wherek,1ER and if ,naps (c, d] onto (a, b]
then
Ji = sign(lc)J i 18
Section 2.
Line integrals
For two points A and B in C we have in particular
Jf=-Jf IA,Bl
(C) LetcE]a,b[
LB,A)
and define paths 7J and 72
by
:= 7(1) for t E [a,cj 72(t) := 7(1) for I E [c,bl Then
If C
Jf=Jf+Jf E [A, BJ' , where A and B are points of C, then in particular
1+
11= 1 tA,Bl
[A,CJ
fi
IC,B)
(d) If {f,j is a sequence of continuous functions on -y' that converges wufonnly on
to f, then
/ /
j
as n —+
(e) If f is defined on a neighborhood of-y and has a prisninve, say F, there, then
Ji = F(7(b)) Ifflirthennore -y is closed then ff =
—
F(7(a))
0.
7
0
Proof: Left to the reader.
We conclude this chapter by some applications of line integrals. They are investments for the future. Definition 11. A subset
oil
A of
is said to be starshaped, if there exists a point a E A such that
the segments
{tx+(1—t)aItE[0,1J}
,
xEA
are contained in A; we say that A is slarshaped with respect to a. 19
Chapter 2.
Holocnopldc and Analytic Functions
Clearly any convex set is starshaped. An interval is a simple, but important special case.
Lemma 12. Let ci be an open subset of C which Is starshaped with respect to a point a E ci.
Letg E C(ci)have:hepropersythat f g=O,wh never
in Il witha
8A
as one of its vertices. Then the fisnction
zEfZ (a,zJ
is holomorphic on Cl and G' =g. Inparticularghasaprhnitive. Proof: Let z E ci. If hE C is so small that [z,z+h) is contained in Cl then we get from our assumption that
G(z+h)—G(z)= J g—Jg= J 9=hJ9(z+th)dt (a,zJ
ts,a+hJ
0
Division through by h and use of the continuity of g at z gives us the lemma. We can now give the promised simpler proof of Proposition 4: Proof of Proposition 4: Let us, as in the earlier proof, for convenience take zo =0. The power series
g(z) converges according to Corollary 1.4 uniformly on compact subsets of B(O, p) , so term by term integration is permitted and yields
1(z)
=
J g(w)dw+ao for zEB(O,p)
I0,zJ
By Lemma 12 it now suffices to prove that the integral of g along any triangle B(O, p) vanishes. But that is easy: Indeed,
/ g(z)dz =
>aitnJ z"1dz = 0
because the integral of z"t vanishes (z"1 has a primitive, viz. zn/n). 20
in
Sec*ioz 2.
Exercises
Lemma 13. Letçó: —.
Thenthefuncuon ,
zEC\7'
is anoiytic, hence holomorphic and
f(")(z)
=
for n = 0,1,2,...
(w
If C\7' then the power series expansion off around zo converges everywhere in the open disc aivund zo with radius dist(zo, As support for our memory we note that the formula for differentiate f under the integral sign.
appears
when we
Proof:
If
—
zol 0? Exhibit a specimen if your answer is yes. 25. Let h : S' C\{0} be continuous, and consider the closed curve 7(t) := h(exp(it)) for t E [0,2ir] (a) Show that h can be written in the form h(z) = z"exp(d(z)) , where n is an integer and h(e") and C(S1) . Hint: Let F be a continuous logarithm of t n = Ind7(O), so that h(c*t) Note that the exponent
a function
=
=
:= F(t) — mt satisfies that
=
and so defines
on S1.
(8) Show that Ind7(O) is even if h is even. In particular, if h is odd then Ind7(O)
(y) Show that
0.
26. We will in this exercise present a proof of the 2-dimensional version of the so-called
Erouwer's Theorem on Invariance of Domain: Let be an open subset of the complex plane, and let f : C be continuous and 1-1. Then f(Q) is open in C. (a) Verify that the theorem is a consequence of the following Lemma. If the function f : B[0, 1] C is continuous and 1-1, then 1(0) is an interior point of f(B[0,1)). We proceed with a proof of the lemma: (fl)Showthatwemayassumethatf(0) = 0,andthatthenr := dist(0,f(S')) >0. (y) Show that it suffices to lead the following assumption to a contradiction: There exists a point zo B(0,r)\f(B[0,1)). (6) Consider now the closed curve'y(t) := for t [0,2irl. Show that 0 = Ind7(zo) = Ind7(0).
(e) Show that F: S' x [0,1]
C\{0}
,
F(z,3) := f(j-f)
given by —
is a homotopy in C\{0} between I Is' and the odd function
h(z) := F(z,1) =
—
(() Use Exercise 25(7) to arrive at the desired conclusion. 41
Chapter 3. The Exponential Function, the Logarithm and the Winding Number
27. (a) Show that there at any given time are opposite ends of the earth which have the same weather, i.e. the same temperature and the same barometer pressure. This is a meteorological interpretation of the 2-dimensional version of Borsuk-Ulam's Theorem. x E R3 I Let S2 = 1) be the unit sphere. 1ff: S2 —+ C is continuous then { there exists an such that f(xo) = f(—zo). E (fi) Prove the Borsuk-Ulam theorem. Hint: Consider the map !i : B[0, 1] —. C given by
h(x + iy) := f(z,y, s/i —
—
y2)
—
x2 — y2)
the closed curve 7(t) := h(exp(it)) in C. (-y) Prove the following : If a balloon is deflated and laid on the floor, then there exist two antipodal points which end up over the same point of the floor. and
Hol(fZ) and assume that the zeros of I all are 28. (Cf Proposition 20) Let f isolated. Let r be a continuous square root of 1. Show that r E
42
Chapter 4 Basic Theory of Holomorphic Functions While Chapter III essentially teeats continuous functions and a bit of topology of the complex plane, this chapter specializes to holomorphic functions and their surprising properties, derived from the Cauchy integral theorem. The contents of this chapter are dealt with in all introductory text books on complex function theory.
SectIon 1 The Cauchy-Goursat Integral theorem The main results of this paragraph are the Cauchy-Goursat integral theorem and
some of its surprising consequences: The integral of a holomorphic function I around a closed curve, the interior of which is contained in the domain of definition of f, is zero. A holomorphic function is analytic, i.e. it can be expanded in a power series. Theorem 1 (Cauchy-Goursat). Let be holomorphic in an open subset of the complex plane. Let a, b.c E Q and with vertices a, b and c is contained in ft Then assume that the triangle
f
Jf(z)dz =0 Proof.
C
a
b
Divide the triangle as in the figure, using the vertices and the midpoints of the segments [a,bJ, [b,cJ and tc,a). We get 4 smaller triangles and as 43
Chapter 4. Basic Theory o( Holomorphic Functions
shown in the figure. Orienting all the triangles in the counter clockwise direction it is a matter of direct verification that
J f(z)dz=> J Denoting the absolute value of the left hand side for a, we have for at least one small triangle call it that
Note that where I denotes length of the path in question. = We can repeat the process to successively obtain triangles such that
=
and
There is a point, say
f is differentiable at
in the intersection niJI
fi > 0 we have from a certain n on, in the entire triangle (*) 11(z) — f(zo) — f'(zo)(z — zo)I
so given
that — 201
The constant function f(zo) and the function z — have primitives, so (by Proposition H.1O.(e)) the line integrals of these along any closed path vanish. Thus where (*) holds: for any
= where we have used that zo for any z and that Iz — zol ( a. Since this holds for any > 0 we conclude that a = 0.
Now,
0
By a bit of ingenuity we can get a stronger looking version of Theorem I Out by allowing the function f to have singularities
Corollarj 2. Let be an open subset of C, and let f be holomorphic in Then fiirther,nore thai (z — zo)f(z) —, 0 as z —,
Jf(z)dz =0 44
Q\ 4
}.
Assume
Section 1. The Cauchy-Cioursat integral theorem
for any triangle
in Il such that zo
Proof:
c
b
a
Let 6 be a small triangle inside of & containing zn in its interior. Subdividing into smaller triangles we get by the Cauchy-Goursat theorem that
Jf(z)dz =Jf(z)dZ Let us now consider a sequence Oi, of equilateral triangles, each with Zn as its center, and with diameters shrinking to 0. By elementary geometry, when we choose By assumption so that = 1/n then = where
lf(z)I
c(z)
—' 0
Let e > 0 be given. Choosing n so large that Ic(Z)I
= The
statement
[i
of the
as z
—'
e for
=
sup
corollary
follows
we get
all z E
since this is true
for
any e >
0.
D
From Corollary 2 we deduce a remarkable and important regularity property: A holomorphic function cannot possess a minor singularity at a point. If has a point singularity it is a serious one. This result is called "The removable singularity theorem" or "Riemann's extension theorem": it
45
Basic Theory of Holomorphic Functions
chapter 4.
Theorem 3 (The removable singularity theorem). Let be a point in an open subset Q of:he complex plane, and let f : Q\{zn} C be holomorphic in 1Z\ (Zn). fin particular 4ff is bounded near then f can Jf(z — zo)f(z) —. 0 as z —+ be extended to a function which is holomorphic in all of Q.
Proof Without loss of generality we take be a triangle, = 0 in the proof. Let contained in Q, with positively oriented (Cf = 0 in its interior, and its boundary Example 111.13) so that (Theorem 111.17(d))
1=
fdw
1
2irij w—z for any z E I
each of the two singularities 0 and z of the function
For any fixed z E
on
satisfies the condition of Corollary 2. Subdividing
into two triangles, each with only
one of the singularities, we get
j
f(w)
f(z)d =
which implies that
1(z)
for any z
=
The right side is analytic and hence holomorphic off extension F may unambiguously be defined by
(1(z)
(Lemma 11.13), so the desired
for z
I.
OA
0 We see that Corollary 2 actually does not strengthen Theorem 1, because the exceptional point is not a singularity after all! The corresponding result is not true in one real variable: The function t is not differentiable at 0. During the proof of Theorem 3 we noticed the remarkable fact that a holomorphic function is analytic. We shall in Theorem 8 below prove that its power series converges in the largest possible disc. 46
Section 1. The Cauchy.Oowsat integral theoitm
As a by-product of "The removable singularity theorem" we note the following example which will be used several times, both in the near and more distant future. Example 4. If f is holomorphic in the open set
g(z) :=
and a E Q then the function for z E
1
forz=a
If'(a)
is holomorphic in all of Theorem 5 (The Cauchy Integral theorem). Let f be holomorphic in an open starshaped subset f(z)dz = 0 for each closed path in Q.
of the complex plane. Then
Proof. When we combine the Cauchy-Goursat theorem with Lemma 11.12 we see so we may refer to Proposition 11.10(e). that f has a primitive in 0
Theorem 6 (The Cauchy integral formula). Let f be holomorphic in an open starshaped set have for any closed path in the formula
Ind7(z)f(z)
Proof Fix z E
of the complex plane. Then we
bra!!
=
z E Q\'y
Then the function for w E
for w=z
tf'(z)
is according to Example 4 holomorphic in fl, so by the Cauchy integral theorem its integral along 'y is 0, i.e.
=0 Recalling the formula (Theorem 111.17(d))
Ind7(z) =
1 1 dw 2irzjI w—z
7
we get the theorem.
47
Chaper 4. Basic Theocy 01 Holomorphic Functions As a special case we note the Cauchy integral formula for a disc:
Theorem 7.
1ff is holomorphic in a neighborhood of the closed disc Bizn, r], then we have for each z E B(
r) the formula
fH' J
w—z
'w—.zo'=r
The purpose of the next chapter is to extend the Cauchy integral theorem and the Cauchy integral formula to more general domains than starshaped. The following remark (which will not be used later) points in another direction.
Remark. There is an extension of the Cauchy integral formula to general functions. It is often called the Cauchy-Greenfornuda since Green's theorem is used to prove it (For a proof, see e.g. [HO;Theorem 1.2.11): Let w be a bounded open domain in C with a smooth positively oriented boundary -y.
If u is C' in a neighborhood
u(z) =
1 u(w)
1
2irz
j7
w—z
then dw
1
—
—I ir
jx—z
dm(x) for z E w
dm denotes Lebesgue measure on C = R2. Note that the last term on the right hand side drops out if u is holomorphic, so that the formula in that case reduces to the ordinary Cauchy integral formula. A version of the Cauchy integral theorem can be derived from the Cauchy-Green theorem (replace u by (w — z)u(w) etc.): where
11 u(z)dz
j
=
_1fOu
j F(x)dm(z)
There is a striking contrast between R- and C-differentiability: There are examplea of functions on R with the property that f is continuously differentiable and f' is not differentiable anywhere. But as we noticed after Theorem 3 any C-differentiable function, i.e. a holomorphic function, is in fact even analytic and hence in particular infinitely often differentiable. In the future we will use the words analytic and holomorphic interchangeably. What is new in Theorem 8 below is that the power series converges in the largest possible disc.
