TheBierstoneLectureson ComplexAnalysisScribedbyOlegIvriiduringSpring2007PrefaceThese notes of my lectures on Complex Analysis at the University of Toronto were writtenby Oleg Ivrii on his own initiative, after he took my course in the spring, 2007. The courseiscross-listedasMAT1001S,oneoftheCoreCoursesinourGraduateProgram, andasMAT454S, afourth-yearundergraduatecoursewhichisthesecondpartof atwo-termsequenceincomplexanalysisinourSpecialistPrograminMathematics. Olegtookthiscourse as a second-year undergraduate.Mylectureswerebasedontheclassical textbooksof LarsAhlfors(ComplexAnaly-sis, Third Edition, McGraw-Hill 1979) and Henri Cartan (Elementary Theory of AnalyticFunctionsofOneorSeveralVariables, Dover1994). Myexpositioncloselyfollowsthesetexts. Oleg has included the problems that I assigned as homework; most of these can befound in Ahlfors, Cartan or other textbooks. Oleg has added a concluding chapter as anintroduction to several fundamental further results. I have included some of these topicswhengivingthecourseonearlieroccasions, andtheyimprovemylecturesaspresentedhere!During my thirty-ve years of teaching at the University of Toronto, I have been trulyfortunatetohavehadsomanyinspiringstudentslikeOleg. Theyarepassionateaboutmathematicsandtheycareforeachotherandtheworldaroundthem. IamgratefultoOleg that he thought my lectures were interesting enough to want to write them up, andI give him my warmest wishes for a very happy and fruitful mathematical career.March 2008 Edward Bierstone2Contents1 SpaceofHolomorphicFunctions 51.1 Topology on the Space of Holomorphic Functions . . . . . . . . . . . . . . . 51.2 Series of Meromorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . 71.3 The Weierstrassp-function . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 The Weierstrassp-function II . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Complex Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 The Elliptic Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.7 Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Interpolation 162.1 Meromorphic functions with prescribed singularities . . . . . . . . . . . . . 162.2 Innite products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Entire functions with prescribed zeros . . . . . . . . . . . . . . . . . . . . . 192.4 The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.5 Normal families and compact subsets of C() . . . . . . . . . . . . . . . . . 223 Automorphisms 253.1 Local Geometry of Holomorphic functions . . . . . . . . . . . . . . . . . . . 253.2 Automorphisms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3 Riemann Mapping Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 Riemann Mapping Theorem II . . . . . . . . . . . . . . . . . . . . . . . . . 293.5 Christoel-Schwarz Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . 3134 ComplexGeometry 334.1 Complex Manifolds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2 Complex Manifolds II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.3 Examples of Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 364.4 Example: y = _P(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.5 Abels Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.6 Abels Theorem II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 AModularFunction 425.1 Analytic Continuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.2 Monodromy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.3 Modular Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.4 Modular Function II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465.5 Little Picard Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476 Problems 497 FurtherResults 557.1 Big Picard Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2 Runges Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.3 Prescribing Zeros in Arbitrary Domains . . . . . . . . . . . . . . . . . . . . 597.4 Prescribing Singularities in Arbitrary Domains . . . . . . . . . . . . . . . . 604Chapter1SpaceofHolomorphicFunctions1.1 TopologyontheSpaceofHolomorphicFunctionsSuppose is open in C, letC() be the ring of continuous Cvalued functions on andletH() be the subring of holomorphic functions.We can topologize C(): a sequence of functions {fn} converges uniformly on compactsets if for every compact setK , {fn|K} converges uniformly.Uniform convergence on compact sets denes the so-called compact-open topology.The fundamental system of open neighbourhoods at 0 isV (K, ) = {f: |f(z)| < , z K}.This is becausefn funiformly on compact sets if and only iff fn lies inV (K, ) forn large enough for any givenK, .In fact,C() is metrizable (even by a translation-invariant metric).Write = iKi and letd(f) = i12i min(1, Mi(f)) whereMi(f) = maxzKi|f(z)|.C() is complete:if {fn} is a sequence of continuous functions which converge uniformlyon compact sets,f= limfn is continuous.We giveH() the subspace topology.Theorem. (1). H() is a closed subspace. (2). Furthemore, the mapf f

