RhodesUniversityDepartmentofMathematicsM3.2ComplexAnalysisJulienLarena20122ForewordsThese lecture notes are intendedfor aone semester third-year course inmathematicsatRhodesUniversity.Mostly,theyfollowthetracksandspiritoftheexcellentintroductorybookby H.A. Priestley, Introduction to complex analysis, OUP, and do not presentanyoriginal result, oranyoriginal pathtoknownresults. Ihavetriedtokeep the references to real analysis as limited as possible, so that this coursecouldbe studiedwithout muchprior knowledge of real analysis. Never-theless, tocut onsome time-consumingproofs, I have omittedtoprovesomeresultsasCauchysconvergencecriterionforsequences, orBolzano-Weierstrasstheorem,asthesetheoremswillbeprovenduringthecourseonreal analysis, andtheirproofsincomplexanalysisareverysimilar, orcanbededucedfromthepropertiesintherealcase. Ifareaderisinterestedintheseproofs,shecanrefertoW.Rudinsbook,PrinciplesofMathematicalAnalysis, McGraw-Hill, thattreatsof propertiesingeneral metricspaces.Also, I did not prove Jordans curve theorem, as it would have been painful,long,andmostlyunnecessary,sincewewilldevelopcontourintegrationforthelimitedclassofnonself-intersectingcontours, forwhichthenotionsofinteriorandexteriorarequiteobvious. I,nevertheless, mentionedtheideaof thegeneral proof, incasesomestudentsmaybewillingtothinkabouttheproblem. Theclassof nonintersectingcontoursmadeof piecesof arccircles and line segments is largely sucient for applications in an introduc-torycourseoncomplexanalysis.iiiI shall recommend to read another excellent book: Visual Complex Analysis,by T. Needham, OUP. It is a wonderful, very graphic, exposition of complexanalysis, withanemphasisonphysical andgeometrical interpretationsofthenotionspresentedinthesenotes. Iencouragethestudentswhowishtodeveloptheirintuitiononcomplexanalysistoreadthisbook.Studentswillingtondmoreexercisesandproblemsthanthosewewillbedoing in class and tutorials can refer to Complexvariables by M. R. Spiegel,inthecollectionSchaumsoutlines,McGrawHill.Finally,letmeemphasizethatanalysisisanewsubjectforthirdyearstu-dents, andadedicatedstudywill benecessaryinordertosucceedinun-derstandingthiscourse. Thisismainlyduetotheintroductionofrigorousproofs. Therefore, studentsmustpayattentiontoproofsandtomethodsusedduringtheseproofs, astheyarethekeystomasteringthetechniquesofComplexAnalysis.Coverillustration: Augustin-LouisCauchyaround1840. LithographybyZephirinBelliardafterapaintingbyJeanRoller.NotationsWeuse N, Z, Rand Ctodenotethesetsofnaturalnumbers,integers,realnumbersandcomplexnumbers, respectively. Whenwewanttoexclude0from one of these sets, we will simply star the set. For example R= R\{0}.Inthesameway, whenwewanttoindicatethatwekeeponlythepositive(or zero) (resp. negative or zero) real numbers, we will write R+(resp.R).Alotof notationsincomplexanalysisaredirectlytransferablefromtheircounterparts inreal analysis. Whenit is thecase, wesupposedthat thereadercoulddothetranslationherself.Also, we use the standard abbreviations such as i for if and only if, e.g.forforexample andi.e. forthatis. Theendof aproof isdenotedbytheusualsymbol: .iiiivContents1 Thecomplexplane 11.1 Thecomplexplane . . . . . . . . . . . . . . . . . . . . . . . . 21.1.1 Complexnumbers . . . . . . . . . . . . . . . . . . . . 21.1.2 ComplexAlgebra. . . . . . . . . . . . . . . . . . . . . 81.1.3 Exercices . . . . . . . . . . . . . . . . . . . . . . . . . 121.2 Geometryinthecomplexplane . . . . . . . . . . . . . . . . . 131.2.1 Lines,circles,andothersubsets. . . . . . . . . . . . . 131.2.2 ExtendedcomplexplaneandRiemannsphere. . . . . 191.2.3 Mobiustransformations . . . . . . . . . . . . . . . . . 231.2.4 Exercices . . . . . . . . . . . . . . . . . . . . . . . . . 261.3 Abitoftopologyinthecomplexplane. . . . . . . . . . . . . 261.3.1 Openandclosedsetsofthecomplexplane . . . . . . . 271.3.2 Convexityandconnectedness . . . . . . . . . . . . . . 321.3.3 Limitsandcontinuity . . . . . . . . . . . . . . . . . . 361.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 391.4 Curves,pathsandcontours . . . . . . . . . . . . . . . . . . . 401.4.1 Denitions . . . . . . . . . . . . . . . . . . . . . . . . 411.4.2 Contours . . . . . . . . . . . . . . . . . . . . . . . . . 421.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 432 Complexfunctions 452.1 Complexseriesandpowerseries . . . . . . . . . . . . . . . . . 46vvi CONTENTS2.1.1 Complexseries . . . . . . . . . . . . . . . . . . . . . . 472.1.2 Powerseries. . . . . . . . . . . . . . . . . . . . . . . . 512.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 542.2 Somecomplexfunctions . . . . . . . . . . . . . . . . . . . . . 542.2.1 Theexponentialfunction . . . . . . . . . . . . . . . . 552.2.2 Complextrigonometricandhyperbolicfunctions . . . 582.2.3 Rootsofunity . . . . . . . . . . . . . . . . . . . . . . 602.2.4 Thelogarithmicfunction . . . . . . . . . . . . . . . . 612.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 622.3 Multifunctions . . . . . . . . . . . . . . . . . . . . . . . . . . 632.3.1 Example1: thelogarithmicfunction . . . . . . . . . . 632.3.2 Branchpointsandmultibranches . . . . . . . . . . . . 652.3.3 Example2: Fractionalpowers. . . . . . . . . . . . . . 672.3.4 Example3: Anexamplewithtwobranchpoints . . . 683 Dierentiation 713.1 Holomorphicfunctions . . . . . . . . . . . . . . . . . . . . . . 723.1.1 DierentiationandtheCauchy-Riemannequations . . 723.1.2 Holomorphicfunctions . . . . . . . . . . . . . . . . . . 763.1.3 Someusefulresults. . . . . . . . . . . . . . . . . . . . 783.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 813.2 Someholomorphicfunctions. . . . . . . . . . . . . . . . . . . 813.2.1 Aresultonthedierentiationofpowerseries . . . . . 823.2.2 Theexponentialfunction . . . . . . . . . . . . . . . . 843.2.3 Complextrigonometricandhyperbolicfunctions . . . 853.2.4 Thelogarithmicfunction . . . . . . . . . . . . . . . . 853.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 873.3 Conformalmapping . . . . . . . . . . . . . . . . . . . . . . . 873.3.1 Conformalmapping . . . . . . . . . . . . . . . . . . . 873.3.2 Someexamples . . . . . . . . . . . . . . . . . . . . . . 89CONTENTS vii4 Integration 934.1 Integrationinthecomplexplane . . . . . . . . . . . . . . . . 944.1.1 Integrationalongpaths . . . . . . . . . . . . . . . . . 944.1.2 Thefundamentaltheoremofcalculus . . . . . . . . . . 994.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1024.2 Cauchystheorem . . . . . . . . . . . . . . . . . . . . . . . . . 1024.2.1 HistoricalCauchystheorem. . . . . . . . . . . . . . . 1024.2.2 Cauchy-Goursattheorem . . . . . . . . . . . . . . . . 1044.2.3 Deformation . . . . . . . . . . . . . . . . . . . . . . . 1154.2.4 Thecomplexlogarithm... again. . . . . . . . . . . . . 1184.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1214.3 Cauchysformul . . . . . . . . . . . . . . . . . . . . . . . . . 1214.3.1 Cauchysintegralformula . . . . . . . . . . . . . . . . 1224.3.2 Cauchysformulforderivatives . . . . . . . . . . . . 1244.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1284.4 Powerseriesrepresentation . . . . . . . . . . . . . . . . . . . 1294.4.1 Integrationofseries . . . . . . . . . . . . . . . . . . . 1294.4.2 Taylorstheorem. . . . . . . . . . . . . . . . . . . . . 1304.4.3 Multiplicationofpowerseries . . . . . . . . . . . . . . 1344.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1354.5 Zerosandsingularities . . . . . . . . . . . . . . . . . . . . . . 1364.5.1 Characterizingzeros . . . . . . . . . . . . . . . . . . . 1374.5.2 IdentityandUniquenesstheorems . . . . . . . . . . . 1394.5.3 Countingzeros . . . . . . . . . . . . . . . . . . . . . . 1444.5.4 Laurentstheorem . . . . . . . . . . . . . . . . . . . . 1494.5.5 Singularities . . . . . . . . . . . . . . . . . . . . . . . . 1544.5.6 Meromorphicfunctions . . . . . . . . . . . . . . . . . 1594.5.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1614.6 Cauchysresiduetheorem . . . . . . . . . . . . . . . . . . . . 1614.6.1 ResiduesandCauchysresiduetheorem . . . . . . . . 162viii CONTENTS4.6.2 Calculationofresidues. . . . . . . . . . . . . . . . . . 1644.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 1685 Applications 1715.1 Someapplicationsofcontourintegration. . . . . . . . . . . . 1725.1.1 Evaluationofrealintegralsbycontourintegration . . 1725.1.2 Someremarksonindentedcontours . . . . . . . . . . 1745.1.3 Integralofrationalfunctions . . . . . . . . . . . . . . 1775.1.4 Integralofotherfunctionswithanitenumberofpoles1805.1.5 Integralsoffunctionswithaninnitenumberofpoles 1845.1.6 Integralsinvolvingmultifunctions. . . . . . . . . . . . 1865.1.7 Summationofseries . . . . . . . . . . . . . . . . . . . 1905.2 TheFouriertransform. . . . . . . . . . . . . . . . . . . . . . 1915.2.1 IntroducingtheFouriertransform . . . . . . . . . . . 1915.2.2 Someapplications . . . . . . . . . . . . . . . . . . . . 194Chapter1Thecomplexplane:Geometry,TopologyandAnalysis12 CHAPTER1. THECOMPLEXPLANE1.1 Thecomplexplane1.1.1 ComplexnumbersAbitofhistoryComplexnumbersrstappearinthemathematical historyduringthesix-teenthcentury, intheworkof GirolamoCardano, ArsMagna(1545), andshortly after, in LAlgebra, by Rafael Bombelli (1572). In these initial works,theauthorswereinterestedinthesolutionof cubicalgebraicequationsoftheform:x3= 3px + 2q.This problemis equivalent tondingtheintersectionpoints of thecubiccurvey =x3andtheliney =3px + 2q. It is oftensaidthat complexnumbersappearedasnecessaryentitiesinrelationtondingrootsof thequadratic equations x2= mx+c. Indeed, this equation admits a pair of realsolutions as long as m2+4c > 0, but no real solution in the case m2+4c < 0.Butwhatistheproblemwiththat?Asimplegraphwillshowthatthereisnointersectionintheplaneinthatsecondcase. Thisisclearlynottrueforthecubic,asthereisalwaysanintersection(cfgure1.1).Figure1.1: Representationofaquadraticandacubicequations.1.1. THECOMPLEXPLANE 3Cardanohadshownthatthisequationcouldbesolved, andtheinter-sectionwasgivenby:x =3_q +_q2p3+3_q _q2p3.Bombelli, inspectingthisformuladiscoveredatroublingpropertiesof thesolution: takep=5andq=2, Then, q2 p30isthelocusof pointsatadistancerfromM. IfthepointOhascoordinates(a, b)intheCartesianplane, suchalocusisthereforecharacterizedbytheequation:(x a)2+ (y b)2= r2.Exercise4. Proveit.This simple form becomes even simpler in the complex plane. Lets notez=x + iythecomplexnumberassociatedwithapointMonthecircle,andletsallc = a +ibthecomplexnumberassociatedwiththecentreOofthecircle. Then,z c = (x a) + i(y b),andoneimmediatelyseesthat|z c|2= (x a)2+ (y b)2. Hence,theequationforthecirclecentredonOandofradiusris:|z c| = r.Note that the same argument shows that the disc centred on O and of radiusrischaracterizedby |z c| r(forthecloseddisc;seebelow).TogofurtherThereexistsanotherusefulcharacterizationofcirclesinthecomplexplane.For(a, b) C2and R, = 0,thepointsassociatedwithzsuchthat:z az b= formacircle, knownasacircleofApollonius. Wewill seethatthisrepre-sentationisveryusefulwhenstudyingconformalmappings.16 CHAPTER1. THECOMPLEXPLANEExercise: Showthat this locus is actuallyacircle; conversely, showthateverycirclecanbedescribeslikethat.We can now turn to the description of circular arcs joining two points Aand Bassociated respectively to the complex numbers a and b. Let Pbe anarbitrarypointonthiscirculararc. Then, asimplegeometrical argumentshows that the angle