48
________
Section 1. The Cauchy-Goursat integral
Theorem 8. Let f be holomorphic in an open subset fl of the complex plane. Then I is analytic in any zo E fI the power series expansion off around zo converges in the largest open disc in around zo. In particular, I is infinitely often C-differentiable and all derivatives f', f",... are holomorphic.
partial derivatives of all orders with respect to the real variables x and y exist and are continuous.
Also I
i.e.
Proof. Combine Lemma 11.13 and Theorem 7 with the uniqueness of power series expansions. For the last statement we note that L =
and
= if'
from which we by induction get ç
OxmOyn
— —
0 It may be remarked that Theorem 8 does not hold for an analytic function of a real variable. To take an example, the function
f(x)
1+x2
for xER
is analytic on all of the real line, but its power series expansion
f(z)= 1 converges only for —1 < z < 1. The explanation is that if we view f as a function of a complex variable 1(z) :=
1+'z2 for zEC
then f is not holomorphic = analytic on all of the complex plane, but only on the domain C\{i, —i} , and the biggest disc around 0 in that domain is B(O, 1). We next present a converse to the Cauchy-Goursat theorem, viz. Morera's theorem, which is often expedient in determining whether a given function is holomorphic. It can be used in situations where a direct resort to the definition - estimation of difference quotients etc. - is hopeless or at least very complicated. For a more thorough discussion of Morera's theorem we refer to the beautiful article (Za]. For the latest news see [GIl.
4°
Chapter 4.
Baiic Theory ci Holomorphic Fimcucns
Theorem 9 (Morera's theorem). Let f : — C be a continuous function on an open subset Q of the complex plane. Assume that each point z E has a neighborhood ç Il with the property that f f( z )dz = 0 for all triangles contained in U1. Then f LS holomorphic in
Proof: Let z E and choose r > 0 so small that B(z,r) c U1. Ii suffices to prove that f is holomorphic in B(z,r) for each z E f has by Lemma 11.12 a primitive F in D(z, r). Being C.differentiable means that F is holomorphic. By
0
Theorem 8 so is its derivative f.
Section 2 Selected consequences of the Cauchy integral formula It is a sad fact from the theory of functions of a real variable that a uniform limit of a sequence of differentiable functions need not be differentiable, although ii is continuous. Indeed, there exists a continuous nowhere differentiable function on 10,1], and (by the classical Weierstrass approximation theorem) it is a uniform limit of a sequence of polynomials.
The situation is quite the contrary for holomorphic functions, as demonstrated by the following surprising result, in which no assumptions are made on the derivatives: Theorem 10 (Welerstrass' theorem). be a sequence of functions which are holomorphic in an open subset Let I i, of the complex plane. Assume that the sequence converges, wufornily on each compact
subset of Q, toafunctionf. Furthermore, the sequence Then f Lc aLso holomorphic in converges, again wufonnly on each compact subset of to
of derivatives
f.
We shall later (Exercise VI11.10) meet a stronger version of Weierstrass' theorem, viz. Vitali-Porter's theorem. The assumption about uniform convergence is important; poiniwise convergence does not suffice. For a discussion see [Dii] and [Za* 111.
Proof
LetzoEQandchooser>Obesosmalltha*B[zo,r]çcl. Accordingtothe Cauchy integral formula (Theorem 7) we have (*)
1
t J
dw
50
forall
of the Cauchy integral formula
Section 2. Selected
and going to the limit n
we get
00
f
1 (**) f(z)=— 2irz
i
j
f(w) w—z
dw forall ZEB(zo,r)
I
which [by Lemma 11.131 shows that f is analytic and hence holomorphic in B(zo,r).
Now f E Hol(fl), the disc being arbitrary. Concerning the last statement we may now assume that f = 0 [if necessaiy we replace f, by — f] By a compactness argument it suffices to show that {f } converges uniformly to ç Q. Differentiating (*) we get such that B[zo, 0 as n — oo on any ball D(zo, [directly or using Lemma 11.131 that
(2dW
J
forall
1w—
z
0
Let now w be any point such that Jw — f(zo)j < R , and let us apply — w and H : For IC — zoI = r we get
to
the two functions F(z) = f(z) IF(C) —
= If(zo) — wI
values f(z) also lie in that disc. Such considerations form the basis for the so-called Principle of Subordination. For more information on that topic we refer the reader to the paper [Mac). Another important application of Schwarz' lemma is to the uniqueness question of conformal mappings, so we will return to Schwarz' lemma during our treatment of the Riemann Mapping Theorem. We have already earlier (Theorem 111.5) observed that any continuous function g satisfying (exp og)(z) = z, i.e. any continuous logarithm, automatically is holomorphic. And we have claimed that an assumption in Proposition 111.20 (that f never vanishes) is superfluous. Both these results axe special cases of our next theorem. Theorem 18. Let g : ci .-. C be a continuous function on an open subset ci of the complex plane, and let f be a non-constaju holomorphic function, defined on an open connected subset of the complex plane containing g(ci). If the composite map f o g is holomorphic on ci, then so is g. Proof: Since holomorphy is a Local property we may in the proof assume that fl is connected. Let a ci be arbitrary. We shall show that g is complex differentiable at a. 56
Section 3. The open mapping theorem
Since f is holomorphic around g(a) we can write
f(w) — f(g(a)) = (w — g(a))k H(w) 0. So there is an open disc D around g(a) where H is holomorphic and H(g(a)) in which H never vanishes. By Proposition 111.21 H has a holomorphic kth root fo on D. So we write f(w) — f(g(a)) = [(w — g(a))fo(w)Jk for all w near g(a)
Here k is an integer 1, fo E HoI(B(g(a), 6)) for some 6 > 0 and fo(g(a)) 0. Replacing w by g(z) and using the abbreviation h for the holomorphic function h = o g — f(g(a)), we get for z in a neighborhood of a, say for z in B(a, €) ç ci,
f
that
h(z) = [(g(z) — so we see that the bracket on the right hand side is a continuous kth root r of the holomorphic function h on the left hand side. There are now two cases: 1) h is identically 0 in B(a, e). In this case g(z) = g(a) near z = a, so g is clearly complex differentiable at a. 2) h is not identically 0 in B(a, e). In this case we know by the Unique Continuation Theorem (Theorem II) that a is an isolated zero for h. According to Proposition 111.20 the root r is holomorphic in B(a, e)\h'(O) , so a is an isolated singularity of r. But r is continuous in B(a, e), so by the Removable Singularity Theorem (Theorem 3) r is actually holomorphic in a neighborhood of a. Dividing the identity r(z) = (g(z) — g(a))fo(g(z)) by z — a and noting that r(a) = 0 we get
r(z)—r(a) =
9(Z)_9(a)f(())
The left hand side has a limit as z .-. a, and, since fo(g(a)) difference quotient of g, i.e. g is differentiable at a. Terminology 19. A function is said to be univalent,
0, so does the
0
it is holomorphic and infective.
Theorem 20.
1ff is univale,u in an open set Il, then f(1l) is open and
is holomorphic on f(1l).
Being open, Cl is a union of open balls. f is injective so it is not constant on any such ball. Hence f(cl) is open by the Open Mapping Theorem (Theorem 14). Since f is an open map, f_1 is continuous, so we can apply Theorem 18 with g = 0
f*
The reader might note that a result corresponding to Theorem 20 is false for functions of one real variable: The inverse of a differentiable function need not be differentiable (Example f(x) = x3). 57
Chapter 4. Bask Theory o( Holomorphic Functions
Theorem 21. Let f be holomorphic in an open set of the complex plane and let E there exists a neighborhood of zo on which f Is univoJetu and only if f'(zo)
Then
0.
Proot Then Let us first assume that f is univalent on the open neighborhood Q of g = ' is holomorphic on f(1Z) by Theorem 20, so we may differentiate the identity g(f(z)) = z . We find via the chain rule that g'(f(zo))f'(zo) = 1, so f'(zo) 0. Let us conversely assume that f'(zo) 0. Then the Jacobian of the mapping
I
f=u+iv : R2—iR2 0 (use the Cauchy-Riemann equations), so by the inverse function theorem f is 0 injective on a neighborhood of zo. is
Remark: Theorem 21 extends to the case of holomorphic functions of several complex variables. An induction proof on the number of variables can be found in the note [Ro].
SectIon 4 Hadamard's gap theorem As application of Theorem 8 we make an unnecessary, but nice, digression back to power series.
Theorem 22 (Hadamard's gap theorem). Let fl > 0 be the radius of convergence of the power series
1(z) = where {nk 1k = 1,2,•• .} is a sequence of natural nwnbers satisfying nk+1 (1 + O)nk
fork = 1,2,••. for some 8 > 0. Then IzI = R is a natural boundary for f in the following sense: If there is a holomorphic extension off to a connected open set Q 2(0, R). then Q = B(0, fl). Proof.
We assume for convenience that R = 1 , and denote the extension of f by F. It suffices to derive a contradiction from the assumption 11(0, 1). Under that assumption Q fl S' 0, say 1 E Q. Choose an integer p 1/8 and consider for w E C the convex combination WI' + WP+l
2
58
Section 4.
Hadamard's gap theorem
We have
zEB(o,1)ccz if IwI1
,and
and
z = 1 if w = 1 so z E fl for all w in a neighborhood of B[O, 1]. The composite function 2
is therefore defined and holomorphic in that neighborhood of B[O, 1J, and so the radius power series expansion around 0 is strictly bigger than 1. Let us of convergence of find the power series expansion of 9S: For w E B(O, 1) we have
=
(*)
+
+
+
The exponent of the last term in the kth bracket, viz. + , is strictly less than the exponent of the first term in the next bracket: Indeed, since p 1/0, we see that Pflk+1 — (pnk
+ nk) p(l + O)nk — (pnk + nk) (p0 — 1)n* > 0
which means that no power of w occurs more than once. We infer that the power series expansion of is gotten by simply omitting the brackets in (9: Indeed that power series converges, at least for anz" and g(z) =
for IzI
0 such that
If(z)I
S
CeTN for all z E ci
Prove
The Phragmdn-Lindelbf Theorem: 1ff is analytic and of exponetufal type in a sector ci of angular opening strictly less than x, iff is also condnuous in the closed sector ?L and if If I < M for some constant M on the boundaiy of 1.1, then (f I < M throughout ci.
34. Let f be holomorphic on an open subset ci of the complex plane. [a,bj —' ci be paths such that = P,Q E ci and let = P 70(b) = 71(b) = Q. :
Let and
Assume furthermore that are fixed end point homotopic in ci, i.e. there and exists a continuous map H : [a, b] x [0,11 ci such that
H(a, s)
P,
H(b,
s) = Q for all
f(z)dz
=Jf(z)dz
s E [0, 11
Show that
35. Let ci be an open connected subset of C. Let
be a sequence of continuous
functions on ci, converging locally uniformly to f E Hol(Q). Assume that none of the f vanishes at any point. Show that I vanishes everywhere or nowhere. 36. Derive Brouwer's fixed point theorem (Theorem 111.11) from Roucbe's theorem. 67
Chapwr 4.
Basic Theory of Holomorphic Functions
37. In this exeivise we will derive a formula which as special cases includes the Poisson inugra!
Je_z2dx =
and
the Fresnel iiuegrais
iirnf con (x2)dx =
and
iirnf sin (z2)dx = To that purpose we consider the meromorphic function
f(z) :=
,
C
E
—
(a) Show, to ease your computations below, that
1(z) + f(—z) =
+
Consider
and
+
—
for each R > 0 and fixed a
A, B, C, D , in which
A=-B=-Re'°, E=
F=
—G
=
68
jO,
=
[
the
following parallelogram
Section 5.
Exercises
y C
D x
A
Note that f is holomorphic on the interior of the parallelogram and on its boundary only has a singularity at z = 0. (b) Show that
I
f(z)dz—.0 and
IB,Cl
J f(z)dz—.0
as
R—'oo
IA,DJ
(c) Show that
J f(z)dz = —ie'° Jexp {it2e2'° }dt IC,D1
and that
J f(z)dz + J f(z)dz =
—&°
r
IF,Bl
IA,G)
fexp {ii2e2to
Hint: Use (a). (d) Show that the integral of f along the small half circle on the figure from G to F as r —' 0+ converges to i (e) Show that R
urn fexp {ii2e2ba}dt =
R—ooj
0
69
1
+
Chapter 4.
Basic
Theory of Holomorphic Funcuons
(f) Derive the Poisson and Fresnel integrals. (gJ Show the following more compact form of the result of (e):
I
=
for a
is taken on the right. where the principal value of The exercise is adapted from the paper [Ra].
70
0
Chapter 5 Global Theory Section 1 The global Cauchy integral theorem In this section we will state and prove a general version of the Cauchy integral formula and the Cauchy integral theorem. Theorem IV.6 states the Cauchy integral formula for a closed path in a starshaped domain. We would of course like to remove the condition of starshapedness, but we must be careful, for if we do so then some kind of restriction on the paths in question is necessary: Take as domain the annulus A := {z C I 1:1 < i} and consider functions which are holomorphic in an open set containing the closure of A. It is simply not true that
f
for
zEA
Kl=1 as
the function f(z) =
demonstrates. On the positive side there exists a version of the Cauchy integral formula in A, viz.