is continu-ous.5(1)meansthatif {fn} H()convergesuniformlyoncompactsets, f =limfn H(). (2)meansthatif {fn} H()convergesuniformlyoncompactsetsandif f=limfn, then {f

n} converges uniformly on compact sets tof

.Proof of (1). Wehavetoshowthatf(z)dzisclosed. ByMorerastheorem, thisisequivalent to showing that _ f(z)dz = 0 for any small loop. Asfn(z) are holomorphic,_ fn(z)dz = 0 holds for alln. Then, _ f(z) = limnfn(z)dz becausefn converges tofuniformly on the image of.Proofof(2). It is enough to show thatf

nconverges tof

uniformly on a closed diskD(z, r)in. Letbeapositivelyorientedlooparoundadiskofslightlylargerradius.By Cauchys integral formula,limnf

n(z) =limn12i_fn()( z)2d =12i_f()( z)2d = f

(z).The limit above converges uniformly as z is bounded away from 0.Corollary. If a series of holomorphic functions n=1fn(z) converges uniformly on com-pact sets of , the sum is holomorphic and can be dierentiated term-by-term.Proposition(Hurwitz). If is adomain(i.econnected), if {fn} H() convergesuniformly on compact sets and eachfnvanishes nowhere in , thenf= limfnis never 0or identically 0.Supposefwas not identically 0. Then its zeros are isolated. In particular around anypointz, there is a small looparoundzfor whichfdoes not vanish on the image of.But as the image ofis compact,fis bounded away from 0. Then,limn12i_f

n(z)fn(z)dz =12i_f

(z)f(z) dz.The left hand side counts the number of zeros offnenclosed within, which is 0. Hence,f(z) also does not possess any zeros inside , i.e f(z) = 0. As z was arbitrary, f(z) vanishesnowhere in .Proposition. If is a domain and {fn} H() are one-to-one and converges uniformlyon compact sets, thenf= limfnis either one-to-one or a constant.Supposefwas not identically a constant, but neverthelessf(z0) = f(z1) = a. Choosedisjoint neighbourhoodsUandV ofz0andz1respectively. Thenf(z) a vanishes at a6point in U, so we can extract a subsequence of {fn} for which fn(z) a vanishes at a pointin U. From this subsequence, we extract another subsequence for which fn(z) a vanishesat a point inV . But then,fn(z) would fail to be injective.1.2 SeriesofMeromorphicFunctionsA series of meromorphic functions n=0fn on an open set C converges uniformly oncompact subsets of if for every compact K , all but nitely many terms have no polesinKand form a uniform convergent series.Wedenethesumonanyrelativelycompact (precompact) openUas n 0,thereisabsuchthat |f(z)| b.The function g(z) =_sin z_2also enjoys these properties. The rst property is obvious.To check the second property, we rst check that sin zvanishes only on the integers.By periodicity, it suces to check that the principal part ofg(z) at the origin is1z2; for aproof, see the calculation below.7Easiest way to see property (3) is by means of the identity | sin z|2= sin2x+sinh y,asy goes to , sin z and henceg(z) 0.Hence, f gisholomorphicinCbecause f andghavethesamepolesandsameprincipal parts;f g is bounded, so f g is a constant by Liouvilles theorem, and by thethird property, the constant is 0.We derive the formula n=11n2=26 .Consider the function_sin z_21z2= n=01(zn)2. It is holomorphic in a neighbour-hood of 0 and asz 0, it tends to2(z 163z3+. . . )2 1z2=1z2(1 13z2+. . . ) 1z2=23.But whenz = 0, the function is 2

n=11n2.Example2Theseries f(z) =1z+ n=0_1zn+1n_=1z+ n=0zn(zn)converges uniformlyandabsolutely on compact subsets of C by comparison with n=01n2.Thusf(z)isameromorphicfunctionwithpolesz=nandprincipal parts1zn(i.esimple poles with residue 1).We notice:f