APB= is constant along the arc. If Pis representedbythecomplexnumberz, letsdenotearg(z a)=andarg(z b)=.Then,itisclearfromthegure1.5,that = . Hence:arg(z a) arg(z b) = [2],or,equivalently,theequationforthearc:arg_z az b_= [2].SomeothersubsetsofthecomplexplaneWecannowdescribequicklyafewsubsetsof thecomplexplanethatwillappearinthenextchapters. Notethatthetermsopen andclosed usedhere agree with the denition that we will introduce later, when we examinethetopologyofthecomplexplane. Wehavealreadyencounteredthenotionofdisc,i.e. thesetofallthepointsthatareatlessthanagivendistancefromonepoint. Letusmakethatnotionmoreprecise. Theopendisccentredona Candofradiusr R+is,bydenition:D(a, r) = {z C, |z a| < r}.1.2. GEOMETRYINTHECOMPLEXPLANE 17Figure 1.5: Geometric construction of the equation for a circular arc betweenAandB.18 CHAPTER1. THECOMPLEXPLANEThecloseddisccentredona Candofradiusr R+is:D(a, r) = {z C, |z a| r}.To put it simply,D(a, r) consists in the union of D(a, r) and its bound-ary,thecirclecharacterizedby |z a| = r,thatwewilldenote(a, r)in these notes. We will also need the punctured disc,D

(a, r),centredona Candofradiusr R+:D

(a, r) = {z C, 0 < |z a| < r}.ItisthediscD(a, r)fromwhichwehaveremovedthecentrea. Another important class of regions is made of the annuli. For (s, r) R+R+,theyaredenedby:A(a, s, r) = {z C, s < |z a| < r}.scorrespondstotheinnerradiusof theannulus, andrtoitsouterradius.Notethatthecases = 0correspondstothepunctureddiscpresentedabove. Wewillalsoneedtocharacterizehalf-planes. Theopenupperhalf-planeisgivenby:+= {z C, Im(z) > 0},and the closedupperhalf-plane is the union of +with its bound-ary:+= {z C, Im(z) 0}.The other half-planes are dened accordingly (Do it for the lower half-planeandthetwoothernaturalones).1.2. GEOMETRYINTHECOMPLEXPLANE 19 Finally, letsintroducesectors, i.e. theregionof thecomplexplanethat is madeof thecomplexnumbers withanargument comprisedbetweentwovalues:S,= {z C, < arg(z) < }.Notethathalf-planesarealsosectors,for = .Togofurther1.2.2 ExtendedcomplexplaneandRiemannsphereTheRiemannsphereItistimetoseeourrstexampleofamappingofthecomplexplane. Con-siderthefunctiongdenedasfollows:C Cz 1/z.Hence, writingz =rei, g(z) =1rei. This means that theunit circle|z| =1ismappedintoitself. ThepuncturedunitdiscD

(0.r)inmappedintotheexterioroftheclosedunitdisc:{z C, |z|>1}, andtheexteriorofthetheclosedunitdiscismappedintothepuncturedunitdiscD