J
J IcI=1 (Exercise:
for zEA
Derive it from the Cauchy integral formula for starshaped domains by
inserting suitable auxiliary segments). In the case of a starshaped domain Q the Cauchy theorem states that
(s) for any closed path 7 in and any function f which is holomorphic in Q. Let now be any open subset of the complex plane, and let be a closed path in Q. If (*) holds
for all f E Hol(cl), then in particular
forall zEC\Q Our general version of the Cauchy integral theorem is that the converse holds. The
crucial necessary condition on the path 7, i.e. that Ind,(z) = means intuitively that does not circle any "hole" in
71
0
for all z
C\Q,
Chapter 5.
Global Thcoy
Theorem I (The global Cauchy theorem and integral formula). Let 0
be a closed path in an open subset Il of the complex plane such that Ind7(z)
broil
z
=
C\ft.
Then we have for any f E Hol(ft) that
(a)
Jf(z)dz=o ,and forall
(fi)
Example 2. Here is a drawing of a path that satisfies the condition of l'heorem 1, but which nevertheless is not null-homotopic [Q = C\{O, 1)]
Proof of Theorem I: (due to [Dii). (a) follows from (8) when we replace I by so from now on we may concentrate on (8). ((— By the formula for the index (Theorem 111.17(d)) (8) is equivalent to
—.
lJf(z)_f(w)dwo z—w
2in
forall
zEft\7'
7
Introducing the function h
h(z,w) :=
x Q —, C given by
for z
1
Lf'(w)
w
(orz=w
it thus suffices to prove that
(oral! zEQ
72
Section 1. The global
integral Ihecwem
We start by noting some of the properties of h and g. h is continuous in all of This is obvious off the diagonal in fl x fl. To prove that h is continuous on the diagonal as well we compute x
1(z) - 1(w)
+ i(z - w)))di = (z
=J
- w)
I
f'(w + i(z - w))dt
which implies that h near the diagonal can be expressed as
f'(w + t(z — w))dt
h(z, w)
= a formula that also holds on the diagonal. This formula proves the continuity of h on the diagonal. So g makes sense and is continuous in fl. By help of Morera's theorem (Theorem IV.9) we will prove that g is even holomorphic in fl
If A is a triangle in fl then
=
=
Now, the function z —, h(z, w) is holomorphic in fl for fixed w [Example !V.4J, so according to the Cauchy-Goursat theorem
,andhence
Jg=O
so according to Morera g is holomorphic in fl. By our assumption on the path the set
:= {z E C\y' Ind7(z) = 0) is an open subset of C satisfying fl U f1O = C For
z E fl fl flo we get:
lJ
g(z)=-J_fh(z,w)dw=_LJ f(Z)d 2ir:
z—w
27r:
= -1(z) Ind,(z) + 27r1
2irz
f(w)
f w—z dw =
w—z
27r1
7
7
The function
go(z) :=
1
(7(w)
I
f(W)dW z—w
w—z
7
73
dw for z
fl0
dw
chapter s.
Theoy
is holomorphic on (b (Lemma IL 13); the computation just made shows that g and go patch together over (1 fl to an unambiguously defined function G on (1 U (bo = C, i.e.
C given by
fcvzEQ
' ' is
for
entire.
The unbounded component of
is by Theorem 111.17(c) contained in Qo, so for
z out there we can estimate C as follows:
..L. sup{If(w)IIwE
IG(z)I =
= which shows that G(:) —+ 0 as z —' 00. But then C vanishes identically by Uouville's
theorem (Theorem 1.11). In particular g is identically 0 on (1, and that fact is exactly the contents of 0
An inspection of the proof reveals that it applies to the case of several paths, not just one. This comes in handy for, say, an annulus as mentioned in the introduction of this section. The precise statement is Theorem 3 (The global Cauchy theorem and integral formula). Let
72,•
be closed paths in an open subset (1 of the complex plane such thai
,
bra!! zE C\Q Then we have for any I E Hol(Q) that
(a)
=
0
,and
)=17,
(/3) 1(z)
Ind7,(z) =
forall Proof: Replace
g
in the proof of Theorem 1 above by
Qo by
go
by
J=I 74
Simply connected sets
Section 2.
The details of the easy modification of the proof of Theorem 1 are left to the
0
Corollary 4. Let f be holomorphic in an open subset of:he complex plane. Let 71,72, ,7N and let n i, closed paths in , and be and mi, m2,••• , nN , OM be integers such that M
N
forall z E
>2njIndi,(z) = 1=1
En,Jf = >2 miff j=1
1=1
Corollary 5. Let f be holomorphic in an open subset which are homotopic in closed paths in
of the conzple.x plane. Let y and p be two Then
/1=/fl
In particular,
is null-homo:opic in Q. then f I =
0.
7
Section 2 Simply connected sets The topological concept of simple connectivity is intimately relaxed to homotopy
(Definition 111.16).
Definition 6. A topological space is said to be simply connected, if each closed cu,ve in it Lx null-homotopic.
It is obvious that any starshaped space is simply connected. R3\{0} is an example of a topological space which is simply connected, but not starshaped. It is intuitively
obvious that R2\{0) is connected, but not simply connected (For a rigorous proof see the remark following Theorem 9). B(0, 1) U B(3, 1) is simply connected, but not connected.
75
chapter 5.
Global Theoiy
Theorem 7. Let fl be a simply connected open subset of C,
be a closed path in fl and let
fEHol(ffl. Thenff=Oand 7
f(z) Ind7(z) =
for all
J
z
E fl\7
0
Proof: Obvious from Corollary 5. Theorem 8.
Let fl be a simply connected open subset of C. Then any f E Hol(Q) has a (holomorphic) primitive on fl. Proof: It suffices to consmict a primitive on each of the connected components of fl, so we may as well from the start assume that fl is connected. Choose zo E fl and
Then
F(z) := Jf(w)dw is according to Theorem 7 independent of the choice of path from zo to z, so F is well 0 defined on fl. Proceeding as in the proof of Lemma 11.12 we see that P = 1. We can now as promised extend Theorem 111.21 and 111.10 to simply connected sets.
Theorem 9. Let fl be a simply connected open subset of C and let f be a never vanishing holomorphic function on Il. Then f has a holomorphic logarithm and a holomorphic square root on Il. Proof: It suffices to consuuct the logarithm and the square root in each of the so we may as well assume that fl is connected. Let F is be a primitive of f'/f in fl; such one exists by Theorem 8. The function I
connected components of fl, constant because
(fe")' = we see that F — w is a holomorphic c= f= logarithm of f. A holomorphic square root is exp [(F — w/2)]. 0 76
Ewcises
Seczion 3.
In passing we note that this gives us a rigorous proof that R2\{O} is not simply connected: Combine Example 11122(b) and Theorem 9. For other results on simply connected sets see the discussion around the Riemann mapping theorem (Theorem Vul.20), Chapter VflI.*4, Remark 1)1.11 and Exercise IX.20.
SectIon 3 ExercIses 1.
Consider the path -y in Example 2. What is
Jz 7
f
dz
J z—f
J
f>Ois small?
when
e _ed
2. Let f be holomorphic in C\{O}
J
Show that
i=Ji
lzl=R
3. Show that the function 1
.
1z11
has a holomorphic square root in any simply connected open subset of the complex plane which does not contain { —1, 1) . Find the possible values of —
z2
dz
f
J y'1—z2 7
is a closed path in such a domain. 4. Let be an open, connected and simply connected subset of the complex plane, and let zo,al,a2,•••,aN E fi be distinct points. (loose €>Oso small that the closed balls B[ai,e),. . . are contained in Cl and so that a' B[afr,E) when j k. For h Hol(Cl\{ai,a2,.. ,aN}) we introduce the constants when
J h(w)dwEC lw—a, l=(
Let
=
be any path in Cl\{aj,a2,. . . ,aN} from
following function of z
fish(w)—
J
I.
w—ai
—•••—
7
77
CN
1
w—aN)
to z. Show that the
Chapter 5.
Global Theory
is independent of the choice of path from zo to z. S. Theorem 9 deals with holomorphic functions, but the result is really a topological
one. The purpose of this exercise is to establish the following topological version of Theorem 9: Theorem 10. be a simply connected open subset of the complex plane and let 1: C\{O} be continuous. Then I has a continuous logarithm on Let
—,
It suffices to prove the theorem for each of the connected components of so we may as well assume that Q is connected. (a) Prove that Ind;07(0) = 0 for all closed curves in be continuous curves such that 7i(0) = 72(0) and : [0,1) —.. (fi) Let and 1072 10,1) —+ C be continuous logarithms of 10 71(1) = 72(1). Let respectively, starting at the same point = Show that = and wo E C such that exp wo = f(zn). Find for any z E a Fix zo curve such that 7(0) = zo and 7(1) = z and take a continuous logarithm of 107 such that = wo. does not depend on the choice of the curve so that we unamShow that biguously may define F(z) := 7(1)
(6) Show that F is a logarithm of f. (e) Show that F o is continuous for any curve (C) Show that F is a continuous logarithm of 1.
78
from (y).
0
Chapter 6 Isolated Singularities We begin the present chapter with a study of functions which are holomorphic in an annulus; we show that they can be expanded in Laurent series in much the same spirit as functions which are holomorphic in a disc can be expanded in power series. The special case of a holomorphic function, defined in a punctured disc is the topic of the remainder of the chapter. The center of the disc is in that case said to be an isolated singularity of the function. We classify isolated singularities into removable singularities, poles and essential singularities. We finally prove the Residue Theorem
and use it to evaluate definite integrals of various types; this is certainly one of the high points of any introductory course on complex analysis. A deeper study of essential singularities can be found in the next chapter in which the two Picard theorems are derived.
SectIon 1 Laurent series Definition 1. A Laurent series is a series of the form (1)
for n E Z
where the coefficients
are conzple.x numbers, and where z is a complex
number different from 0.
If the two series (2)
=
and
(3)
both are convergent with sums S_ and
respectively, then we will say that the Laurent
series (1) converges and that its sum is S_ + If at least one of the series (2) and (3) diverges, then we will say that the Laurent series (1) diverges. The concepts of absolute convergence and uniform convergence are defined simBy definition the convergence of the Laurent series (1) means convergence of the two series (2) and (3) which are power series in 1/: and z respectively. So it is to be expected that results, analogous to those for power series, hold for Laurent series.
79
Isobted Singularities
Chaixer 6.
Theorem 2.
Let
and R:= rn—co
Assume
that 0 < r
1 —-———
2irz
[ f(w) (itt'
j lwI=r'
w—z
r' be treated as follows: g(z) =
2irs
f
1(w) dw = ! 27rzz J w—z
jf
fwlr' I
I
1(w)
dw
Iu'l=r'
fl
F
I
J
n=o
lwl=r
11
00 1
,
f(w)wdw j lwIr
which exhibits a Laurent expansion of g. We conclude that f = G + g has a Laurenc
expansion in the annulus {z E C r' < Izi
0
such that (z —
a is called the order of the pole. It equals the order of the zero of
1/f. If a pole has order I we call it a simple pole. An example of a simple pole is A more general example of a simple pole is a quotient of the form 1(z) = g(z)/h(z) where g and h are holomorphic near z = a, h(a) = 0, h'(a) 0 and g(a) 0. 82
Section 2.
The classificauon of isolated singularities
The third and final case is the one in which urn If(z)I as z in R U {oo}. We say that f has an esseiuial singularity at a.
a
does not exist An example is
f(z) = exp(z_'). The beautiful relation between the coefficients of the Laurent expansion of f and the behavior of f at the isolated singularity is taken up in Exercise 1 below. Definition S. Let a be an isolated singularity of the holoniorphic function f : B(a, r)\{a} By the residue of f at a we mean the complex number
C.
J f(z)dz I
e is any number in the interval 10, r(. By Corollary VS the residue off at a is independent of the choice of e. where
The next sections will show that residues are important for the evaluation of definite integrals. So we must be able to compute residues. The simplest instance occurs when I has a removable singularity at a. Here Re.s(f; a) = 0 by the Cauchy integral theorem (Theorem IV.5). Another particularly simple case - which even occurs quite often - is the case of a simple pole. This is singled out as a special instance in the following recipe for computing the residue at a pole.
Proposition 6.
(a) 1ff has a pole of order n < oo at z = removable singularity w z = a. and
Res(f;a)=
1
a
a then the function (z —
{(z—a)
(b) In particular, 4ff hasasimple pole atz =
a then
Res(f; a) = {(z — (c) 1ff and g are holomorphic near z = a,
Res(L;a) = g
g(a) =
0
and g'(a)
0
•
then
g(o)
Proof: We prove only (b) and (c) and leave the easy generalization (a) to the reader. (b) The definition of a simple pole implies that g(z) := (z — a)f(z) remains bounded near z = a, so that g has a removable singularity at z = a. Now, let's write
83
ChapW 6.