(z) = 1z2

n=01(z n)2= _sin z_2=ddz( cot z).Hence, f(z) cot zis constant, which is 0 as both functions are odd (to see thatf(z) is odd, group terms with +n and n). So,1z +

n=12zz2n2= cot z.1.3 TheWeierstrassp-functionLet be a discrete subgroup of C with generatorse1, e2linearly independent over R,i.e = {n1e1 +n2e2 : n1, n2 Z}.Vectors e

1, e

2areanothersetof generatorsforif andonlyif e

1, e

2arearelinearcombinationof e1, e2withdeterminantof matrixof coecientsinvertibleintheringofintegers, i.e 1.8A function f(z) has as a group of periods if f(z+n1e1+n2e2) = f(z) for all n1, n2 Z.We denep(z) =1z2 +

w,w=0_1(z w)2 1w2_.Note that the Weierstrassp function depends on the choice of .Wecheckthatitconvergesuniformlyandabsolutelyoncompactsubsets,forthisweestimate:1(z w)2 1w2 = 2zw z2w2(z w) = . . .We can restrict our attention to the compact subset |z| r. As nitely many terms dontmatter, we only need to estimate whenw is suciently large, say |w| 2r. We continue: =z(2 zw)w3(1 zw)2 r 52w314 =10r|w|3.We want to show that w,w=01|w|3< . We denePn = {n1e1 +n2e2 : max(|n1|, |n2|)}.We letkbe the distance between the origin to the closest point ofP1. As the number ofpoints inPn is 8n, we see that

w,w=01|w|3=

n=1

wPn1|w|3 0.By the implicit function theorem (for real-valued functions), we see that we can solvey1, y2 asC1functions in terms ofz1, z2; x1, x2 near (x0, y0, 0).The equationdz=fxdx +fydyshows that thatdyis a linear combination ofdx, dzwith holomorphic coecients (here we are using thatfy = 0). This means the solution isholomorphic.15Chapter2Interpolation2.1 MeromorphicfunctionswithprescribedsingularitiesWenowshiftgearstoconstructingmeromorphicfunctionswithprescribedsingularities,i.e with given poles and their singular parts.Over the Riemann sphere, the only meromorphic functions are rational functions; butover the complex plane, we have a wealth of meromorphic functions such as tan zor sec zbut innity is a limit of poles.Theorem (Mittag-Leer). Suppose {bk} C are given and distinct with limkbk = andPk(z)bepolynomialswithoutconstantterm. Thenthereisameromorphicfunctionin C with polesbkand principal partsPk(1zbk). The most general meromorphic functionwith these poles and primitives parts isf(z) =

k=1_Pk_1z bk_pk(z)_. .+g(z)wherepk(z) are polynomials chosen such that the series convergences andg(z) is an entirefunction.We can assume that nobk = 0. The functionPk(1zbk) is holmorphic in |z| < bkso wecan expand it as a convergent power series in that disk.Letpk(z) be the sum of the terms of order up tomkwheremkis chosen such thatPk_1z bk_pk(z) 12kon |z| |bk|216We show converges uniformly and absolutely on |z| rfor anyr. Chosenn such that|bk| 2r ifk > n. It suces to compare

k=n+1Pk_1z bk_pk(z)with 12k. Thus ismeromorphicin Cwithpoles {bk}andprinciplepartsPk(1zbk).Given any other function with prescribed singularities, the dierence of it and is holo-morphic and conversely, given any holomorphic functiong(z), the function + g(z) alsohas the desired singularities.2.2 InniteproductsWesaythat n=1bnconvergestop Cif nobnare0andthepartial products pn=b1b2. . . bn have a non-zero limit p. Actually, this is too restrictive: we should allow a nitenumber of terms to be 0, but after removing these few terms, the partial products shouldhave a non-zero limit.Anecessaryconditionforconvergenceisbn=pnpn11. Wewill insteadwritetheproduct as (P) n=1(1 +an), so the condition requires thatan 0. To the product, weshall associate the sum (S) n=1 log(1 + an) where we take the principal branch of log,dened on the complement of the negative real ray with log 1 = 0.Proposition. The product (P) and the sum (S) converge or diverge simultaneously. Justaspnis then-th partial product of (P), we will writesnfor then-th partial sum of (S).If snconverges, thensodoes pnas pn=esn. Conversely, supposepnconvergestosomep =0. Fixadeterminationoflog p=log |p| + i arg p. Choosedeterminationsoflog pn = log |pn| +i arg pn such that arg p < arg pn arg p +.Thensn = log pn + 2i knfor integerskn. Forsnto converge, we need to know thatkn stabilize. Indeed,2i(kn+1kn) = log(1 +an+1) log pn+1 + log pn,= log(1 +an+1) (log pn+1log p) + (log pnlog p).The three terms log(1 + an+1), log pn+1 log p, log pn log p are small whenn is large.As the left hand side can take upon a discrete set of values, it must be 0.17We say that the product (P) converges absolutely if (S) does, but the latter is equivalentto (S