(0, 1).Now,itisclearthatifoneconsidersapointarbitrarilyclosedtotheorigin,its image will be a point with an arbitrary large modulus, and if one considersa point such that |z| +, its image will be arbitrarily closed to the origin.Therefore,even if the mapping is not dened at the origin,it seems that itsbehaviour around the origin is very regular, and we would like to extend themapping to the origin,so that it can be a mapping of C into itself. In other20 CHAPTER1. THECOMPLEXPLANEwords, bytakingalimit, wewouldliketosaythattheoriginismappedintoapointatinnity(|1/z| +),and,conversely,apointatinnitywould be mapped into the origin. This can be done very naturally by addingasinglepointtothecomplexplane, inordertomakeitcompact(wewilldenethis properlylater). Theideaof completingthecomplexplaneinthatwayisduetoRiemannandfoundalotofremarkableapplicationsingeometry. Let us embed the complex plane C into the Euclidean space R3byidentifyingthecomplexnumbersz= x + iywiththepointsofcoordinates(x, y, 0)inthisspace. Letusdenoteby(u, v, w)thecoordinatesin R3. TheRiemannsphereisdened,in R3,astheset: = {(u, v, w) R3, u2+v2+w2= 1}.Itisaspherethatintersectthecomplexplaneontheunitcircle |z| =1.LetN=(0, 0, 1), thenorthpoleof theRiemannsphere. Wearegoingtoconstruct thestereographicprojectionof theRiemannsphereontothecomplexplane. Letsconsiderapointmof , withcoordinates(u, v, w),suchthat u2+ v2+ w2=1. Theline(Nm) is generatedbythevector

Nm=(u, v, w 1); inotherwords, anypointMonthislineissuchthatthereexistsMforwhich

NM=

Nm. IfthispointMisinthecomplexplane,M= (x, y, 0),andonends:x = uy = v1 = (w 1).Itisclearthatwhenw=1, i.e. whenm=N, thissystemisnotsatised.Inalltheothercases,weseethat = 1/(1 w),and:x =u1 wy =v1 wwith w2= 1 u2v21.2. GEOMETRYINTHECOMPLEXPLANE 21Inotherwords, switchingbacktocomplexnumbers, wehaveconstructedamappingf:\N CfromtheRiemannspherewithoutitsnorthpoleontothecomplexplanethatisdenedby:(u, v, w)fu +iv1 w.Onecanshowthatthisapplicationisbijective, i.eone-to-one, andcontin-uous(Showit!). LetusseewhathappenswhenmapproachesN. Inthatcase, wapproaches 1 and the image point Min the complex plane is sent tohigherandhighervalueofitsmodulus. Moreprecisely,anopenneighbour-hoodof NontheRiemannsphereismappedintotheexteriorof anopendisconthecomplexplane. Thesmallertheneighbourhood,thefarthertheboundaryof theexteriorisfromtheorigin. Roughlyspeaking, wewouldliketosaythateverypointatinnityisanimageofNbyanextensionoff. Notethatitdoesnotmatterwhichdirectionweconsider,thepointsatinnity areall imagesof N. Thus, wewill deneanewpoint, whichwewill denote , andwewill addthispointto CanddenetheextendedcomplexplaneC= C {}byconstructinganapplication: Csuchthat:(u, v, w)_u+iv1wifw = 1 ifw = 1.If wenowcomebacktotheapplicationg(z)=1/zthatweintroducedatthebeginningof thesubsection, wecanextendit intoanewapplication g:C C,thatisone-tooneontheextendedcomplexplane:z g1/z,withtherule:1= 0and10= .This means that we are now allowed to divide a non-zero complex number byzero. Morespecically, thefollowingalgebraicrulesapplyintheextended22 CHAPTER1. THECOMPLEXPLANEFigure1.6: StereographicprojectionandtheRiemannSphere.complexplane:a = a = and a/= 0, a Ca.= .a = and a/0 = , a C+= .= = Remark 2. Notethatsomeoperationsarenotdened, suchthat ,0/0or /.Whyworkingintheextendedcomplexplane?Thetwomostimportantthingsthataregainedbyextendingthecomplexplanearethefollowing: Intheextendedcomplexplane,linesandcirclescanbeuniedintoasingleclassofobjects,calledthecirclines.1.2. GEOMETRYINTHECOMPLEXPLANE 23 It is now much easier to study in details the behaviour of functions atinnity,sinceitisonlyapointinC.Firstofall,consideracircleonthatpassesthroughN. Itsimagebythestereographicprojectionisalineonthecomplexplane. So,inessence,the point can be viewed as belonging to any line inC. Now, take a circleonthatisparalleltothecomplexplane. Then,itsimageinthecomplexplaneisclearlyacirclecentredon0. ItcanbeshownthatanycirclethatdoesnotpassthroughNonprojectsontoacircleon C, andthateverycirclein Ccanbeconstructedinthatway. Hence,wecanregardlinesinCascirclesthrough . WewillthencalllinesandcirclesonCcirclines. Ifwerememberthealgebraicparametrizationsgivenpreviously, weseethatcirclinescanbedescribedbytheequation:z az b= , > 0,wherealinecorrespondsto = 1.Theother interestingresult of this extensionlies inthepossibilitytotreattheinnityasanormal point. wewill seeinthenextchaptersthatthis can allow us to talk about the intersection of curves at innity,or theirbehaviourthere.1.2.3 M obiustransformationsTonishthissection, wearegoingtointroducealargeclassofmappings,calledMobiustransformations. ThesearemappingsofContoitself thattransformcirclinesintocirclines. Thesetransformationshaveawiderangeofapplicationsinbothalgebraandgeometry.Denition 4. A Mobius transformation Mis a mapping ofConto itselfoftheformM(z) =az +bcz +d,where(a, b, c, d) C4andad bc = 0.24 CHAPTER1. THECOMPLEXPLANEProposition3. Mobiustransformationsarebijective.Proof. Letusrstrememberwhatbijectivemeans: afunctionf:D Fisbijectiveiitisinjective(one-to-one)andsurjective(onto).fisinjectivei((z1, z2) D, z1 = z2, f(z1) = f(z2)).fissurjectivei(w F, z D, w = f(z)).LetusstartwiththeinjectivityofM,andletusproveitbycontradiction.Let (z1, z2) C2, z1 =z2. Then, let us supposethat M(z1) =M(z2).Thisimplies, afterabitof algebra, that(ad bc)z1=(ad bc)z2. But,ad bc =0, hence, z1=z2, whichcontradicts our hypothesis. Hence,z1 = z2 M(z1) = M(z2). TheMobiustransformationsarethusinjective.Toprovethesurjectivity, wepickupw C. Then, wehavetondthez Csuchthatw=M(z). Thisisequivalenttow=(az + b)/(cz + d),orz=(dw b)/(cw + a). OnecantheneasilycheckthatthiszisinC.Hence,Missurjective.Beinginjectiveandsurjective,Misbijective.Proposition4. Theinverseof aMobius transformation, M(z) =(az+b)/(cz +d)istheMobiustransformation:M1: z dz bcw +a.Proof. Thisfollowsdirectlyfromtheconstructionusedtoprovedthesur-jectivityinthepreviousproof.Example1. HerearesomeexamplesofMobiustransformations: z zei, R: anticlockwiserotation z z +a, a C: translation z 1/z: inversion z Sz, S R+: Stretching1.2. GEOMETRYINTHECOMPLEXPLANE 25Exercise5. ShowthatanyMobiustransformationscanbedecomposedinasequenceoftheprevioustransformationsasfollow: T1: z z +dc; I: z 1/z; R : z z exp_i arg_bcadc2__; S: z bcadc2z; T2: z z +ac.LetusnowseetheeectofaMobiustransformationofcirclines. LetCbe a circline of equation |z|/|z| = . Let f(z) = (az+b)/(cz+d) be aMobius transformation. So, if w = f(z), we know that z= (dwb)/(acw).Hence, wecansubstituteforzintheequationofthecirclineCtonditsimageundertheMobiustransformation:w f()w f()= c +dc +difc +d = 0andc +d = 0or,|w f()| = a +bc +difc +d = 0andc +d = 0or,|w f()| = a +bc +difc +d = 0andc +d = 0.(1.3)Notethatc + dandc + dcannotbothbezerobecausead bc =0, bydenition. Thus,weseethat:Proposition5. TheimagebyaMobiustransformationof acirclineisacircline.26 CHAPTER1. THECOMPLEXPLANE1.2.4 Exercices1. Characterizeandrepresentthesetsofthecomplexplanedenedby:(i) S1= {z C, |z 1| < 2};(ii) S2= {z C, |z +i| < 1};(iii) S3= S1 S2;(iv) S4= S1 S2;(v) S5= {z C, arg(z) = /4};(vi) S6= {z C, |z| 0(dependingonz)suchthatD(z, ) S.Roughly speaking, that means that one can always go around any z Sinanydirectionwithout leavingS; thedistance that is permittedwillvaryfromonepointtoanother, dependingwhetherzisfarorclosetotheboundary(thecloserzistotheboundary,thesmallerthepermitted).Thefollowingpropertiesapplytoopensets:Proposition6. (i) If S1,..., Snforn Nareopenset, thenS=S1... Snisalsoopen.(ii) If Sjforj J(whereJissomecountableindexset)areopensets,then