Isolaled Singularities
and observe that the first term on the right hand side is holomorphic across the singularity z = a by the Removable Singularity Theorem. So
f(z)dz=
J
J Iz—aI=
I
-±_=O+g(a)2,ri
J Iz—aI=
proving (b).
(c) Since f/g has a simple pole (if 1(a) 5&
f(a) =
0)
or a removable singularity (if
0) at z =a we get from (b) that
\g
= /
g(z)
z—a
f(z)
=lim (g(z)
f(a)
— g(a))/(z —a)
g'(a)
0 As a final example of how to compute residues we consider the function exp which has an essential singularity at z = 0. Since the series
fl\
00 1
—n
converges uniformly on compact subsets of C\{O) we can integrate term by term and find
Res(exp
=
=1
= Of course, by the same method we see around z = a has a Laurent expansion
1(z) has
Res(f;a)
Section
more generally
that any function f
which
-
=
a_i.
3 The residue theorem
The statement. The residue theorem generalizes both the Cauchy integral theorem and the Cauchy integral formula. Our version of it runs as follows:
84
Section 3.
The residue theorem
Theorem 7 (The Cauchy residue theorem). Let be an open subset of the complex plane and let
that Ind.,(z) =0 for all z E
be a closed path in fl such Let I be holomorphic in Q except for finitely many
isolated singularities at,
E fl\-y. Then =
The hypothesis on -y is clearly satisfied if -y is homotopic in to a constant curve, say if is simply connected. In most examples 'y will be a simple closed path traversed counterclockwise, and Ind,(z) will be 1 or 0 according to whether z is inside or outside
Proof. Choose closed paths -yr, in circles around such that each the singularity once and does not contain any of the other singularities of f. More precisely choose them such that
Ind,,(z) =0 forall
and
j=1,2,.,N
and
Ind.7,(aj)=6,k for j,k=1,2,.",N Then N
for all z E C\(Q\{al,a2,...,aN}) j=1 so
by the Global Cauchy Theorem V.3 (or better its Corollary V.4)
fi = Dividing
through by 2,ri we get the Residue Theorem.
0
The Cauchy Residue theorem can be used to evaluate definite integrals. How successful one is depends largely on correct choice of contour and representative function. The techniques will be illustrated below by various examples that will provide us with recipes for evaluation of definite integrals of different types.
Example A. An integral of the form
JF(cos 0, sin 0)dO 85
Chapter 6.
Isolated Singularities
is by the substitution z = exp (19) convened into a complex line integral as follows:
J
0
1:1=1
As a non-trivial example we evaluate
7
sin2O
a+
for
dO
con 9
0
where the restriction on a is chosen so that the denominator in the integrand does not
vanish anywhere. Using the above substitution we find
( sinO
J0
1
1
2i J
a+cos9
(z2 —1)
2
z2(z2+2az-i-1)
1:1=1
where the integrand, i.e. the function —
1(z)— has
a pole of order 2 at z =
0
(z2_1)2 z2(z2+2az+1)
and simple poles at
z=—a+ Va2_i and z= —a— with corresponding residues —2a, 2v'a2 — 1 and — 2v'a2 — 1 , where we to avoid ambiguity agree to take 'Va2 — 1 as that branch of the square root of a2 — 1 which is positive for a > 1. (Cf Example 111.22(c)). To apply the Residue Theorem none
of the poles
h(a) :=
—a
—
—1
Ih(a)I = 1
,
0 , we
then g(a)
—a
+ Va2 —
1
h(a) and g(a) are mots of the equation —2a. Now if, say, is the complex conjugate of h(a) , so
may belong to the unit circle Izt
+ 2az + 1 =
and g(a) :=
1
.
Since
have h(a)g(a) =
1
and h(a) + g(a) =
= h(a) +h(a) = h(a) +g(a) But this means that a is real and in the excluded cases a E [—1, 1)
— 1 .
=
—a
is purely imaginary, and that happens only
Since Ih(a)I never 86
takes the value 1 and is > 1
The residue theorem
Section 3.
for a > 1 we see that Ih(a)I > 1 for
Ia
all a considered, and so
< 1 for all
By the Residue Theorem
a E C\[—1, 1]
sin6 9d9 =
+
i} =27r{a_ Va2_ i}
=
The method of this example can be applied to any integral of the form 2r
f
P
and Q(cost,sint)
Q
0 for all t E
[0, 27!J.
Example B. In this example we want to evaluate integrals of the form f R(x)dx where 1? is a rational function. We state our result as a proposition: Proposition & Let P and Q be polynomials in one variable with Q has no zeros on the real line. (I) If4tz > 0 then
P <deg Q and assume that
R
IP(x) I —e" dx R-.ooJ Q(x) urn
—R
exists, and
iirnf wherethe
=
Q(w)=O}.
ranges over
(u) If deg(P) < deg(Q)
— 1 .
then 00
[
J Q(x) -00 converges
absoiwely, and
J
.!.dx = 87
Chapter 6.
Isolated Singularities
> 0, Q(w) =
where the sum ranges over {w E C
0}.
Proof (1) Note that there exist two positive constants A and B such that
Q(z) Choose
of Q
R>
Izi
when Izi B
so large that the triangle in the picture below contains all the zeros
in the upper half plane.
y
The distance from the origin to each of the two skew sides of the triangle is
for z E [R iRJe u
hR1 —R)
With the usual parametrization of the segment y := [R. ii?]
,
i.e.
= [R,iR](t) = R(it + 1 —1) for 0 t < 1 we find
J
=
R,iRJ
/
j
—
0
=
(i -
-+0 as R
In exactly the same way we see that the integral along the other skew side [iR, —RJ of the triangle tends to zero as R 00. The result (i) is then an immediate consequence of the Residue Theorem. 88
Section 3.
The residue theorem
(ii) Proceed in the same way except for noting the stronger estimate
P(z)
A
large.
0 Example C. As an example of an integral involving a multivalued function we consider
f
J0
P(x) dx x°Q(x)
assume that 0 < a < 1 and that P and Q are polynomials in one variable such that = exp(a Log x) deg P < deg Q and such that Q has no zeros on [0, oo[. Of course, in the integral. We
It turns up to be a good idea to introduce that branch of the logarithm which is defined in := C\[0, oo[ and has its imaginary pan in the open interval 1O,2ir[. So for the remainder of this example, if z E fl we let
logz = + iO(z) , where 0 < 9(z) 0 is chosen so large, and r and €, r > e > 0, are chosen so small that the area inside contains all the zeros of Q. denote the two horizontal paths
where
for r— 0 small the integral
f
J
Ix-alt 92
Section 4.
9. Let m and
Exercises
be integers 0. Find
n
1 —
2iriJ(
m
1
z—a )
z—a)
z—a+————
dz when
IzI=1
Deduce Walks' formula 2w
11
(2rn)!
2ir
(rn!)2 1)
10. Show that 2w
t
1
2ir
2rcose+ r2 =
1—
Ii
when r E R\{1,—1}
—
1)
11. Show that 2w
27r
dO
J 1 + acosO =
—
for a2
00
0
13. Show that w
dO
J
when a>0
o
14. Show that 00
I
for
-00 15.
Show that w
for bER\{0) 0
93
Chapter 6.
Isolated Singularities
16. Pinpoint the error in the following reasoning : Let be the half circle in the upper half plane with center 0 and radius R> 0. Since the function (called Sinus Cardinalis)
:= S1flZ —
Ssnc(z) is
in all of C), it follows from the Cauchy integral theorem
entire (i.e.
with 1(z) = (Sincz) that
J f(x)dx =
iimf f(x)dx = iimJ —
Estimating sinus by 1 we get
= so that
R-.oo
(Stncz)dx=0 jI. 00
2
-00 Show
that the correct value of the integral isn't 0, but
17. Show for n =
that
J
=
—
—
1)!
18. Show that 00
f
z3sinx
00
19. Show that 00
fxsinx dx=—e
I Jx4+1
.
fl
2
0
20.
Show that
7 Jcoshx
cosh(E
Hint: Consider the contour of height 94
Section 4.
Exercises
0
-R
R
21. Show for n = 2,3,••• that
[dx J by help of the following contour in which P = Rexp(2iri/n)
p
Compute the more general expression
I
X
1+
dx where
—1 b
then f is a polynomial of degree < m. As an important application of Liouville's theorem we derive the fundamental theorem of algebra which asserts that every algebraic equation over the field of complex
numbers has a root. More generally that it takes any complex value and thus shows the validity of Picard's little theorem for polynomials. The fundamental theorem of algebra is important because it implies that any polynomial can be written as a product of linear factors. Corollary 4 (The fundamental theorem of algebra). Let an, ai,••• , a,, C where n 1 and a,, 0 , and let p denote the polynomial
p(z) :=
an
+ aiz f... + a,,z" for z E C 99
Chapter 7.
The Picaid Thcorcms
C such thai p(zo) =
Then there exists a
0.
Although the fundamental theorem of algebra looks like an algebraic result, being
concerned with polynomials, it also involves the completeness of the reals: It is obviously false if C = R + iR is replaced by Q + iQ, where Q denotes the rational numbers. For a discussion and many references see p.117-118 of the monography [Bu].
Proof: The proof goes by contradiction, so we assume that p never vanishes. Then the function f(z) := 1/p(z) is entire and converges to 0 as Izi —' 00. In particular f is bounded and so a constant by Liouville's theorem. But then p is constant, contradicting
0
thatthedegreeofpis1.
The Casorati-Weierstrass' theorem is an immediate consequence of Picard's big theorem. We give a direct proof though, since it is quite simple. In the Russian literature the theorem is attributed to Y.V. Sokhotskii. For a detailed history of it see [NeuJ. Theorem 5 (Casorati-Weierst,uss' theorem). Let f have an essential singularity at E C. Then for any w and —. w as n cc. sequence such that z,, —.
C there exists a
Proof: The proof goes by contradiction: If the assertion were false then there E C and an > 0 and a neighborhood Q of zo such that
would exist a
If(z)—wnle forall The function
g(z) := is then analytic in cZ\{ Zn) and
g.
1
1(Z) —
,z E
bounded (by lIE),
so zo is
a removable singularity for
Thus
is either analytic at z =
0), or has a pole at z = zo (if g(zo) = 0). In zo (if g(zo) any case it contradicts our assumption about being an essential singularity. 0
SectIon 2 Picard's two theorems Our key to the proof of Picard's big theorem is the following remarkable result on the range of functions which are analytic in the unit disc.
100
two theorems
Section 2.
Theorem 6 (Bloch.Lassdau's theorem). There exists a positive number A with following property: Let f be any function. 1. Then the range off contains holomorphic in the unit disc and such thai If(0)I
a disc of radius
A.
The surprising feature of the theorem is of course the existence of the universal constant A in spite of the vast class of functions involved. For our purposes any
positive constant wiil suffice. Remark. Let
us for
I E 4' := {gE Hol(B(O,1))IIg'(o)I 1) put
L(f) :=
sup
{r >
f(B(0, 1)) contains a disc of radius r}
The theorem above then tells us that Landau's constant L :=
inf{L(f))
The exact value of Landau's positive. Our proof of Theorem 6 reveals that L constant is not known, but it has been ascertained that 0.5 L 0.56. is
6 is an immediate corollary of an even more surprising quantitative discovery on the range of an analytic function: Theorem
Theorem 7 (Block's theorem). 1. Let f be any function which is holomorphic in B(O, 1) and has lf'(o)I Then there exists a subdisc D1 of B(0, 1) such that 1(D1) contains a disc of radius 0.43 and f is
univalent on
D,.
Remark. Let D(f) be the supremum of all r > 0 for which there exists a region D on which f is univalent and such that f(D) contains a disc of radius r. Then Bloch's theorem tells us that Bloch's constant
B := inf{B(f)I 1€ Hol(B(0,1)) and If'(O)I 1) is bigger than 0.43. The exact value of Bloch's constant is not known, although it is
known that 0.43 B 0.47. For interesting gossip on Andre Bloch's life see the essay [Cal and the follow up [CFJ. 101
Chapter 7.
The Picard Theorems
Proof of Theorem 6: We may assume that If'(O)I = 1 . We will first treat the special case where I is holomorphic on a neighborhood of the closed disc B(O, 1). The (unction a : (0, 1) [0, oo[ given by
a(r)
:= (1 — r)
for 0
sup
r :S 1
continuous. Since a(0) = 1 and a(1) = 0, there exists a largest number s E (0, 11 such that a(3) = 1. Choose E B(0, 1) such that 1(1 = 3 and If'(OI = sup {If'(z)I} IzIs Consider for R := (1 — s)/2 the function is
F(z)
2(f(Rz +
F is holomorphic on B(0, 1), F(0) = IF'(O)I
for
() —
1
0, and
= 2RIf'()l
= 2Ra(:)
=
1
Furthermore, since a(r) < 1 when 3 < r, we get I
Rsup
{
I
R
R
=
l_s_R0(3+R)
IZnI>IZn+'f>"PO Consider for fixed n the function (
which is holomorphic in B(O, 1),
omits the values 0 and 1 and has M . By Schouky's theorem there exists a constant C, depending only on the bound Iv!, such that
C for all (
13[0.