) n=1an converging absolutely as limz0log(1+z)z= 1.Now, suppose thatfn(z) are continuous complex-valued functions dened on an openset C. We say that the product n=1fn(z) converges uniformly (or absolutely) on acompact setK if log fn(z) does.Theorem. Supposefn(z) are holomorphic in and n=1fn(z) converges uniformly andabsolutelyoncompact subsetsof . Thenf(z)= n=1fn(z)isholomorphiconandf= f1f2. . . fq

n>q fn(z).If Z(f) denotes thezeros of functionf, thenZ(f) = Z(fn). Additionally, themultiplicity of a zero offis the sum of that offn.Corollary.Under the same hypothesis, the series of meromorphic functions n=1f

nfncon-verges uniformly on compact subsets of and limit isf

f .LetUbearelativelycompactsubsetofandgq(z) = exp_

n>q log fn(z)_.Bytheproduct rule:f

(z)f(z)=q

n=1f

n(z)fn(z) +g

q(z)gq(z)=q

n=1f

n(z)fn(z) +

n>qf

n(z)fn(z).The latter equality is justied as logn>q log fnconverges uniformly to a (branch of) log gqand thus can be dierentiated term-by-term (takeq large enough so thatfn have no zerosinUforn > q).ExampleWe develop an innite product for sin z. The productf(z) =z

n=1(1 z2n2) convergesuniformly and absolutely on compact subsets of C (because z2n2does).Dierentiating logarithmically, we nd that1z +

n=12zn21 z2n2=1z +

2zz2n2= cot z =sin

zsin zby Example 2 in Section 1.2. It follows thatf(z) =c sin z. But as limz0f(z)z= 1 andlimz0sin zz= , we see thatc =1. Hence, sin z = z

n=1(1 z2n2).182.3 EntirefunctionswithprescribedzerosAnentirefunctionf(z)withnozerosisoftheformeg(z)forsomeentirefunctiong(z).Indeed, the formf

(z)f(z) dz counts the number of zeros, but since there arent any, it is exact.So, it is the derivative of an entire functiong(z). Thenddz_f(z) eg(z)_ = f

(z) eg(z)f(z) f

(z)f(z) eg(z)= 0,sof(z) = const eg(z)but the constant can be absorbed intog(z).An entire function with nitely many zeros {ak} (these not be necessarily distinct) canbe written asf(z) = zmeg(z)n

k=1_1 zak_.Here, weputthezerosatz= 0separatelyinthefactorzm. Now, weexaminethecasewhen an entire function has innitely many zeros.Theorem(Weierstrass). Suppose {ak} C, limkak = ,aknot necessarily distinct.There exists an entire function with prescisely {ak} as zeros. Any such function is of theformf(z) = zmeg(z)