jJ Sjisopen.28 CHAPTER1. THECOMPLEXPLANEProof. (i) Let z S and consider k> 0 such that k {1, ..., n}, D(z, k) Sk. Let= min(1, ...k). Then> 0(thisiswherethefactthattheSkarenitelymanyiscrucial),andclearly, k {1, ..., n}, D(z, ) D(z, k) Sk,hence,D(z, ) S.(ii) Thispointistrivial: everySjisopen, and z jJ Sj, j J, z Sj; then z jJ Sj, l J, >0, D(z, ) Sl}. Since, trivially,Sl

jJ Sj,thisendstheproof.Examplesofopensets:(i) Theemptyset isopen(theconditionforittobeopencannotfail);(ii) Cisopen;(iii) a C,D(a, r)isopen;(iv) {z C, |z a| > r}isopen;(v) {z C, s < |z a| < r}isopen;(vi) S,isopenProof. Thersttwoaretrivial.(iii) LetsstartwiththediscD(a, r)fora Candr> 0. Letz D(a, r)and Rsuchthat: 0 0, D(z, r) C\Sz S, r > 0, D

(z, r) S= Nopointof C\SisalimitpointofS AnylimitpointofSisinS1.3. ABITOFTOPOLOGYINTHECOMPLEXPLANE 31Thisshowsthat(i)and(ii)areequivalent. (iii)isequivalentto(ii)becauseSistheunionofSanditslimitpoints.2. Letusconsiderthesecondproposition. Letz CsuchthatforeveryopensetV containingz,V S = . Since r> 0, D(z, r)isanopenset containingz, wehave, inparticular, that D(z, r) S = . So,eitherz S,andandthisistriviallytrue,orz S,andthisimpliesthatD