I
In particular
2r't
0.
Examples 13. (1)
)tR(z) := is
2R R2
a metric on B(0, R) with constant curvature (ii)
o(z) is
— 1z12
—1.
9
1+IzI 2
a metric on C with constant curvature +1.
The formula in the following lemma expresses that the curvature is a conformal invariant, i.e. is preserved under holomorphic mappings, and so it is a natural object
that ought to be studied. Lemma 14. Let f : be a holomorphic map between two open subsets ci and ci' of the complex plane, and let p be a metric on ci'. Then fS(p) := pof is a pseudo-metric
onfland
K(z,f(p))=n(f(z),p)forali zEIlforwhich f'(z)
0
Proot It is obvious that f'(p) is a pseudo-metric, so left is the formula for the curvature, i.e. that —
poflf'12
—
p(f(z))2
or in other words that
of) + Log If'I }(z) =
Log p)(f(z)) 1f1(z)12
Here the term If'I) on the left hand side vanishes, because Log 11,1 (locally) is the real part of a holomorphic function (branches of log f'), so it is left to show the formula
f)}(z) = (t,.F)(f(z)) If'(z)12 a
formula that may be left to the reader. 109
The Picaid Theorems
Chaptcr 7.
Theorem 15 (Ahlfors' lemma). A1 is the biggest pseudometric on B(0, 1) among the ones with curvature at most —1.
Proof: Let us for arbitrary but fixed r EjO, 11 consider the function v := p/A, Ofl 1 , i.e. that p A, because we then obtain the B[0,r]. It suffices to show that v desired conclusion by letting r —. 1 Since v is non-negative and v(z) —. 0 as Izi —' r , we see that the continuous function v attains its supremum in at some point zo in the open ball B(0, r). We shall show that in < 1. If m = 0 we are done, so we may as well assume that p(zo) > 0. is a maximum for v and hence for Logy, we get that 0 2 (A Log Since Now,
0? (ALogv)(zo) = (A Log p)(:a) — (ALo9A,)(:o) —ic(zo,p)p(zo)2 + K(zfl,A,)A,( Zn)2
=
—
A,(zo)2
By the assumption on the curvature of p. we find 0 2 p(Zo)2
p(zo)/Ar(zo) S
1
—
A,(z0)2
,
50
in =
0
.
PropositIon 16. There exists a metric p on C \ { 0, 1) with the following two properties: (i) p has curvature at most —1.
(ii) P 2 cc for some constant c > 0. Proof: Using the Laplacian in polar coordinates we find by brute force computations
for any a E Rand z
C\{0} that
A(Log(l + Izl°)) = a2
2
(1 +
)
Furthermore A(Log IzI) = 0 because Log Izi locally is the real part of a holomorphic function (branches of log z). Combining these two facts we get for any a, E R and Z
C\{0) that 1
1
t Since the Laplacian A =
+
1z10)
82/0x2+82/8y2 has constant coefficients it satisfies chat
(AF)(z — so
(1
J
= A(w
F(w
—
—
in particular forz E C\{1} 1
A=a
J
110
2
IzhI (1 + I:— 110) 2
Alternative ircaimcnt
Seclion 4.
We
claim that for suitable o > 0
> 0 a positive multiple of the metric
,
1+IzIl+Iz—it can
be used as the desired metric p. By a straightforward calculation 2$
Z
ic(z, r) = — 0 2
(1
+
Iz
Iz — lIa+2$_2
(1+ Iz so
C\{0,1}
for
lz—1I
Izi
-
Iz12'9
(1+ Izr)2
the curvature is everywhere negative. Thking a = 1/6 and fi = 3/4 we find that (1)
Inn
r(z)
and (2) limlt(z,r)=—oo when z—.O,loroo
IzI—oo
Combining (2) with the fact that the curvature is negative we find that there exists a has curvature constant k > 0 such that ic(z, r) —k on C\{0, 1}. Then p := at most —1, which is the first property. From (1) we see that urn
xI—.oo c(z)
0
which implies the second property.
Theorem 17 (Scholiky's theorem). To each M > 0 and r E JO, 1[ there exLus a constant C > 0 such that the following implication holds: If I E Hol(B(0, 1)) , 11(0)1 M and the range of f omits 0 and 1 , then sup {If(z)I) C. IsIr Proof.
Let p and c be as in Proposition 16. Then f(p) is a pseudo-metric on
B(O,1) with curvature at most —1 (Lemma 14), by Ahlfors' lemma f(p)
f'(u) c'f(p)
0. (a) Show that f'(A?)
(fl) Show that f(A,) =
0.
(y) Derive Liouville's theorem. 4. In this exercise we give a proof of Picard's little theorem, based on Ahifors'
lemma and the metric p from Proposition 16(i).
So let f : C
holomorphic.
on any ball B(0,R). (a) Show that f'(p) (fi) Show that f(p) = 0. ('y) Derive Picard's little theorem. 112
C\{0, 1) be
Chapter 8 Geometric Aspects and the Riemann Mapping Theorem We start this chapter by studying geometric aspects of analytic maps, in particular
of the so-called MObius transformations. We finish by proving the Riemann mapping theorem which states that simply connected regions contained in C but different from C, are conformally equivalent.
SectIon
1
The Rlemann sphere
We have earlier studied analytic functions near isolated singularities. To handle such singularities it is aesthetically pleasing and often even convenient to introduce the Riemann sphere = the extended complex plane. As an example consider the map which has an isolated singularity ax z = 0 and whose image does not contain z 0.
map into a homeomorphism of the extended complex plane. Definition I. The extended complex plane consists of the complex plane plus an point 00 (called infinity). We give the topology which is defined by keeping the topology on C and by letting the sets (C\K) U (oo} , where K runs thmugh the compact subsets of C, be the neighborhoods of Co. The sets
,O 2= 2 1—lal 1—lol
unless a =
0
0
Suppose H is not onto B(0, 1) and let b E B(0, 1)\H(fl) . The function A(b, H) does not vanish in Q, so it has a holomorphic square root 4, in Q: (4,(z)J2 = .4(b. Clearly 4' E,7. Recalling that H(a) = 0 and IH'(a)I = , we compute that
k/(a)I = 124
Section 6.
Now,
Exercises
also belongs to J, and we find that
:=
l+IbI
>'l
which is a contradiction. Hence H is onto.
Step 5. The uniqueness. The uniqueness of H is an easy consequence of the uniqueness statement in Example 0 14 above.
SectIon 5 PrImitives We have seen in Theorem V.8 that every holomorphic function on a simply connected domain has a primitive there. There are open sets for which this fails. For example the plane punctured at the origin has not got this property : The function l/z defined on C\{O} does not possess a primitive [Because if it did, the line integral of l/z along the unit circle would be zero, but it is 2,rij. It is a most remarkable fact that a connected open set has this property if and only if it is homeomorphic to the open unit disc.
Theorem 22. For any non-empty connected open set ci the following are equivalent: a) Every holomorphic function on ci has a primitive. b) ci is homeomorphic to the unit disc.
Proof: a) => b) If ci = C, the map z
(i —
L.
a homeomorphism of C onto the open unit disc. If ci C then the hypotheses of Theorem 21 are satisfied: Let f be a non-vanishing holomorphic function on ci. By assumption 1,/f has a primitive, say qS. It follows that f = const Adjusting 0 by adding a suitable constant we see that f has a holomorphic logarithm, and in particular is
a holomorphic square root. By Theorem 21 ci is conformally equivalent to the unit disc.
b) =>
a)
If ci is homeomorphic to the unit disc then it is simply connected, and so we may appeal to Theorem V.8. 0
SectIon 6 Exercises 1. When
is a Möbius transformation T a projection, i.e. when is T2 = T? 125
Chapcr 8. Gcomcuic Aspects and the Ricmann Mapping
Theorem
I to 00, the unit circle (except I) to the imaginary axis. > 0) 3. Find a MObius transfonnation that maps the semi-disc {z E 11(0,1)1 onto the positive quadrant. 4. Find a MObius transformation that maps the open right half plane onto 11(0, 1) in such a way that = 0 and > 0. 5. Let and g be holomorphic mappings of 11(0,1) onto a subset of the complex plane. Assume that f is a bijection of 11(0, 1) onto Q and that 1(0) = g(0). Show that 2. Show that the MObius transformation which takes —ito 0, 0 to I and
takes
f
g(D(0,r)) ç 1(11(0,1)) for any r E10,1[ 6. Let f: 11(0,1) B(0, 1) be holomorphic and fix more than one point. Show forall z E 11(0,1). that 1(z) = z 7. (a) Why can't you display an analytic function f: 11(0, 1) 11(0,1) such that = 3/4 and f'(i/2) = 2/3? (b) Can you find an analytic function f : 11(0, 1) —. 11(0, 1) such that f(0) = 1/2 and f'(O) = 3/4 ? How many are there? —' be analytic. Assume be the open right half plane, and let 1: 8. Let . Show that If'(a)I < 1 that f(a) = a for some a E 9. Let f,g : 11(0,1) —' C be holomorphic functions, g univalent. Assume Show that If'(O)I g(B(0, 1)) Ig'(O)I and that 1(0) = g(0) and 1(11(0,1)) equality sign implies 1(11(0, 1)) = g(B(0, 1)). 10. Prove Vitali-Porter's Theorem: Theorem (Vitali 1903, Porter 1904). Let fi be a connected open subset of C and let {f,, } be a sequence of holomorphic } is locally uniformly bounded in Il. and Assume that the sequence functions on that has a limit as n —. oofora set of z E f with an accumulation point in fl. exists locally uniformly in Then lim
f(i/2)
.
Hint : Montel's theorem. 11. Let {f,,) be a locally uniformly bounded sequence of holomorphic functions in a connected open subset of C, and assume that I,, is zero-free in Q for each n. Show that urn I,, = 0 locally uniformly in Q if there exists a zo Q such that
=
0
Hint: Apply the maximum principle to 1/f,, and refer to the Vitali-Porter theorem. 12. In this problem we generalize a pan of Schwarz' lemma.
bea bounded, connected, open subset of C, leta E a holomorphic function with f(a) = a. Let
be
and letf :
—'
We put for
that
(a) Show If'(a)I
that there exists a constant
K, depending on
K. 126
and a but not on 1'
such
Section 6.
(48) Show that If'(a)I 1 Assuming that f'(a) =
Exercises
Him: Compute 1 , prove that f(z) = z for all z f(z)—z=crn(z—a) m +(z—a) rn-fl h(z) .
E
Hint: Write
compute the coefficient cm. (8) Assuming that lf'(a)I = 1 , prove that f is a bijection of Q onto Q. Hint: Find a sequence of integers oo as k oo and a function g such that and
/ j(a)
—41 and
is a conformal equivalence of the strip 13. Show that z the plane cut along the negative real axis. 14. Find a conformal equivalence of
>
0 and
0,
> 0 has
so f has no zeros on the positive
semi-axes.
Let R> 0
and
consider the quarter circle
z = Re',
0 9
On this quarter circle f is
f(Re*9) = R sufficiently large the parenthesis has positive real pan, so a continuous argument function for f along the quarter circle is 9 —, 89
+ Arg
}
where Arg denotes the principal determination of the argument.
Consider now the curve indicated on the following figure 130
Seclion 2.
Rouche's theorem
y
iR
x
We have observed that > 0 when t 0 . So as z varies from iR to o along the imaginary axis the argument of f increases by Arg 1(0) — Arg.f(iR) = —Arg f(iR) , and so the argument increment is in absolute value less than ir.
The argument of f is constant (viz, equal to 0) from 0 to R. Finally from R to iR along the circle it increases by where < ir. So along the curve + = + on the figure the argument off increases by 4,r + something , where But the argument increment is an integer multiple of 2,r, so here it is exactly 4,r. By the argument principle there are exactly 2 zeros of f (counted with multiplicity) inside the contour, and hence (since R> 0 can be arbitrarily large) in the first quadrant.
SectIon 2 Rouché's theorem The following clever lemma will be useful in the proofs of Rouch6's and Runge's theorems. It is an integral representation theorem like the Cauchy integral formula. In contrast to the earlier results it asserts the existence of a curve with certain properties, not that every curve has those properties. The power of integral representations has already been demonstrated in connection with the Cauchy integral formula; it led to local power series expansions.
Lemma 4. Let Il be an open subset of C and let F set of piecewise linear closed paths 7i, 'Y2,' 131
Cl be compact. Then there exists a finite , y,
in Cl\F such that for any function
Chapter 9. Meromophic Functions and Runge's Theorems
IE =
0
and 1=17
aEF j—1
Proof: Consider
the following grid (i.e. set of squares) on R2
j range over the integers, and where the positive integer N is chosen so large that any square of the grid having a point in common with F is contained in We orient each of the squares the counter clockwise direction. Denote by Qi, Q2, Qm the set of those squares in the grid that have a point in common with F. The union of these Q1 contains F that much is clear. Note also that if for example a vertex of a square belongs to F, then all the four squares having this point as a vertex are included in the system ,Qm. A similar remark applies to a point on an edge. Let L1, L2,. .. , L, denote the boundary segments of this system of squares, i.e. L1, L2,. .. , L, are those edges of the system which are edges where i and
,
-
of exactly one square. Then L, must for each j be contained in Q\F
.