k=1__1 zak_ pk(z)_.Here, pk(z) arefactors whichguaranteethat theproduct converges. Theproductconverges if the sum of logarithms of its terms converge, so we look at the expansionlog_1 zak_ = zak 12 z2a2k 13 z3a3k. . .Thus, a good form forpk(z) = exp_zak+ 12 z2a2k+ +1mkzmkamkk_.Weneedtoshowwecanchooseintegersmktomakethattheproductconverge, i.eweneed to make k_log_1 zak_ +zak+12 z2a2k+ +1mkzmkamkk_ converge. We denote thegeneral term of this sum bygk(z).Actually, we would need that < Imgk(z) < , so we know that we took the principalbranch of logarithm and not some other branch (this would be automatic because we needgkto tend to 0 for convergence).19Supposer > 0, for |z| r, we consider terms with |ak| > r:gk(z) = 1mk + 1_zak_mk+11mk + 2_zak_mk+2. . .Thus,|gk(z)| 1mk + 1_r|ak|_mk+1_1 r|ak|_1It suces to choose {mk} so that (r|ak|)mk+1converges for everyr> 0 (note;however,thatthesumhaslesstermswithgreaterr); thentheproductconvergesuniformlyandabsolutely in |z| r. Clearly,mk = k suces.Corollary. EverymeromorphicfunctionF(z)denedontheentirecomplexplaneisaquotient of entire functions.Indeed, let g(z)betheentirefunctionwiththepolesof F(z)asitszeroes. Then,F(z)g(z) is an entire functionf(z), soF(z) =f(z)g(z).Corollary.Suppose ak C, limkak = ,bk C,mk N. Then there exists an entirefunctionf(z) such thatakis a root of ordermkoff(z) = bk.Letg(z) be an entire function with zero of ordermkatak. Write:f(z) = g(z)h(z) = bk +g(z) _h(z) bkg(z)_Now chooseh(z) to be a meromorphic function with poles {ak} such that at eachak, theprincipal part ofh(z) was the principal part ofbkg(z).Atz=ak, g(z)hasazeroofordermk, h(z) bkg(z)isholomorphicwhichshowsthatf(z) = bkhas a root of order at leastmkatak.To make sure that f(z) = bk has a root of order exactly mk at ak, we make two changes:(1) we change the order ofg(z) atakfrommktomk + 1 and (2) we change the principalpart ofh(z) atakto bebkg(z) +1zak.202.4 TheGammaFunctionConsider the innite product:g(z) = z(1 +z). .f1(z)

n=2_ _1 +zn__1 1n_z. .fn(z)_.We want to show it converges uniformly and absolutely on compact subsets of C. For thispurpose, we examine the sum oflog fn(z) = log_1 +zn_+z log_1 1n_.=_zn z22n2 +z33n3 . . ._+z(1n 12n2 13n3 . . ._If |z| r andn > 2r, we have the estimatelog fn(z) 2 r2n2 _1 rn_1< 4r21n2.Hence the associated sum converges by comparison with 1n2.Letgn(z) = nk=1fk(z). As_1 12__1 13_. . ._1 1n_ =12 23 n1n=1n, we seegn(z) = z(1 +z)_1 +z2_. . ._1 +zn_ nz=z(z + 1) . . . (z +n)n!nzThe functiong(z) has zeros at 0, 1, 2, . . . , all of which are simple.Also,g(z)g(z+1)= z. Indeed,gn(z + 1) =(z+1)(z+n+1)n!nz1, sogn(z)gn(z+1)=znz+n+1 zasn . Additionally,g(1) = limngn(1) = limnn+1n= 1.Dene the Euler gamma function (z) as1g(z). It is meromorphic with simple poles at0, 1, 2, . . . and satises (z+1) = z(z), (1) = 1, from which we can see (n+1) = n!.We now prove the identity (z)(1 z) =sin z:gn(z)gn(1 z) =_z(z + 1) . . . (z +n)n! nz__(1 z)(2 z) . . . (n + 1 z)n! nz1_=n + 1 zn zn

k=1_1 z2k2_.BytheExampleinSection2.2, asn , theinniteproductconvergestosin zandn+1zn1, henceg(z)g(1 z) =sin zz.21We now give an alternate innite product for1(z). We writegn(z) = zn

k=1__1 +zk_ezk_ez(1+12++1nlog n).This suggests thatg(z) = z

k=1__1 +zk_ezk_ ez.The convergence of the innite product can be established as before (we can again comparewith 1n2). This alternate expansion shows that 1+12 + +1nlog n converges; its limitis, the Euler gamma constant.The logarithmic derivative of (z) is