(z, r) S = foreveryr > 0,whichisthedenitionofalimitpoint. Hence, zis in the closure of S: z S. To show the converse, letusproceedbycontradiction: letz Sandsupposethatthereexistsan open set Vsuch that z Vand V S= . The fact that Vis openguaranteesthatthereexistsanr>0suchthatD(z, r) V . Hence,thereexistsr>0, D(z, r) S= , whichcontradictsthefactthatz S. Thisshowsthesecondproposition.3. Toprovethethirdproposition,itisenoughtoprovethattheclosureofSisSitself (bythe rst proposition):S=S. Toprove it bycontradiction, let ussupposethat thisisfalse: let z Ssuchthatz S. Then, r >0, D(z, r) S= . Ontheother hand, sincez S, there exists an wsuch that w D(z, r) S(second propositionappliedtoSinsteadofS). Hence,w S,and,D(z, r)beinganopensetcontainingw, thesecondpropositiongivesthatD(z, r) S = .Thatisthedesiredcontradiction.Herearesomeexamplesofclosedsets: All thesectors, whendenedwithweakinequalities , ratherthanstrictones 0 is closed because it is the complement of D(a, r){z C, |za| > r} that is an open set. Of course,D(a, r) is also the closure32 CHAPTER1. THECOMPLEXPLANEofD(a, r).Remark3. Becareful: somesetsareneitheropennorclosed!Considerforexample,for0 < a < brealnumbers,S= {z C, |z| [a, b[}.One can note that for any r > 0, D(a, r) S; since a S, S cannot be open.Ontheotherside, foranyr>0, D(b, r) S = , sothatD(b, r) C\S;henceSisnotclosedeither.Finally, let us introduce the last notions of this subsection: those ofboundedandcompactsets.Denition8. A set S C is bounded i (M R+, z S, |z| M). LetS C. Scompact Sboundedandclosed.Examples of compact sets inCare: circles |z a| =r, closeddiscsD(a, r),butalsolinesegments[a, b]where(a, b) C2.Tocomebacktotheextendedcomplexplane, deningopendiscsinCshouldntbeaproblemnow: whenthecentrez C,weusetheusualwaydescribedabove,andwhenz= ,wesimplywrite,forr > 0:D(, r) = {z C, |z| > r} {}.This denition can be made very precise using the stereographic projection:D(, r)istheimagebyhomeomorphismoftheopendiscaroundNon(forthecanonicaltopologyof). Moreover,onecanshowthat,sinceiscompact,Cisalsocompact.1.3.2 ConvexityandconnectednessInthis subsection, we will be interestedincharacterisingthe shape ofsubsetsofthecomplexplane.ConvexandpolygonallyconnectedsetsDenition9. LetS C. Sisconvexi,_(a, b) S2, [a, b] S_.1.3. ABITOFTOPOLOGYINTHECOMPLEXPLANE 33Figure1.9: Examplesofaconvexandanon-convexsets.Thismeansthat, foranytwopointsof S, thelinesegmentjoiningthetwopointsiscontainedinS. Forthisreason, itisobviousthatasetliketheunionoftwonon-intersectingdiscsisnotconvex. Itthesameway,thecomplexplane,fromwhichonehasremovedalinecannotbeconvex.Let us look at two examples.C\R and C\[0, +[ are clearly not convex,buttheyareneverthelessquitedierent; whereasinthecaseof C\R, thetwohalf-planes+andarestrictlydisconnected,itisnotthecaseforC\[0, +[: in this case, two points with positive real parts cannot be joinedby a straight line segment, but they can clearly be joined by a nite series oflinesegmentsthatavoid[0, +[. Thisillustratethefactthatconvexityisnotsucienttocharacterizeasubsetof C. Oneshouldintroduceaclassofsetsforwhichpolygonal routescanbeemployedtojoinpointsinthesets.Letusrstdeneapolygonalrouteprecisely.Denition10. Let (z0, z1, ...zn1, zn) Cn+1, forn N. Apolygonalroutefromz0toznistheset:[z0, z1] [z1, z2] ... [zn1, zn].34 CHAPTER1. THECOMPLEXPLANEFigure1.10: ApolygonalroutebetweenaandbinS.Thisallowsustocharacterizesubsetslike C\[0, +[:Denition11. AsubsetS Cispolygonallyconnectedi_(a, b) S2, (z1, ..., zn1) Cn1, [a, z1] [z1, z2] ... [zn1, b] S_This means that for any two points a and b of S, there exists a polygonalroutefromatobthatliescompletelyinS.Itisclearthateveryconvexsetinpolygonallyconnected. Anyannulus,onthecontrary,ispolygonallyconnectedbutnotconvex.Denition12. Asubset GofCis connectediit cannot bede-composed into the union of non-empty open sets G1and G2such thatG1G2= . In other words, if G is connected and G1 G and G\G1arebothopen,then,necessarilyG1= GorG1= .1.3. ABITOFTOPOLOGYINTHECOMPLEXPLANE 35 Anon-emptyopenconnectedsubsetof Ciscalledaregion.Theorem3. LetG Cbeanon-emptyopenset. Then, GisaregioniGispolygonallyconnected. Inparticular,anynon-emptyopenconvexsetisaregion.Proof. First,supposethatGisaregion. Leta Gand:G1= {z G, apolygonalroutefromatozinG}.G1is then the subset of G that is polygonally connected to a. We will write:G2=G\G1. It is clear that G1 = , becausea G1. Theideaof theproof istoshowthatbothG1andG2areopen. SinceGisaregionitisconnected, sothisimpliesthatG=G1. Gisopen, so, foranyz G, wecanndar > 0suchthatD(z, r) G. LetwbeanelementofD(z, r). Byconstruction[z, w] D(z, r) G. Ifz G1, thenthereis, bydenitionapolygonal routeinGfromatozinG1, andtheadditionof [z, w] tothisroutegivesanewpolygonalroutefromatowviaz,sothat,w G1. ThismeansthatD(z, r) G1, andG1isopen. Ontheotherhand, if z G1(z G2), that means that there is no polygonal route from a to z, so clearlyno polygonal route from a to zvia w. Hence w G2. Then,we have shownthat D(z, r) G2,proving that G2is open. G1and G2are thus both open.Itfollows, byconnectednessof theregionG, sinceG1 = , thatG1=G,andtherefore,Gispolygonallyconnected.Conversely, suppose that Gis non-empty, openandpolygonallycon-nected. Inordertoprovetheresultbycontradiction, wewillsupposethatGis not aregion. Inother words, wesupposethat thereexist twodis-jointnon-emptyopensetsG1andG2suchthatG=G1 G2. Considera G1andb G2. SinceGispolygonallyconnected, wecanconstructapolygonal routebetweenaandbinG, P=[z0, z1] ... [zn1, zn] with36 CHAPTER1. THECOMPLEXPLANEFigure1.11: ProofthatGisaregioniitispolygonallyconnected.a = z0andb = zn. Then,atleastoneofthelinesegments[zk, zk+1]issuchthat zk G1andzk+1 G2. Apoint of this segment canbedescribedbyz(t) =(1 t)zk+ tzk+1witht [0, 1]. SinceG1 G2= , foreacht [0, 1], eitherz(t) G1, orz(t) G2. Moreover, sinceG1andG2areopen, if z(t) G1(resp. z(t) G2), then >0, z(t + ) G1(resp.>0, z(t + ) G2). LetS=sup{t [0, 1], z(t) G1}. Itisclear, fromwhatwejustsaid,thatq ]0, 1[(becauseG1isopen,sothereisnecessarilyanelementofthesegmentcloseenoughtozktostillbeinG1). Butwecaniterate this process! Consider z(q). Since G1is open, there exists> 0 suchthatz(q + ) G1, incontradictionwiththedenitionofq. Ifz(q) G2,sinceG2isopen, wecanalsonda >0suchthat, for all ssatisfying0 < q < s q,wehavez(s) G2. Thisagaincontradictsthedenitionofq. Hence,weseethatGhastobearegion.1.3.3 LimitsandcontinuityIt is time to start investigating genuine notions of complex analysis. We willbeginwiththeconceptsoflimitsandcontinuity.1.3. ABITOFTOPOLOGYINTHECOMPLEXPLANE 37AfewdenitionsWe rst have to dene what we will call sequences, as well as some propertiesofthesesequences.Denition13. A sequence (zn)nNis a one-to-one relation betweenthenaturalnumbersn Nandcomplexnumberszn. Inotherwords,itisanorderedlistof complexnumbers. Pleasenotethat, insomecases, wewillneedtodeneasequenceonasubsetof NratherthanNitself. Asequence(zn)nNis boundedithereexists MRsuchthat,n N, |zn| M. Asequence(zn)nNconvergeswithlimita Ci > 0, N N, n N |zna| < . Asequence (uk)kNis a subsequenceof the sequence (zn)nNithere exists natural numbers (ni)iNwith i N, ni< ni+1, such thatk N, uk= znk.Theconvergenceofasequenceissimplythefactthat,fornbigenough,themembersof thesequenceaccumulatearbitrarilyclosearoundagivencomplexnumbera,thatis,forthisreasoncalledthelimitofthesequence.Wewillusethefollowingnotationtodescribealimit:limn+zn= a.Wecannowdenelimitsandcontinuityforcomplex-valuedfunctions.Denition14. Letf: S CbeafunctiondenedonasubsetS C. Let a S. Then,the limit of fwhen ztends to a,noted limzaf(z)existsandisequaltow Ci > 0, > 0, (z S, 0 < |z a| < ) |f(z) w| < .38 CHAPTER1. THECOMPLEXPLANE Leta S. fiscontinuousatai > 0, > 0, (z S, |z a| < ) |f(z) f(a)| < .Note, inthedenitionof thelimit, that |z a| >0, i.e., thelimit isdeterminedbywhathappenstothefunctionasitapproachesa. fmaynoteven be dened at this point a, so its value there is of no importance for theconceptof limit. Itisdierentforthenotionof continuity. Nevertheless,one sees that a function is continuous i limzaf(z) exists and equals f(a).Obviously, afunctionis saidtobecontinuous if it is continuous at eachpointofitsdomain.Theoperationsonlimitstranslateeasilyfromthoseinthecaseof realanalysis,andwewillusethemwithoutfurtherproofs. Wesimplylistthemhereforfunctions(similarresultsholdforsequences).Proposition8. Letf:S C Candg:T C C, Letz0 S T,andsupposethat:limzz0f(z) = Aand limzz0g(z) = B.Then:limzz0(f+g) (z) = A+Blimzz0(fg) (z) = ABlimzz0_f(z)g(z)_=ABifb = 0..Also, itisintuitive(andeasytoprove)thatacomplexsequenceandacomplexfunctionconvergeitheirrealandimaginarypartsconvergein R.We also list a few important results that we will use later without proofs:theseproofswouldbelongandtime-consumingwithoutbringinganythingdecisive to the subject of these lectures. Moreover, they canbe easilyadaptedfromtheircounterpartsinrealanalysis.Theorem4. Anyboundedsequencein Chasaconvergentsubsequence.1.3. ABITOFTOPOLOGYINTHECOMPLEXPLANE 39Acorollaryof thistheoremisthatanyinnitecompactsubsetSof ChasalimitpointinS. ThisisknownastheBolzano-Weierstrasstheorem.Thisleadstotheimportantconvergencetheorem:Theorem5. Cauchyconvergencetheorem.(zn)nNconvergesi > 0, N N, (m, n) N2, m, n N, |zmzn| < .Anotherusefulresultis:Theorem6. LetS Cbecompact,andf: S Cacontinuousfunction.Then: fisbounded,i.e.:M> 0, z S, |f(z)| M fattainsitsbounds,i.e.:(z1, z2) S2, |f(z1)| |f(z)| |f(z2)|.Finally,weconcludethissectionbyalast,unproventheoremfromrealanalysis,thatisadirectconsequenceoftheintermediatevaluetheorem:Theorem7. Let[a, b] Randf: [a, b] Zacontinuousfunction. Then,fisconstant.1.3.4 Exercises1. Determineifthefollowingsetsofthecomplexplaneareopen, closedorneitheropennorclosed:(i) S1= {z C, |z| < 1, 0 < arg(z) < /3};(ii) S2= D(i, 2) D(2, 1);(iii) S3= {z C, Im(z) = 1, 0 < Re(z) < 1}.2. Findthelimits,iftheyexist,ofthefollowingcomplexsequences:40 CHAPTER1. THECOMPLEXPLANE(i) un= (i)n;(ii) un=(i)nn;(iii) un=1n+i;(iv) un=n22in+4in2+6n+3i.3. Find,iftheyexist,thefollowinglimitsofcomplexfunctions:(i) limziz2+z+i2iz3+z+3;(ii) limz1+3iarg(z)z;(iii) limzi/214z2+1.4. Determinewhetherornotthefollowingfunctionsarecontinuousatthegivena C:(i) f(z) = z2+iz + 2,a = i;(ii) f(z) =arg(z)z2+1 ,a = i;(iii) f(z) =z+1z2,a = i.1.4 Curves,pathsandcontoursThis section is an introduction to a key object in complex analysis: contours.Indeed,inordertodevelopthetheoryofintegrationinthecomplexplane,wewillneedtogobeyondthesimpledescriptionofcurvesassubsetofthecomplexplane, aswehavedonesofar. Wewill needtoconsiderthemasroutefollowedbyamovingpoint;thisisexactlywhatwehavedoneintheproof of Theorem 3, when we had to consider the points of a segment whoseendpointswereintwodierentopensets. Forthat, theroutefollowedbythemovingpoint is describedbyareal parameter, thecomplexnumbercorrespondingtothepointbeingafunctionofthisrealparameter.1.4. CURVES,PATHSANDCONTOURS 411.4.1 DenitionsLetusbeginbydeningtheconceptsofcurvesandpaths.Denition15. Let (a, b) R2and [a, b] be a closed bounded interval of R.A curve with parameter interval [a, b] is a continuous function : [a, b] C. Itsinitial pointis(a), anditsnal pointis(b). isclosedi(a) = (b).issaidtobesimplei (s, t) ]a, b[, s = t, (s) = (t).For such a curve , we will denote the image of [a, b], i.e. {(t), t [a, b]}by. ThecurveissaidtolieinasetSi S.Remark 4. [a, b] isacompactsetof R. Sinceisitsimagethroughthecontinuousfunctionitisthusalsocompact(in C). Thisimpliesthatisaclosedset.Onesees that, inthenotionof curve, thereis naturallyembedded, anotionof orientation: theimageof thecurveis describedinaparticulardirection,from (a) to (b). Of course,for any curve ,there is a curve withthesameimagebuttheoppositeorientation:()(t) = (a +b t), fort [a, b].Alinesegmentbetweentwopointzaandzbisclearlytheimageinthecomplexplane of the curve : [0, 1]Csuchthat t [0, 1], (t) =(1 t)za +tzb.One sees that it is fairlysimple tojoincurves together toformnewcurves: if 1 and 2 are two curves dened respectively on [a1, b1] and [a2, b2],such that 1(b1) = 2(a2), their join can be dened as : [a, b] C, with:(t) =_1(t) ift [a1, b1],2(t +a2b1) ift [b1, b1 +b2a2].Then, a polygonal route as those used in the previous section is the imageofthejoinoflinesegments.42 CHAPTER1. THECOMPLEXPLANEDenition16. Afunctionf : [a, b]Cis saidtobe dierentiable att [a, b]ilimh0g(t +h) g(t)hwitht +h [a, b], exists.Itisdierentiableiitisdierentiableatanyt [a, b]. If itexists, thislimitisnotedg