Since each
square is oriented, each segment is oriented, too. Now let f be holomorphic in ci. Since ci we get by Cauchy's theorem that (1)
ff(z)dz=0
for
1 0 there exLus a rational function P/Q with poles only in S such
f
P(z)
<eforallzEF
Proof: In the proof it will be convenient to interpret expressions of the form where P is a polynomial, as polynomials in z. 135
Functions and Runge's Theorems
Chapter 9.
Letting set
_V1,_V2,••
denote the open set in question there is according to Lemma 4 a finite of oriented segments in Q\F such that
forall
(5)
zEF
7
so the theorem will be proved once we show that each summand on the right hand side of (5) can be approximated uniformly on F by finite sums of the type where the P's are polynomials and the c's belong to S. Let us just treat the first summand;
the other terms may be handled in exactly the same way. The line integral
J can be uniformly approximated by Riemann sums
where b,E-y'.
b_z
Again let us just treat the first term (bi —
, the rest being handled similarly.
does not meet F so b F. Let E S as b, and assume first that c 00. Join b and c by a curve contained entirely in this component. Since this is disjoint from F, there is a positive distance between this curve and F. We can find points We drop the subscript and write 6 for b1.
be in the same connected component of CQO\F
b
=
co,
,c,,, =
c
on the curve such that
< dist(curve,F)
(6) 2Icj
Then writing z—b = z—ci +ci —bwe get = (z
—
civ' >(b — ci)k(z
the series converging uniformly for (7)
(z —
E F by (6). So we may choose N so that
z
b)'
—
N —
>(b— ci)k(z — k=O
is uniformly small for z E F. We can use the same argument on each term of the sum
in (7) to replace
by C2. Namely (z —
= (z
—
C2+
—
cj)_k
which can be expanded in a series that - because of (6) - converges uniformly for E F. In other words we can find a polynomial in (z — c2)' which approximates (: — uniformly on F. This procedure can be repeated until we reach c = Cm. 136
Section 3.
Runges theorems
The argument is thus complete except when the point c chosen in S is oo. To treat where 6 belongs to the unbounded component of C\F. this case consider (z — Choose din the same component such that ffl < for all z E F. From what we approximates (z — have done above there is a polynomial P such that occurring in the polynomial can be each (z — uniformly on F. Since f expanded in powers of z/d, i.e. there is a polynomial which approximates (z — uniformly on F. The argument is thus complete.
0
For the next result (Runge's polynomial approximation theorem) and for the discussion of the inhomogeneous Cauchy-Riemann equation we need a fact from point set topology:
Proposition 9. Let be an open subset of the complex plane. Then there exists an increasing of compact subsets of Q such that sequence K1,
(a)K1UK2U••• (b)
ç mt
for all n =
1,2,
(c) Every compact subset of is contained in for some n (d) Every connected component of COO\K,I contains a connected component of
= The last statement (d) says that K, has no holes except the ones coming from the
holes of
(make a drawing!).
Proof: The cases fl = 0 and = C arc trivial, so we shall concentrate on the remaining case in which 0. We will show that we for n = 1,2,... may take
K, := This K, is a closed and bounded subset of C, so it is compact. It is now almost
obvious that (a), (b) and (c) arc satisfied, so left is only point (d). Here we begin by noting that if a connected component of , say U, intersects a connected component of say V, then U contains V: Indeed, since CQO\KII the connected components of COO\KII cover in particular they cover V. They arc open and disjoint, so by the connectedness of V we get U V. Hence (d) boils down to proving that each connected component U of intersects To do so we notice that
=
U B(z, !) 137
UB(oo, n)
Chapter 9.
Meromorphic
Functions and Runge's Theorems
Let w E U. Then w lies in one of the balls, say w E B(zo, The ball is a connected subset of COO\Kn, so B(z0, , so U intersects component. Now, zo E U fl
ç
U
,
where
U being a connected
0
Theorem 10 (Runge's polynomial approximation theorem). Let bean open subset of C. Every holomorphic function on can be approximated uniformly on compacts by polynomials
is connected.
Proof: We will first show that the condition is sufficient. Proposition 9 tells (c) that it suffices to approximate the given holomorphic function f on any given (d) and so it that COQ\KTh has only one connected component. That one contains in Runge's theorem on approximation by contains oo. Take now S = {oo} and F = rational functions (Theorem 8) to get polynomials to approximate f uniformly on is not connected. Then there exist two Now for the necessity: Suppose that We may suppose that closed, disjoint, non-empty sets F and F1 with union 00 E F1 so that F is a compact subset of C. Clearly
is open. Put 0 =11 U F. F is a compact subset of the open set 0. Let be paths in O\F = as in Lemma4. Let a E F. The function z —,
(z
holomorphic on Q. Assume now that all holomorphic functions on are uniform limits on compacts of polynomials. Then we can in particular approximate the function z —. (z — 0)_i . These by a sequence {Pk) of polynomials on the union of the paths , polynomials are of course holomorphic in 0 and hence by the first pan of the conclusion of Lemma 4 we have =
0
'dz =
0
, which in the limit as k
oo
becomes
.1
0
But that contradicts the second pan of Lemma 4.
Remark 11. If Q be an open subset of C, then
is connected dJf is simply
connected.
is connected. Any function f which is holomorphic on may by Runge's polynomial approximation theorem be approximated uniformly by polynomials on compact subsets of So if is a closed path in Q we conclude Proof: Suppose first that
138
Section 3.
that ffdz =
0
Runge's theoscins
which implies that f has a primitive (use the proof of Lemma 11.12)
and hence that ci is simply connected (Theorem V11L22). Assume conversely that is not connected. We then want to show that ci is not simply connected. We assume it is and arrive at a contradiction as follows: Let F and F1 be closed, disjoint, non-empty sets with union . Without loss of generality we may suppose that oo E F1 so that F is a compact subset of C. Clearly
=I1UF isopen.
PutO:=f1UF. Fisacompact subset of the open set 0.
a)'
be the closed paths in 0 \ F =
Cl from Lemma 4. Let a E F. Since z (z — is holomorphic in Cl we get according to The Global Cauchy Theorem (Theorem V.7) that
•1
0
which contradicts the second part of Lemma 4.
Remark 12. Let K be a compact set. Then the unbounded component of C\K is precisely those a E C for which the function z (z — a)' is a uniform limit on K of polynomials (This follows from the proof of Theorem 8). Runge's theorem on polynomial approximation is not the last word in these matters.
Clearly, if a function f on a compact set K is the uniform limit of a sequence of polynomials then f E C(K) fl Hol(Int(K)) , so this is a necessary condition for approximation by polynomials. Mergelyan's theorem states that it is also sufficient:
Theorem 13 (Mergelyan's theorem (1952)). Let K be a compact subset of C with C\K connected. Let I C(K)flHol(Int(K)) Then there exists a sequence of polynomials converging to I wuformly on K. Note that when K has empty interior then the only condition remaining on f is that C(K). So the classical approximation theorem of Weierstrass for an interval is a particular case of Mergelyan's theorem. For a comprehensible proof of Mergelyan's theorem the reader can consult Chapter 20 of [Ru]. Another extension of Weierstrass' approximation theorem is provided by
I
Theorem 14 (Carle,nan's theorem (1927)). C(R) then there exists to every continuous function e: R function F with the property that
1ff
11(x) — F(x)I (
where we, again for the sake of simplicity, use the notation G(w)
F(w)/w. Intro-
ducing the definition of the Cauchy-Riemann operator we find by a small computation that OG
so Green's theorem for the annulus
=dzv(G,zG)
E(z,n)
(ZLZk where all
0
we get
>
—
akizlk
=
=
>
—
—0) =
which is (15).
1:
Andthenforanyrn0,n (16) IE(z,n)m —
ii
2mIE(z,n) —
where we have used the identity
and
the fact (evident from (16)) that 1E(z,n)i 2 for fri 155
1
E(1,n))
and has
Chapter 10. Representations of Mcromorphic Functions
9—k
(17) IE(z,m +k)—
With these preliminaries to a side we can prove Weiers:rass' factorization theorem:
Theorem 9 (Weierstrass' factorizadon theorem). be a sequence ofthffereiupoiiusfrom Q Let Q be an open subset of C • let be a sequence of positive integers. with no cluster point in Q. and let finally m for at There exists a function g E Hol(fl) such that g has a zero of order suchthatgvanishes nowhere else in Il. k = 1,2,..., and
Proof: not cluster in
Since the sequence {zk} does
there
exist a E
and r > 0 such
B(a,r)
fl and such that B(a,r) does not intersect the set {:k}. Let us for convenience assume that a = 0 and r = 1. Under the transformation : —. 1 the open then goes into an open set V. If := 1. set E V and {ck} that
ç'.
does not cluster in V (Recall that 0 does not belong to V). Choose for each k a point a& E C\V closest to ck. Since {ck} does not cluster in V we get (18)
tim
— CkI = 0
k—oo
[Indeed, if for some 5 > 0 and subsequence {k,) we have that ak. — ci,I 5 , where tends to c (recall that {ck} is bounded), then c must belong to C\V. But by choice we have of lc&,
— ci Ica. —
a contradictionj.
Now put
(19) f(z) :=
for
zEV
If F is any compact subset of V. then by (18) for all large k, 2Ich — aal < z
uniform convergence of the infinite product in (19) now follows from
(17) and Lemma 4. By assumption B(0, 1) we have
so {zIlzl> 1) —
z—ak Using (17) and (19) we see that
(20) If(z)I
is
whenever
V. Since Iakl land lckI 1, Iz 3
bounded for lzl 3 156
The r-function
Section 4.
at 2k and f(1/z) for a 0. Then g has a zero of order Now define g(z) vanishes nowhere else in Q\{O} . (20) says that g is bounded in B(0, 4)\{O). 0 is therefore a removable singularity of g and g can hence be defined to be analytic at 0 as well. If g(0) 0, g is the required function. If 0 is a zero of g of order n, then 0 fulfills all the requirements.
Section 4 The r-function Taking
=
—n
in (12) we find that
(21) g(z) := etzz[1
(i +
is entire with simple zeros precisely at a = 0, —1, —2,.... Here is a constant chosen so that g(1) = 1. is called the Euler constant. The reciprocal of g is meromorphic in C with simple poles at 0, —1, —2,... This is the famous F-function (gamma-function): (22)
and it is of utmost importance. The F-function was introduced by Euler in 1729 (See [Ds] for a readable account of its history). Let us derive some of the most important properties of the F-function. Since q is entire, F has no zeros. Writing
(23) g(z) = and
+
1)e'H { (i +
exp
}
replacing a by a + 1 in (21), the following computation is valid for any a g(z) _e zg(z+1) —
(
z
1
z
+!)} =
n(n1+
1)} fi
}
Using that
=1 we find further that (24)
157
Chapter 10.
Representations
of Meromoiphic Functions
Thus g(z) = czg(z) for z O,—1,—2,•••, where c is a constant. By continuity this holds for all z E C. To find the constant c we note from (21) that lim
z
z—O
=
1
Since also g(1) =
1 we see that the constant is 1. Translating from g to 1' we find the functional equation for r:
(25) F(z+1)=zI'(z) for Since
=
1,
(25) leads to
(26) r(n-l-1)=n! for which explains why the r-functjon is also called the factorial function. We proceed by deriving another fundamental relation for the r-fiinction, called the formula of complementary arguments. Usc (21) with z and —z and multiply to get
g(z)g(—z) =
_z2[J
Appealing to the Euler formula (10) for sine and to (25) the above relation reads in
terms of 1'
(27) r(z)r(1 — z)
=
for z
Z
(27) is the formula of complementary argwnents.
In particular since r'(x) > 0 for z > 0, we get taking z = 1/2 in (27) that (28)
The reader might in other connections have encountered the Mellin transform (= which is the function F9 Euler integral) of a complex valued function g on defined by the
:= Jtz_19(t)dt wherever the integral converges. is of special interest for us here because this The particular case of g(t) = equals the r-function. This result is called The Euler integral formula
(29) F(z) = Jt1_1e_tdt for
158
>
()
Section 4.