(z)(z)= 1z +

n=1_1n 1z +n_.Taking derivatives again, we ndddz_

(z)(z)_ = n=01(z+n)2. This is positive whenz is realand positive, so log (z) is convex.A theorem of Bohr and Mollerup says that is the only meromorphic function whichislogarithmicallyconvexonthepositiverealaxiswhichsatisesthefunctionalequation(z + 1) = z(z) and (1) = 1.2.5 Normalfamiliesandcompactsubsetsof C()Suppose C is open. We have seen thatC() is metrizable and thatH() is a closedsubset ofC() with the induced metric.Recallthatametricspaceiscompactifandonlyifeverysequencehasaconvergentsubsequence.We say that S H() is a normal family if every sequence in Shas a subsequencethat converges uniformly on compact subsets of ; in other words, S is normal if its closureis compact.A set S H() or ofC() is bounded uniformly on compact subsets of if for everycompactK , there is a constantM(K) such that |f(z)| M(K) whenz Kfor allf S.22Proposition.If S H() or of C() is compact, then S is closed and bounded uniformlyon compact sets.The fact that Sis closed follows fromH() being Hausdor. For a xed compact setK, the mapH() R :f maxzK|f(z)| is continuous and hence it takesKto somebounded set of R.The converse is only true inH(). It follows from the following theorem:Theorem(Montel). AsetoffunctionsS H()isanormalfamilyifandonlyifitisbounded uniformly on compact sets.We prove it using two lemmas.Lemma. ThemappingH() H() :f f

takesasubsetthatisboundeduniformlyon compact sets to another subset bounded uniformly on compact sets.Itsucestoshowthatforanyz0 , thereisasmall diskD centeredatz0such that S

is bounded uniformly onD. ChooseD with radiusr/2 such that the diskDcentered atz0with radiusr is contained in . Letbe the positively oriented boundaryofD.Supposez D. By Cauchys integral formula, f

(z) =12i_f()(z)2d. SupposeMissuch that |f(z)| Monfor allf S. As | z| r/2, |f

(z)| M2 4r2 2r =4Mr.Lemma. LetDbetheopendiskwithcenterz0andradiusRand S H()beboundeduniformlyoncompactsubsetsof D. Thenasequence {fn} Sconvergesuniformlyoncompactsetsifandonlyifforeveryk,thesequenceofderivativesat z0, {f(k)n(z0)}con-verges.For the only if direction, we simply need to know that the map f f

is continuous(we dont need the hypothesis).Nowweprovetheifdirection. Itsucestoshowthatforanyz0 , thereisasmall diskD centered atz0such that for anyf S, {fn} converges uniformly onD. LetD(z0, R) be a closed disk in . We pickD = D(z0, r) with radiusr < R.Letr n0, |ankamk| are simultaneously small forall 0 k l. Then {fn(z)} converges uniformly as desired.We are now ready to prove Montels theorem. Suppose S H() is bounded uniformlyon compact sets and {fn} S. We wish to show {fn} has a convergent subsequence.TakeacountablecoverofbydisksD(zi, ri) . Letji: H() Ctakef f(k)(zi). It suces to nd a subseqenceN N such that limnN ki(fn) exists for allk, i.We constructNas follows: We reindexkiasl. By the rst lemma, for eachl,l(fn)isbounded. Thenwediagonalize, i.ewerstchooseN1 NsothatthelimnN1 1(fn)converges; nextwechooseN2 N1sothatlimnN2 2(fn)converges, thenN3 N2solimnN3 3(fn) converges and so on.We then takeNto be the diagonal sequence: N11, N22, N33, . . .24Chapter3Automorphisms3.1 LocalGeometryofHolomorphicfunctionsWesaythat afunctionf is conformal (biholomorphic) if f is holomorphicandhas aholomorphic inverse. Iffis a holomorphic function not identically constant, either of thetwo possibilities hold:(1)Supposew0=f(z0)andf

(z0) = 0. Bytheinversefunctiontheorem, f1existsand is holomorphic in a neighbourhood ofw0 and sofis conformal atz0.(2) Suppose insteadf