(t)andiscalledthederivativeofgatt.Acurveissmoothifithasacontinuousderivativeforall valueoftin[a, b].Denition17. Apathisthejoinofnitelymaysmoothcurves.Remark5. Note that any curve is certainly the join of nitely many smoothcurves, withtherequirementthatthepiecesdontintersect, whereas, inapath,theycanintersectasmanytimesastheywant.1.4.2 ContoursItisclearthatcirclinesareaspecial caseofpaths. Wehaveseenthatthelinesegmentbetweenzaandzbissimplythecurve: [0, 1] Csuchthatt [0, 1], (t)=(1 t)za + tzb, thatisalsoapath. Moreover, acirculararc centred on a C, of radius r > 0, between two angles and such that0 2,anddescribedclockwise(resp. anticlockwise)istheimageofthepath(resp. )suchthat(t) = a +reitfort [, ].Acirclinepathisthejoinofnitelymanypathscorrespondingtolineseg-mentsorcirculararcs.Denition18. Acontourisasimple,closedcirclinepath.Toputitshort,theimageofacontourismadeofnitelymanycirculararcs and line segments that do not cross each other. A contour is said to bepositively oriented i, as t increases, its image is described anticlockwiseroundanypointinsideit.1.4. CURVES,PATHSANDCONTOURS 431.4.3 Exercises1. Giveaparametrisationofthefollowingcontours:(i) thesquareABCD, withA= 1 i, B= 1 + i, C=1 + i,D = 1 +i;(ii) thearcofcirclecenteredon0,betweenA =3 +iandB= 2i;(iii) the contour made of the segment [R, R],together with the pos-itivesemi-circlebetweenRand R.2. Considerthethreecurves:1(t) =22(1 +i)tfort [0, 1]2() = eifor _4, 34_3(s) =22(1 +i)sfors [0, 1].Characterizetheimage: 1 2 (3).44 CHAPTER1. THECOMPLEXPLANEChapter2Complexfunctions4546 CHAPTER2. COMPLEXFUNCTIONSWearenowenteringintheheartofcomplexanalysis. Beforestudyingthe class of functions that are of most interest in complex analysis, i.e. holo-morphicfunctions, inthenextchapter, wewill concentrateinthepresentchapteroncomplexseries,andacertainnumberofcomplexfunctionsthataredenedthroughtheirseriesexpansions. Thelastsuchfunctionthatwewill present is the complex logarithm, and it will be the occasion to introducetheconceptofmultifunctionanditsphenomenology.2.1 ComplexseriesandpowerseriesAs you remember, many real functions f: R R can be expressed as powerseries:f(x) =+

n=0cnxn,the cns being real constants, at least on a given interval x ] R, R[, whereRiscalledtheradiusof convergenceoftheseries. Inrealanalysis, onehasafewcriteriatodeterminetheradiusof convergenceof aseries, butnoclearreasontounderstandthespecicvalueofthisquantityforagivenseries. Considerforexamplethetworealfunctions:F(x) =11 x2andG(x) =11 +x2.Byusingthegeometricseries:11 y=+

n=0ynfory ] 1, 1[,andmakingthechangesof variabley=x2forFandy= x2forG, onendsthat:F(x) =+

n=0x2nG(x) =+

n=0(1)nx2n.2.1. COMPLEXSERIESANDPOWERSERIES 47Boththeseserieshavearadiusof convergenceR=1(youcanapplytheratioortheroottests). InthecaseofF, thereasonforthatcaneasilybeunderstoodinreal analysis: thefunctiondivergesatx= 1, sothatitsradius of convergencecorresponds totherst singularityinthefunction;thedivergenceintheseriesexpansionistheresultofagenuinedivergenceinthefunctionitself. ButwhataboutG?itisperfectlyregularatx = 1,andyet, its series expansionis onlyvalidin] 1, 1[. This caneasilybeunderstoodifwenowgotothecomplexplane. ConsiderG(z) = 1/(1 +z2)tobetheextensionof Gtocomplexvariables. Then, itisclearthatGisdivergenti1 + z2=0, inotherwords, atz= i. Thedistancebetweenthecentreoftheexpansion, 0, and iisexactly1. Andagain, theradiusofconvergenceoftherealseriescorrespondstothedistancetothenearestsingularity, butthistimeinthecomplexfunctionthatgeneralizestherealfunctiontocomplexvariables! Thisisageneral andverypowerful result,andwewillseeinthefollowingofthiscoursehowcomplexseriesaremucheasiertodeal withthanreal series, andtoacertainextend, howalotofresultsfromrealanalysisaremuchmoreunderstandablewhenweconsidertheminthecomplexplane. Inthissection,wewillproperlydenecomplexseriesandcomplexpowerseries, andexploreafewoftheirproperties. Wewillseeinthenextchapterhowtheyarekeytocomplexanalysis.2.1.1 ComplexseriesDenition19. Let (cn)nNbeacomplexsequence. Thecomplexseries

cnof generictermcnisthesequenceof thepartial sumsof cn: sN=

Nn=0cn. The series

cnis said to converge i the sequence sNconvergeswhenN +. Thelimitisthen: s =

+n=0cn.Asinthecaseofsequences,andforthesamereasons,acomplexseriesconvergesiitsrealandimaginarypartsconvergein R.Example2. Hereareafewcomplexseries:48 CHAPTER2. COMPLEXFUNCTIONS(i)

(i)n;(ii)