The F-function
To derive it we use Prym's decomposition (30)
Jtt_1e_tdt =
+ Je_ftt_Idt = P(z) + Q(z)
/
The first integral on the tight, i.e. Q(z) is entire. Expanding e_t and integrating term by term we find that the second integral equals (31)
The right hand side of (31) converges uniformly in any compact set not containing the points z = 0,—i,--2,•.., and we see that
(32) F(z) is a meromorphic function in the entire complex plane with simple poles at = 0, —1, —2,.... Ii also equals the integral in (29) for > 0. Using this, routine integration shows that
(33) zF(z) = F(z + 1) when 3tz >
0
But then (33) must hold in the entire z-plane minus the points 0,—i, —2,.•• (The Unique Continuation Theorem). We know that r satisfies the identity (33) in form of (25) and that F never vanishes. Thus from (22) we see that
(34) 0 :=
= Fg
is entire and has period 1, i.e. G(z+1) = G(z). Since F(1) = F(1) =
1,
G— 1
vanishes at the (real) integers. Now, sin wz has period 2, has simple poles at the integers and vanishes nowhere else. Comparing 0 with sin irz we see that
0(z)—i
(35) H(z)
srnirz is entire and has period 2. We will prove that H is zero by showing that H is bounded
in the period strip 1 3tz (except for the factor exp (—x2))
the following estimate of the integrand
2,*— 1
(1
+
< exp
—
1)} exp
e so the integrand above is bounded uniformly in n by the integrable function e exp (— z2). 160
Section 5.
The Minag-Leffier expanston
We next show that the integrand converges poiniwise, so x is fixed for a moment. Using the power series expansion of the logarithm we find for n 00 that
x\
/
x
/1
x2
and so log
{(i +
= (2n — 1)log (i +
+
= (2n —
=
—
—
+
—
Finally, by the dominated convergence theorem 00
2J
e_Z2e_X2ds
=
as n —.
00
00
0 As the end of this paragraph we mention the famous dupilcanon formula of Legendre
(see the exercises for a proof):
=
(38)
1r(z)r(z +
Section 5 The Mittag-Leffier expansion The theorem of this paragraph is analogous to the Weierstrass theorem on zero sets in §4. Instead of zeros we deal with poles. Before we state and prove the theorem we remind the reader of the following fact from the theoiy of Laurent expansions:
Theorem 11. Let f be holomorphic on the punctured disc B(O, R)\{O}. Then f may in exactly one way be written in the form
1(z) = g(z) + h
for z
B(O, R)\{O}
where g E Hol(B(O, R)) and where h is an entire function such thai h(O) = 0.
Proof This was seen during the proof of Theorem VU.
0
More generally, if a E C is an isolated singularity of a holomorphic function f then f can be written in the form (39)
f(z)=h(_!_) +g(z) 161
Chapcr 10.
Representations of Meromorphic Functions
where h is an entire function with h(O) = 0, and where g is a holomorphic function on the domain of definition of f and a is a removable singularity for g. The decomposition (39) is unique. The first term, i.e. the function z —' h((z — aY') is called the
principal part offal a. Example. The function (Cf formula (9) above) 2
sin2 has
for each n
E
1
=
— n)2
N the principal part (z — n)2
at
n.
=
z
However, if we are given a sequence of principal parts, then their sum will normally not converge. Mitiag-Leffler's theorem says that a meromorphic function with the prescribed principal pans nevertheless exists:
Theorem 12 (Miltag-Leffler's theorem (1884)). be an open subset of C. Let pi, be a sequence of different points from without cluster point in Il. and let P11P2,... be a sequence of polynomials without constant term.
Let
Then there exists a function which is meromorphic on fl, has
of poles in
is Pk((z
and whose principal part at
} as its set
PLY')for all k = 1,2,...
—
Remark. If we only specify that the desired function should have poles at P2 of given orders mi, then it can be found by help of Weierstrass' theorem on zero sets from §4: Indeed, if f is a holomorphic function with zeros at p1,P2,••• of orders then 1/f satisfies the requirements.
Proof: LCtB(pk,rk),k=1,2,.. be
and choose for each k
disjoint balls in
a function
4'k E
r&))
such that
is identically I on the smaller ball Bk :=
so
it satisfies
f
£ the requirements of the theorem except that it is not holomorphic. To
function
is smooth on
.
.} and reduces on B& to fk.
remedy that we consider the function F
g is clearly
on
f
for z E cZ\{pi,p2,. . for z E And on the punctured disc Bk\ {p,, } the function
f reduces to a holomorphic function (viz 1k). so Of Ofk 9 — — — oz uz 162
Secdon 6. The
on Bk\{pk} . Thus
and p-fimctions o( Weiers*rass
g
The inhomogeneous Cauchy-Ricmann equation = g has a solution u E C°°((l) (Theorem IX.16). by the veiy definition of g. Now, h := — u is holomorphic in On Bk\{pk} we have h = 1k — u. As observed u E Hol(Bk) , so h has the desired 0 principal pan
f
Wcierstrass' factorization theorem (Theorem 9) told us how to construct a holomorphic function whose zero set is a prescribed sequence of points. The next result is that what can be done with zeros can be done with any sequence of values:
Coroilar, 13 (Gerasay'z interpolation theorem). be an open subset of the complex plane. Let {zl,z2,...) be asubset of Cl w2,••• be a sequence of complex numbers. without limit points in Cl, and let Then there exists a function f E Hol(Cl) such that f(zk) = Wk for k = 1,2,...
Proof: By Weierstrass' Factorization theorem (Theorem 9) there is a holomorphic function 0 fork = 1,2,... By the 10 with simple poles at the zk, i.e. fo(za) = 0 and Mittag-Lefficr theorem there is a function h E Hol(Cl\{zi,z2,. .}) such that .
Wk
fo(zk)
1
z —
holomorphic in a ball around zk for each k. Now, f := foh is holomorphic on Cl, we find that because the zeros cancel the singularities, and for z is
f(z) = fo(z)h(z) = fo(z){h(z) —' 0
+
— ft,(zk)
z — Zk
}+
z — Zk
= Wk
sof(zk)= Wk fork= For a generalization of Corollary 13 see Exercise 17.
Section 6 The (- and p-functions of WeIerstrass We have examples of periodic holomorphic functions. E.g. exp, cos and sin. Liouville's theorem rules out that a nonconstant holomorphic function can have two independent periods. We shall now give an example of a doubly periodic meromorphic function, viz Weierstrass' p-function. Let w and w' be non-zero complex numbers with non-real quotient, i.e. they are linearly independent over the reals. Let G be the group generated by w and w', i.e. G consists of the complex numbers of the form g = mw + nw' where m and n are integers. If w and w' are periods of a given function then so is every g E G. 163
Chapter 10. Representatiom of Metamorphic Functions
By Minag-Leffler's theorem there is a meromorphic function with principal part g)1 at each element g E G. However we do not have to appeal to that theorem, since an example is given by the so-called Weierstrass (-function (Weierstrass zeta(z —
function):
(42)
To prove that (is such an example we prepare a lemma. Lemma 14. 1
(43)
>
where ImI+JnI1. If
Proof: = k either
ImI + ml
k/2; if, say lml
or frzl is
+ nw'l =
k/2 then
ka
=
+
where a = Now there axe at most 4k pairs (m,n) such that lml + ml = k. Thus
>2
191
k=1 ImI+1n1b
0
which proves (43).
Let us now return to the right hand side of (42). Consider the disc B(O, 1?). There arc only finitely many g G satisfying R, as is clear from f.ex.. (43). Since 1
1
z—g
g
z
(z—g)g
g
if
=
,
,and and Izl
IgI
R
the series
>2 IgI2R
(1 z
1
9
g
z g
converges uniformly in B(O, R) and represents a holomorphic function in this disc. It follows that ((z), given by (42) is meromcrphic in C with simple poles precisely at
the points of G. 164
SectIon 7.
Exercises
Differentiation of (42) termwise (which is permitted) leads to the
p-
function (Weicrstrass pc-function):
(44) p(z) = .-('(z) =
+
{ (z—g)
2 —g1
}
The p-function is meromorphic in C with double poles precisely at g E G. Let us now show that the p-function is doubly periodic with periods w and Differentiating
(44) we get
(45) p'(z) =
2
2
—
— —
g)
— g)
For any h E G the series for p' is unaltered by a change from z to z + h because C is a group. In other words, p'(z + h) = p'(z), implying that
(46) p(z+h)—p(z) =C(h) where C(h) is a constant, perhaps depending on h. But the series for p shows that is an even function : p(z) = p(—z). Now it is immediate from (46) that
C(h+g)=C(h)+C(g) forall h,9eG Using these facts we find that
But C(h) + C(—h) C(O) = 0, so we must have C(h) = 0 and so p(z + h) = p(z). We record this in the final statement of this section: p and p' are doubly periodic with periods andw', and = —p.
Section 7 Exercises 1. Show that 00
2
3
2. For which z E C will the infinite product :[•J
(i
+
converge? Show that the value of the product is (1 — Hint:
(1— z)J1[ (i +
=
165
1—
Chapter 10. Representations of Metamorphic Functions
N
3. Assume that the limit urn
N—co
0. Show that the product
exists and is
fl
n1 4. Assume that the infinite product fi (1 +
converges. Discuss convergence
of 5. Let
the sequence
} of complex numbers satisfy that
(i) Show that the infinite product ( a so-called Blaschke product)
B(z) := a function which is holomorphic in B(O, 1). Find its zeros. as above with the property that every point on the unit (ii) Find a sequence is a cluster point of circle Iz I = I 6. We shall in this exercise present another way of dealing with the entire function A from §2 (Herglotz' trick):
A(z)=ircot(irz)—
/1
00
2z
(a) Show that A' satisfies the functional equation
+A'(!j_!)}
A'(z) =
(8) Use the functional equation to show that A' is a constant. (Hint: The maximum modulus principle).
7. Show that 2
by help of the partial fraction expansion of the cotangent.
8. Show that
i-ri COslrz=Ilcl— ( 9. Let f
a quotient I
be
4z 2
(2n—1) 2
meromorphic on an open subset Cl of C.
Show
that f
= g/h, where g and h both are holomorphic in Cl. 166
can be written as
SectIon 7.
10. Let t
Exercises
R\{O} . Show that ,T.2
Ir(it)I = 11.
(i) Show that Euler's constant equals
f
1
1
(ii) Derive Gauss' formula N!NZ
r(z)= lam
for
N—co
and from it deduce Wallis' product
2•2 4.4
21.3 3.5 5.7 (iii) Prove the Legendre duplication formula
= 22z_1r(z)r(z ÷ 12. Find the residue of rat z = 0. More generally at z = 13. Let C be the following contour
—n
for n =
y
C
-ì
Prove Hankel's formula
r(z) = For which z E C is Hankel's formula valid?
14. Let fl and B = {b1, b2, . . .} be as in the Mittag-Leffler theorem. Let fk for k = 1,2,••• be entut functions with fk(O) = 0. 167
Chapter 10. Representations of
Functions
Show that there exists an f Hol(Q\B) such that the principal part of f at bk fk((z—bk)) for all k = 15. With the notation of §6 show that
a(z) :=
is
(l(z)2)}
—
is entire with simple poles exactly at the g e G. The logarithmic derivative of is Weierstrass' (-function. a is called the sigmafiuzctlon of Weierstrass.
Show the following generalization of Proposition 2: Let be a sequence of reals with 11k > 0 for all k. Let cl,c2,••• be a sequence of complex numbers from the open unit disc, and assume that 16.
:" for IzI 1 write 00
(6) As said before the third term on the right in (6) is entire; the second term is by our estimate on a,, meromorphic in the complex plane with simple poles at —(n — 1) if a,, 0 , n = 1,2,.... And the first term has a simple pole at z = 1. Writing g(z) = 1/r(z) we may define ( on the complex plane by
(7)
Since
g is entire with zeros at 0, —1,
.
the right hand side of (7) is meromorphic
on all of C with at most one pole, namely one at z = 1. Since g(1) = 1 ,the function ((z) — 1/(z — 1) is entire. 0 Riemann discovered a remarkable functional relation for the (-function from which many of its properties can be deduced. 170
Section 1.
The Ricmann zeta function
Theorem 2 (The functional equation for the zeta-function). The following relation is valid: (8) ((1 — z) = (2ir)
or written in a more synvne:ric form ((1 —
(8')
Proof: Integrate by parts in (2) for
1
to obtain 00
00
((z)F(z)
=J
di =
/
so that
(9)
-
z((z)I'(z) =
Now from the formula (8) of Chapter X we get (10)
The function 1 —
1
1
=
4sin2
001
1
+
t(it +
(it — 2,rn)2
sin2 z Z2
/ \
sin z \
I
Z/\
sin Z
like z2/3 near z = 0. Hence the first integral in (9) is holomorphic for
—1.
for
+
behaves like z2/3! near z = 0, so 1—
behaves
j2
The last integral is in fact entire. Thus (9) holds for
0 such that B[zo,R+e] ç (1, and let us for the sake of simplicity assume that zn =0. From the Poisson formula we get for any z E B(0,R-4-e) that
V,,.p(Z)
and
= / VRP((R +
so, from Property (iv) of the Poisson kernel, the estimates
/
+
1
1
:
/
,#\R+e+IzI
191
Chapter 12.