(z0) = 0. We may assumez0 = 0,f(z0) = 0; otherwise we workwithf(z + z0) f(z0). We may writew = zpf1(z) withf1(0) = 0 for somep 2. Then,w = (zg(z))pwhereg(z) is some determination of thep-th root off1. Thenw = pwhere = zg(z) is a holomorphic coordinate change, i.e a map of type (1).Thus up to a conformal change of coordinates,every holomorphic function is of formp+f(z0). The numberp 1 is called the ramication index offatz0.Lemma. A non-constant holomorphic mapping is open.We know that in a small neighbourhood of f(z0), every value suciently close to f(z0)appears exactlyp times (as this is true forp+f(z0)), sofis open.Corollary. Iff H() and injective, thenfis a homeomorphism. Furthermore,f1isholomorphic.Since (2) cannot be injective,(1) must always be the case,hencefis a local homeo-morphism. But asfis injective, it is a genuine homeomorphism.25Remark. Evenif f

(z) =0forall z , thisnot guaranteethat f isinjective. Forinstance,f(z) = ezis one-to-one on any stripa < Imz< b withb a < 2.Remark. While the complex plane C and the unit diskD = {z : |z| < 1} are homeomor-phic; theyarenotbiholomorphic, foranyfunctionf: C DisconstantbyLiouvillestheorem.Noticethatif f, g:

aretwobiholomorphicmaps, theng1 f : isbiholomorphic. Henceall biholomorphicmaps

aregivenbycomposingaxedholomorphic map with an automorphism of .ExamplesThe automorphisms of the Riemann sphere are the fractional linear transformationsz az+bcz+d. The automorphisms of the complex plane are ane mapsz az +b.The automorphisms of the upper half plane are fractional linear transformations withreal coecients andad bc > 0.3.2 AutomorphismsWe classify automorphism groups of some domains in C.Theorem. Aut C are ane mapsw = az +b witha = 0.Suppose w = f(z) is an automorphism of C. Either, (1) f(z) has an essential singularityat innity, or (2) it is a polynomial (being a rational function with only pole at innity).Therstcasecannothappen: ononehand, Im{|z| 1}shouldbedisjoint(as f(z)isinjective), butas f isopen, therstsetisopen; buttheessentialsingularity at innity implies the second set is dense which is impossible.We investigate the second case further. Supposefis a polynomial of degreen. Then,f(z) = w hasn distinct roots except for special values ofw (namely, whenf

(z) = 0). Asfis injective, we see thatn = 1.We concludef(z) = az + b. Obviously,a = 0. Ifa = 1,f(z) is a translation; ifa = 1,fhas a unique xed pointz =b1a.26Theorem.Aut S2are fractional linear transformations w =az+bcz+d,adbc = 0 (the inverseisz =dwbcw+a). (Note: wtakes innity toacifc = 0 and to innity ifc = 0).Obviously, thefractionallineartransformationsformasubgroupof G Aut S2. Wewant to show that actuallyG = Aut S2.The subgroup of Aut S2which xes the point at innity is precisely Aut C. As fractionallineartransformationsacttransitivelyontheRiemannsphere,forT Aut S2,wecouldconsidera fractionallinear transformationSwhichtakesT() . ThenS Tmustbe ane which shows thatTis a fractional linear transformation.LetH+betheupperhalf-planeandDtheunitdisk. Thetwoarebiholomorphic:indeed, the mapz ziz+iwhich takes H+toD: it mapsi to the origin and points 0, 1, to 1, i, 1 respectively.Theorem. Aut Dare fractional linear transformations of the formw = eiz z01 z0zwith |z0| < 1.The above transformations are automorphisms, they take the unit disk into itself andtheir inverse is of the same form. We now show that there are no other automorphisms.SupposeT Aut D. LetS=eizz01z0zwithz0=T1(0)and=arg T