(i+1)nn2;(iii)_12_nein/3.Letuslistaseriesofresultsaboutconvergentseries.Theorem8. If

cnconverges,then:(i) limn+cn= 0;(ii) M> 0, n N, |cn| M.Proof. (i) UsetheCauchyconvergencecriterionforthepartialsums:n

i=0ciconverges > 0, N N, (m, n) N2, m > N, n > N, |m

i=0cin

j=0cj| < > 0, N N, (m, n) N2, m n > N, |m

k=nck| < .Inparticular,form = n: > 0, N N, n N, n > N, |cn| < ,whichisexactlytheexpressionofthefactthatcnconvergestowards0.(ii) The second point is a direct consequence of the rst one: if there existsannsuchthat |cn|>M, thentheirisanobviouscontradictionwiththerstpoint.Proposition 9. Let

anand

bnbe two convergent complex series.Then,foranyk C,

(an +kbn)isaconvergentseries,and+

n=0(an +kbn) =+

n=0an +k+

n=0bn.2.1. COMPLEXSERIESANDPOWERSERIES 49Proof. Foreaseof notation, letuswriteA=

+n=0anandB=

+n=0an.Then,fork C:n

i=0(ai +kbi) (A+kB) =_n

i=0aiA_+k_n

i=0biB_.Thetriangleinequalitythengives:n

i=0(ai +kbi) (A+kB)n

i=0aiA+k_n

i=0biB_.Let>0. Sinceboth

anand

bnconverge, thereexistsN>0suchthatforalln > N:n

i=0aiA 0, N> 0, m > n > N,m

i=0|ci| n

j=0cj=m

k=n|ck|< Since n N, |cn| > 0,wehave |

mk=n|ck|| =

mk=n|ck|. So,usingtherstinequality,wehaveshownthat: > 0, N> 0, m > n > N,m

i=0cin

i=0ci< .50 CHAPTER2. COMPLEXFUNCTIONSThisprovesthat

ni=0ciisaCauchysequence. Henceitconverges.Wecannowproveacertainnumberofcriteriathatwill beveryusefulininvestigatingtheconvergenceofcomplexseries.Proposition11. Comparisontest.Let

bnbeaconvergentreal serieswith n N, bn 0. Let(an)nNbeacomplexsequence.If k > 0, n N, |an| kbn,then

anisabsolutelyconvergent,andhenceconvergent.Theproofisevidentandislefttothereader.Proposition12. Ratiotest(aka: dAlembertstest):Let

cnbeacomplexseriessuchthat:limn+cn+1cn= l exists.Then:(i)

|cn|,andhence

cn,convergesifl < 1;(ii)

|cn|divergesifl > 1;(iii) thetestisinconclusiveifl = 1.Thetestisinconclusiveifl = 1.Proof. Weconsiderthecomplexseries

cnsuchthatlimn+cn+1cn=lexists,andwenotevn= |cn+1/cn|.(i) Letussupposethatl1. Thenfornbigenough, all thevnswill bearbitrarilyclosetol. Thismeansthatthereisan < 1suchthatforthereexistsN Nforwhich, if n>N, vnN. For 1provesthat

|cn|diverges.Proposition13. Roottest.Let (cn)nNbe acomplex sequence suchthat limn+n_|cn| =l exists.Then:(i) ifl < 1,then

|cn|converges(sodoes

cn);(ii) ifl > 1,then

|cn|diverges;(iii) ifl = 1,thenthetestisinconclusive.Proof. The proofis analogue to the one forthe ratio test: in the case l < 1,thereareN,n_|cn| . So,|cn| n. Thecomparisontestendstheproof.2.1.2 PowerseriesNow that we have set the general context for complex series,we turn to theparticularcaseofpowerseries.Denition 20.A power series is a complex series of the form

cn(za)n,wherea C,(cn)nNisacomplexsequence,andz C.52 CHAPTER2. COMPLEXFUNCTIONSIt is clear that, uptoaredenitionZ=z a, onecouldworkwithpowerseriesof theform

cnZn, buttheintroductionof z awill makemoresenseinthefuturedevelopment. Itisimportanttonotethatpowerseriesaredierentfrompolynomials: polynomialshaveonlynitelymanyterms.Example3. Hereareafewexamplesofpowerseries:(i)

(i)n(z 2)n;(ii)

1n2zn;(iii)_n2+ 2n + 1_(z i)n.Notethatpowerseriesarefunctionsofthecomplexvariablez. Hence,theirconvergencewill dependonthevalueofz. Thisiswhyweintroducethenotionofradiusofconvergence.Denition21. The radiusofconvergence, R R{+}, of the powerseries

cn(z a)nis:R = sup{|z a|,

cn(z a)nconverges}.WewillwriteR = +iftheseriesconvergeseverywhere(foranyvalueofz). Inotherwords,Ristheradiusofthebiggestopendiskaroundaonwhich

cnznconverges;outsidethecloseddisk,theseriesdiverges,anditcanconvergeordivergeonthediskitself. Thisjustiesthetermradiusofconvergence. Actually,wehavedenedtheradiusofconvergenceintermsofordinaryconvergence ofthe series

cnzn,butwehave a strongerresult.Thisisthetopicofthefollowinglemma.Lemma 1. Let

cnznbe apower series withradius of convergence R.Then:(i)

cnznconvergesabsolutelyonD(0, R);(ii)

cnzndivergesforanyz C\D(0, R)(i.e. for |z| > R).2.1. COMPLEXSERIESANDPOWERSERIES 53Proof. (i) Letz D(0, R). Then, bydenition |z| 0suchthat n N, |cnwn| M. Moreover, |cnzn| = |cnwn|.zwnMzwn. Now, since |z/w| randsupposeforacontradictionthat

cnzn. Then, thereexists M>0suchthat, for anyn N,|cnzn| < M. Now, the region for which |z| > R is open; son we can pickupwwithR |w| < |z|. Then: |cnwn| = |cnzn|wzn M|wzn. Thegeometricseries

|wznconvergesbecause |w/z| < 1. Hence,bythecomparisontest,

cnwnconverges, whichcontradictsthedenitionofR.Itremainsnowtogiveafewcriteriatondtheradiusof convergenceofpowerseries. Thesecriteriaaredirectconsequencesoftheratioandroottestsforseries.Proposition14. Let

cnznbeapowerserieswithradiusofconvergenceR. Then:R1= limn+n_|cn| = limn+cn+1cn.Ofcourse,thelastformulaonlyappliesifthecnsarenon-zero.Proof. Simplyappliestheratioandrootteststothepowerseries.Finally, let us note that we coulddene series of functions that arestandardcomplexseriesdependingonavariablez C, butarenotpowerseries. Forexample:(i)

z2n1+nz;54 CHAPTER2. COMPLEXFUNCTIONS(ii)

_ziz+2i_n.Theconvergenceof suchserieswill, again, dependonthevalueof z, but,ingeneral, onecannotspeakof aradiusof convergenceanymore, becausethe subset of the complex plane where they converge is no longer necessarilyadisk, butcanbemorecomplicated. Canyoudeterminethesetof zforwhichthetwoexamplesaboveconverge,usingtheratioortheroottests?2.1.3 Exercises1. Considerthegeometricseries:

zn,forz C. ProvethattheseriesconvergesinsideD(0, 1),andthat:z C, |z| < 1,+

n=0zn=11 +z.2. Usingthepreviousresult,provethat:z C, |z| < 1,+

n=0(1)nzn=11 z.3. Findtheradius of convergence, andthediskof convergenceof thefollowingpowerseries:(i)

(i)nn(z 2i)n;(ii)

n2+3n+2in2+n+1(z 3)n;(iii)

(1 +i)n(z 1)n.2.2 SomecomplexfunctionsNowthatweknowthemainpropertiesofpowerseries,wearegoingtousethemtobuildsomefundamentalfunctionsofcomplexanalysis.2.2. SOMECOMPLEXFUNCTIONS 552.2.1 TheexponentialfunctionTherst, andcertainlyoneofthemostimportantcomplexfunctionistheexponential function. There are manywaytointroduce it, andwe willchoosetodeneitthroughitspowerseriesexpansion.Proposition15. Considerthecomplexpowerseries

znn!. Ithasanin-niteradiusofconvergence.Proof. Letusapplytheratiotesttotheseries:zn+1/(n + 1)!zn/n!=|z|n + 1.Thisratiotendsto0foranyvalueofz,hence,theseriesconvergeswithaninniteradiusofconvergence.Since this series reduces to the Taylor expansion of the exponential func-tionwhenz is restrictedtobe areal number, we propose the followingdenitionDenition22. We call complexexponential function, andwe noteez= exp(z)thelimitoftheseries

znn!:z C, ez=+

n=0znn!.Herearethefundamentalpropertiesoftheexponential:Proposition16. (i) e0= 1;(ii) ez+w= ezew;(iii) z C, ez= 0.Proof. (i) Theresultisobviousifoneputsz= 0intheseriesexpansion.56 CHAPTER2. COMPLEXFUNCTIONS(ii) Bydenition, wehave