Harmonic Functions
which by the mean value property reduce to
R+e+IzI Vn,p(O)
R+e—IzI R+e+
R+e—
These inequalities show that converges uniformly in B[zo,R) The argument actually shows that if converges in a point z, then it converges uniformly in any closed ball B[z,R) in fl. Thus the set {z E converges) is both open and closed in Q, and so by connectedness it equals or the empty set. But zo belongs to the set. We have now shown that converges uniformly on any closed ball in Since any compact set in Q can be covered by finitely many such balls the convergence is uniform on compact subsets of Since is continuous, so is the limit function u. Let E Cl and B[zi,R] ç Cl. Letting It —. 00 in the formula
= / un(zi + we
—
get for z in the ball B(zi,R) that u(z) =
Ju(zi +
—
It now follows from Theorem 8 that u is harmonic in B(zj,R). But
was arbitraiy
0
inQ.
SectIon 3 Jensen's formula If f is holomorphic and without zeros in an open set containing the ball B10, II] then log Ill is harmonic inside the ball, so by the mean value property log 11(0)1 =
In the case where f has zeros Jensen's inequalny
log 11(0)1
flog fr(Re1')
holds as a consequence of Jensen's formula (Theorem 11 below). Jensen's inequality also follows from the theory of subharmonic functions (next chapter). 192
Secüon 3.
Jensen's (otmula
Jensen's formula tells what the difference is between the right hand side and the left hand side of Jensen's inequality. It is useful tool for providing information on the zeros of analytic functions on a disc, when restrictions on their sizes axe imposed.
Theorem ii (Jensen's formula). Let Pt,P2,
f be ineroinorphic in an open set containing B[0, R]. Let Z2," , and ",Pm be the zeros and poies respectively off in 11(0, R) (counted with multi-
Assume none of them Lc zero. log If(Rei#)I is integrable over Then 4'
=
log 11(0)1 + log
Proot Let
and
Jiog f (Ret') k4' be
• ,Zf,1 and
the zeros and poles
respectively of f in the closed ball BED, RJ, counted with multiplicity. The function
is holomorphic and zero free in a neighborhood of BED, RJ, so
flog
log IF(0)l = Putting
z =0, we find from the expression for F that IF(0)l =
so
log IF(0)I is the left hand side of the desired formula.
we find that
I
so
11
=
Ii
taking logarithms we get log
= log If(&'9) I + j=m+1 193
—
II
Il
—
Chapler 12.
Harmoinc Functions
To finish the proof it now suffices to show that — ir,ir[ and that
—.
log
—
is incegrable over
(Now, this is a book on complex function theory, so we shall here evaluate the integral by help of the Cauchy theorem. But the integral can actually be computed by elementary means [AGR),[Yo]). is continuous is locally integrable, and the function (1 — Since x log and never 0 on (.—lr, irj , the function +1og141
is integrable over ) — ir, 7r[. lb show that the value of the integral is 0 we note that z •—*
Log z z—
1
is analytic in the right half plane, so chat its integral along any circle I r E JO, 1[ , is zero by the Cauchy theorem, i.e.
0=!
J
—
= r, where
—f
Taking real parts we get
We will apply the dominated convergence theorem to this for a sequence The desired result is then an immediate consequence, once we dominate the incegrands by an integrable function. By differentiation we see that the function 1—
ret*
1 —
> 2
0,so 1_ret
194
1+r
1
Section 4.
Exercises
and so
log2 > log
log Ii —
—
6i431
— Iog2
0 Section 4 Exercises Show that u(Q) is open, unless u is constant. Hint: 1. Let u be harmonic in Connected subsets of the real line are intervals. 2. Derive the maximum principle for holomorphic functions from the strong maximum principle for harmonic functions. 3. Let f be a continuous, bounded and real valued function on the real line. Define
forx E Randy>Oafunctionuby
lIt (
u(z + iy) :=
00
(z—t)7+y2
dt
f
V
0
fory=0
(1(x)
(a) Show that u is bounded and harmonic in the upper half plane. Show that u is continuous in the closed upper half plane. 1).
4. Let u be harmonic in fl and identically 0 in an open, non-empty subset of (This property of harmonic functions is called the Show that u = 0 in all of
weak unique continuanon property). Hint: The unique continuation for holomorphic functions.
5. Define the map u : {x +iy E dy >0)
by
u(z) := the angle under which the interval [0,1] is seen from z. Show that u is harmonic. Hint: Consider the function Log((z — 1)/z). 6. Show that the function
u(x,y) :=
x2 + y2
is hannonic on R2 except at the origin. 7. Is harmonic for some 8. Show that the polynomial u(x, y) :=
an I
x is harmonic on ft2. Find
Hol(C) such that u =
Same problems with u(x, y) sin x cosh y 9. Show the following formula for r E [0,1[ and 0 1 rcos0=—
E
ft 2d4
27rj
Hint: Solve the Dirichlet problem when the boundary function is f(x + iy) = 195
x.
Chapler 12.
Harmonic Functions
10. (Liouville's theorem for harmonic functions) Let u be harmonic on ft2 and assume that u is bounded from below. Show that u is a constant. Hint: Property (iv) of the Poisson kernel. See also [Bo] and [Chi. 11. Show that the Dirichlet problem on {z E CO < Izi < 1) with the boundary function 1(z) = 0 for IzI = 1 , 1(0) = 1 has no solution. 12. Let u be harmonic in the annulus (z E dR1 < Izi u(xo) there exists a neighborhood U of zo such that
M > u(x)forallx
E
U.
A continuous real valued function is clearly usc. Note for later usc that an USC function is Borel measurable - indeed, the set {z E ClIu(z) 0 is arbitrary we
—
+
that
see —
—
z'i
lnteivhanging z and zi we get Ju,,(z)
tt,,(z')I
—
—
z'i
proving the continuity. Next we note that
u(y)
which implies that uN(z)
—
u(y) — (n +
— zI ,
i.e.
— zI
that {uN} is a decreasing sequence.
Since
u,,(z) = sup {u(y) — nly — zI) u(z) — nfz I
—
zI
= u(z)
it is left to prove that limuR(z) u(z) , i.e. for any prescribed K > u(z) that K for sufficiently large n. For that we use that u is usc : There is a ball B(z,6) such that u(y) < K for all y E B(z,6). Now, = suplu(y)— nly — zI)
I
sup lu(y)—nly—zI), sup {u(y)—nly—zI)
We shall finally deal with the general case in which u is no longer necessarily bounded from above. Choose an increasing homeomorphism (-oo,OJ 1-00,001 (E.g. = —exp(—t)). The function u is usc and bounded from above, so the construction in the special case treated above yields a sequence { v,, } of continuous real valued functions with the properly that as n —' 00 for all Z E 51. It .1 suffices to verify that —oo < vR(z) < 0 for all z E 51, because we may then choose
:=
0
200
Section 2. Irnroductoy propenies of suthannonic functions
The left inequality is trivial because
is finite valued. To settle the other inequality
we use the definition of v,,, i.e.
= sup
—
I
nIy
—
zI}
< 0, and by upper semi-continuity of
By hypothesis u(z) < oo , so a
ou
there exists 6 >0 such that
forall yEB(z,6) Thus
va(z)=max
sup
sup
I,—zI6
0. Since u is usc, then the function 9 u(z + re'9) is Borel measurable and bounded from above. Thus u is essentially a negative function. The value of the integral may a priori be —oo; as we shall see in Proposition 9, it is actually finite. There we shall also see that the inequality (b) holds for all r such that B(z, r) ç not just for sufficiently small r.
A harmonic function is subharmonic by the mean value property. 201
Chapser 13.
Subbxmonlc Functions
asaximu principle). Theorem 6 (The If there is a zo E Let u be subhannonic in
such that u(zO) = supu(z), then sEQ
u is a constant.
u
In particular, is compact, u subharmonic on 0 thmughow Il.
in
and
0 on Oft, then
The maximum principle is one of the crucial propenies of subharmonic functions.
= u(zo)} is closed in il since u is usc. We will
Proof: The set M := {z E
showthatMisopenaswell,solctzEMbearbitrary. Claim: + rei) = u(z) for all 0 E ft and r E JO, R5( We prove the claim by contradiction. If it is false then there are
E [0, 2ir[,
E JO, R3[
and e > 0 so that
0. 0 A subharmonic function need not be C2. But it can be approximated by smooth subharmonic functions:
Theorem II. Let u be subharmonic in Q. Let w be an open subset of Q with the property that > 0. Then there is a decreasing sequence u ... of subharmonic functions in such thai u(z) for all z E Q. Proof: The standard mollifying method (convolution) works. It goes as follows: Then the function Let 0 < R
0 (same proof).
We will show that f is holomorphic in any ball in ci; for simplicity of notation we assume that the ball is B(0, 1) and that B[0, 11 fl. We assume funhermore that
11(2)1< 1 for z E D[0,1]. Note that then, 0 on B(0,1). 210
Applications
Section 6.
Now, for any z E B(O, 1] we have by Theorem 7 that
+
u(z)
0 we get by letting e decrease to 0 that
At those points z E B(O, 1) where f(z)
u(z) < However, u is continuous and the points where f(z)
0 are dense in B(0, 1)
(Proposition 10), so the inequality holds for all points z E B(0, 1). Our arguments work in the same way for —1. i.e. with —u instead of u. Thus
u(z) = ..i_Ju(ei)Pi(eiz)do which shows (Theorem Xll.8) that u is harmonic. In particular u
1)).
Replacing f by if we get v E C°°(B(0, 1)), so f E C°°(B(O, 1)). f satisfies the B(0, 1)11(2) 0). and hence on all 0 of B(0, 1). So f is holomorphic. Cauchy-Riemann equations on the dense set {z
Hardy Spaces Definition 17. Let f be holomorphic in the unit disc.
p E)0,00( ,
f is said to belong to the Hardy space
2r
sup O(r0). continuous on {(x,y) E (fly? 0) and 0 on =0). {(x,y) E Then there exists a harmonic function U on fl such that U = u on {(x,y) E >0). Hint: Show that
(fix '
fory?0 1 —u(x,—y)
for y 0 . Define v on (1 by for Z
v(z) := I.
E
B(0,r)
for z E cl\B(0,r)
u(z)
Show that v is subharmonic on fl. 217
,
where
Chapter 13.
11. We
have seen that hi
general statement: If a 1 and h1, h2,•
,
is
Subharinonic Functions
subharmonic if h is harmonic. Show the following are harmonic, then
(Ihii° + lh2Ia + ... + is subharmonic. 12. Use the procedure from the proof of Radô's theorem to show the theorem on an isolated singularity for a harmonic function (Theorem XII.5).
218
Chapter 14 Various Applications SectIon 1 The Phragmén-Llndelöf principle The weak maximum principle (Corollary IV.16), which says that an analytic function never exceeds its maximal boundary value, does not hold in general for unbounded domains. For example, if fl is the lower half plane, so that Of 1 is the real axis, then 1(z) := exp(iz) is bounded on Ôfl but not on Il. So we need extra
restrictions on the function and the domain to get maximum principles for unbounded domains. The so-called Phraginén-Lindelof principle is a way of finding such maximum principles. Our version is the following.
Theorem I (Phragmén-lindelUJ's principle). be afunction in
Let fl be a connected open subset of C and let which Is bounded and no: Identically 0.
1ff E
fl Hol(1l) andthe constant M
(1)If(z)I Mforailz E
(1 Hol(1Z)
that
0
,and
—+ 0 as z —+ 00 In Il for each jixedq > 0,
(2)
then If(z)I Mforallz E Proof: The case M = 0 is a consequence of the case M > 0 , so we may assume Consider for > 0 the continuous function that M >0. Let B >0 be a bound of
u,,(z) :=
for z E
which is subharmonic on (1 (Example Xll.17(8)).
choose R >
0
By the assumption (2) we may
so large that
fcrall zEIlsuchthal IzIR Since Offl fl B(0,R)) c E fl IzI = R} we get by the maximum principle for subharmonic functions (Theorem XIL6) that
uq(z)_<max{M,M}=M for zEflflB(O,R) This inequality is also true outside of the ball B(0, R) by our choice of R, so
u,,(z) M for z E fl, i.e. :S M for z E fl
0. However, the zeros of are isolated in Il (Theorem IV.1l), so the desired inequality holds by the continuity of f everywhere in 11. 0 219
chapter 14.
Applications
The connectivity assumption was only used at the very end of the proof. We may delete that assumption if we in retwn require that 4' vanishes nowhere. Corollary 2. Let C be an open subset of the complex plane. Let I e C(?i) fl
be
bounded. 11111
MonOQ,then Ill
Proof: Wemayofcourseassumethatfl Lete>0. By continuity of f there exists a point zo e such that If(zo)I < M + e and then there is even a ball B(zo,rJ ç such that Ill M + e on that ball. Applying Theorem I to the set and the function 4'(z) := (z — :o)_1 we get that
If(z)I M + e for all z E Q\B(:ø, r) This inequality also holds for z E D(zo, r] by the very construction of that ball, so
0
f(z)IOwasarbitrary.
Corollary 3 (The three lines theorem). Let := { z E C 0 < < 1). Let be bounded and holomorphic on 1 and continuous on the closure
(1)
of Q and pw
M(r) :=
f
sup —oo