(z0). Letf(z)=S T1(z). Thenf(0)=0and |f(z)| N{f} is compact, by the Arzela-Ascoli theorem, we see that it is uniformlyS2-equicontinuous. Let 0, for weworkin1/r = {z : d(z, ) > 1/r and consider terms withr|ak bk| < 1:gk(z) = 1mk + 1_ak bkzk bk_mk+11mk + 2_ak bkzk bk_mk+2. . .Thus,|gk(z)| 1mk + 1_r|ak bk|_mk+1_1 r|ak bk|_1Just like in the proof of the regular Weierstrass theorem, we need to choose mk so that

(r|ak bk|)mk+1converges for everyr > 0. Again,mk = k suces.Remark.Unfortunately, if {ak} is unbounded, it is not always possible to choose a satellitesequence. Forinstance,let = H+andak=ik. Forthis {ak}asatellitesequencedoesnot exist.59We continue the construction when {ak} happens to be unbounded. Pick a point a Cwhich happens not to be one of the designated ak, nor is one of their accumulation points.The mappingh(z) =1zawill ensure that {h(ak)} is bounded.Letf(z)beaholomorphicfunctionwith {h(ak)}aszeros. Byconstructionf(z)=f(h(a)) has the desired zeros; trouble is that it may not be a holomorphic at a (and it maynot be possible to choosea outside of ). But actually, it is holomorphic ata!But limzGm_akbkzkbk_ = 1 and as product converges absolutely and uniformly nearinnity, we see that actuallyf() = 1, i.ef(a) = 1.Corollary. Foranydomain C, thereexistsaholomorphicfunctionf(z) H()which cannot be continued analytically to any larger domain.Let f(z) be a function whose zeros accumulate to every boundary point of (actually,this is easier said than done).It is even true that any domain C is a domainofholomorphy which means thatthere is a holomorphic functionf(z) H() such that for any pointz0, the power seriesexpansion off(z) converges only inside .For instance, the principal branch of log z (which is dened on C without the negativereal ray) cannot be extended anywhere on this ray, but power series expansions do convergebeyond this ray.7.4 PrescribingSingularitiesinArbitraryDomainsWe wish to construct a functionf(z) meromorphic in a domain with poles {bk} (accu-mulating only at the boundary of ) and principal partsPk(1zbk).Plan: write = iKi as the increasing union of compact sets. Let Li = Ki+1\Ki andBibethebkswhichlieinsideLi. ByRungestheorem, existsarationalfunctionpk(z)with poles lying outsideKi+1 such that |Pk(1zbk) pk(z)| < /2ionKi. We want to takef(z) =

k1#Bi_Pk_1z bk_pk(z)_.The contribution from thebks onLjtoKiwithj>i is less than/2j,hence the seriesconverges uniformly and absolutely implying thatf(z) is meromorphic.60Wehavetoinsurethatapplicationof Rungestheoremisvalid; forthisweneedtotakeK1largeenoughsothateachboundedcomponentofC\K1containsaboundedcomponent of C\ (then this will also be true ofKj,j 2).Lemma. Givenacompact subset K, it ispossibletochooseacompact K1 containingKsuch that every hole (compact component) ofK1contains a hole of .If = C, we can letK1 be giant ball, so in what follows we consider = C.Let r = {z : d(z, ) r} where r is some number less than d(K, ). Alternatively,we may describe the complement of r:C\r =_zC\D(z, r).In particular,ris closed. We would takeK1= r,but rmay not be compact;so wetake be a giant ballBcontainingKand setK1 = r B.Let ZbeaboundedcomponentofC\K1. AsC\Bisaconnectedandunboundedchunkof C\K1, itcannotintersectZ; forif Zhascommonpointswith C\B, Zwouldhave to contain C\Bwhich is absurd. HenceZ _C\K1_\_C\B_ = C\r.Each D(z, r) which makes up C\r is connected; so is either contained or disjoint fromZ (that is why we have shown Z C\r), this tells us that Z is the union of disks D(z, r)which have center inZ, i.eZ =_zZ(C\)D(z, r) =_zZ\D(z, r).AsZis non-empty, the above representation tells us thatZintersects C\, but then weare done: supposeZintersects C\ at a pointp and ifSis the connected component ofp in C\, then as C\K1 C\, thenZwould have to contain all ofSas desired.61