+n=0znn!=limN+

Nn=0znn!, andthis se-quenceconverges. Hence, itsproductwiththesamesequencewithzreplacedbywalsoconverges, anditdoessotowardstheproductofthetwolimits. Inotherwords:ezew= limN+N

n=0N

p=0znn!wpp!= limN+_1 +z +z22!+...zNN!_[1 +w +w22!+... +wNN! ]= limN+_1 + (z +w) +(z +w)22!+... +(z +w)NN!_= ez+wThe third line was obtained by using Newtons binomial formula: (z +w)N=

Nk=0n!(nk)!k!zkwnk.(iii) Usingthersttworesults,wehaveezez= 1,henceez= 0.Note that the coecients in the series expansion of the complex exponen-tial are real numbers,so,if one restricts zto be a real number,one recoverstheusualrealexponentialfunction. Thisleadstothefollowingproperty: ifwewrite: z= x +iywith(x, y) R2,wehave:|ez| = ex.Thisnecessarilyimpliesthat y R, |eiy| = 1.Proof. Toprovetherststatement,letuswrite:|ez|2= ezez= eze z= ez+ z= e2x= (ex)22.2. SOMECOMPLEXFUNCTIONS 57Thesecondlineresultsfromtheexpansionof theexponential andof thefactthatcomplexconjugationiscontinuous, implyingthatonecaninvertconjugationandlimit. Thisgives |ez| =ex, sincebothsidesarepositivereal numbers. Finally, since |ez| = |ex+iy| = |exeiy|, wehavetriviallythat|eiy| = 1(rememberthat z C, ez= 0).Remark6. We have not yet made the link between the complex exponentialintroducedhereandtheoneweintroducedfornotational purposesintherst chapter, whilediscussingEulers formula. This linkwill comesoon,oncewehavetreatedcomplextrigonometricfunctions.Before commenting briey on the geometry of the mapping generated bythecomplexexponential, itwill beuseful tointroduceanothercharacteri-zationoftheexponential.Proposition17.z C, ez= limn+_1 +zn_n.Proof. Usingthebinomialtheorem:_1 +zn_n=n

p=0n!p!(n p)!_zn_p=n

p=0n!p!(n p)!npzpHence, it is enough to prove thatn!(np)!nptends to 1 when n tends to innity,becausethen, werecovertheseriesexpansionof theexponential. Indeed,wehave:n!(n p)!= n(n 1)...(n p + 1) npwhenn +.So,itisclearthatn!(np)!np 1whenn +.Thegeometryofthemappinginducedbytheexponential canbechar-acterizedasfollow:58 CHAPTER2. COMPLEXFUNCTIONS Avertical lineL= {z C, z=x + iy, x=a R}ismappedintoacircleofcentre0andofradiusex. Anhorizontallinel= {z C, z=x + iy, y=b R}ismappedintoalinethrough0makingananglebwiththerealaxis.2.2.2 ComplextrigonometricandhyperbolicfunctionsAsfortheexponential,weshallusepowerseriestodenecomplextrigono-metric and hyperbolic functions. We will see that there exists a strong dual-itybetweentrigonometricandhyperbolicfunctionsoftherealvariables,inthe sense that cos(z) (resp. sin(z)) reduces to the real cosinus (resp. real si-nus) on the real axis, and to the real cosh (resp. real sinh) on the imaginaryaxis.Denition23. Let z C. Wedenethetrigonometricandhyperboliccomplexfunctionsas:cos(z) =+

n=0(1)nz2n(2n)!cosh(z) =+

n=0z2n(2n)!sin(z) =+

n=0(1)nz2n+1(2n + 1)!sinh(z) =+

n=0z2n+1(2n + 1)!.The fact that these series converge can easily be realized by applying theratiotesttoeachof them. Thiswill alsoshowthattheyhaveaninniteradius of convergence. It is now apparent,by a simple reorganization of thetermsofthetrigonometricseries,that:z C, eiz= cos(z) +i sin(z).2.2. SOMECOMPLEXFUNCTIONS 59Thisimplies,inparticular,Eulersformula: R, ei= cos() +i sin().Also, we can see that de Moivres formula follows straightforwardly from thedenitionsintroducedabove. Equivalently,onecanwritethetrigonometricandhyperbolicfunctionsintermsoftheexponential. Forz C:cos(z) =eiz+eiz2; sin(z) =eizeiz2i;cosh(z) =ez+ez2; sinh(z) =ezez2.Byasimplecomparisonof theseformulae(or, equivalently, usingtheseriesexpansions)oneseesthat:z C, cos(iz) = cosh(z)andsin(iz) = i sinh(z).Restrictingtheserelationstotherealandimaginaryaxisleadtotheprop-ertiescitedintheintroduction.Additionpropertiesof complextrigonometricandhyperbolicfunctionsareexactlyidenticaltotheircounterpartsforrealvariables. If(z, w) C2:cos(z +w) = cos(z) cos(w) sin(z) sin(w)sin(z +w) = cos(z) sin(w) + sin(z) cos(w)cosh(z +w) = cosh(z) cosh(w) + sinh(z) sinh(w)sinh(z +w) = cosh(z) sinh(w) + sinh(z) cosh(w).Thesepropertiesfollowdirectlyfromtheexpressionsofthetrigonomet-ric andhyperbolic functions interms of the exponential. Note that thesimilar relations for substractions comefromthesymmetryproperties ofthefunctions:cos(z) = cos(z) ; sin(z) = sin(z)cosh(z) = cosh(z) ; sinh(z) = sinh(z).60 CHAPTER2. COMPLEXFUNCTIONS2.2.3 RootsofunityNowthatwehavedenedproperlythecomplexexponential andtrigono-metricfunctions, weareequippedtostudyanimportant set of algebraicequationsForz Candn N, letusconsidertheequationzn=1. Bywritingz= rei,itisimmediatethatr = 1(|zn| = |z|n). Hence,thepointszsatisfyingtheequationarelocatedontheunitcircle. Then,wehave:_ei_n= 1 ein= 1 cos(n) +i sin(n) = 1 cos(n) = 1andsin(n) = 0 n = 2k, k {0, 1, ..., n 1}. (2.1)In the last line, note that kis smaller or equal to n1. This comes from thefactthat the argument ofa complex numberis denedup to a factorof2:k ncorrespondstovaluesofthatleadtoz= eiequivalenttotheonescovered by k {0, 1, ..., n1}. Hence, the equation zn= 1 in C has exactlyn solutions, called the nth rootsofunity: z= e2ki/n, k {0, 1, ..., n1}.Byconstruction,theyarelocatedontheunitcircle,atangles2ki/n,withk {0, 1, ..., n 1}: theyformaregularn-goncentredat0andwithonevertexatz=1. Ifnisodd, theonlyreal rootisz=1; ifniseven, therearetworealroots: z= 1. Thecasen = 6isdepictedongure2.1.We will see in this course that any complex polynomial equation of ordern admits exactly n roots (not necessarily distinct); this is the fundamentaltheoremofalgebra. Thereexistsmanywaystoprovethistheorem,andwewillproveitinthelastchapterofthesenotes. But,atthemoment,youwillbeabletoprovethatapolynomialwithrealcoecients,ofordertwo,hasalwaystworootsinthecomplexplane.Exercise6. Letaz2+ bz + c = 0for(a, b, c) R3andz C,witha = 0.2.2. SOMECOMPLEXFUNCTIONS 61Figure2.1: 6throotsofunityformingaregularhexagon.Provethatthisequationhasexactlytworoots,givenby:z1,2= b2a 12a_b24ac ,if b24ac 0z1,2= b2a i2a_|b24ac| ,if b24ac < 0(2.2)2.2.4 ThelogarithmicfunctionBeforediscussingthecomplexlogarithmindetails, weneedtocomebacktoour denitionfor the argument of acomplexnumber. The fact thatthecomplexexponential is periodicwithperiod2implies that therealnumbersuchthatforz C,z= reiisnotunique. Thisisthesourceofonesubtletyofcomplexanalysisthatisnotpresentinrealanalysis: many-valuedness of complex functions. For any z C, we will all theargument62 CHAPTER2. COMPLEXFUNCTIONS(insteadofanargument)ofztheset:arg z = { R, z= |z|ei}.In real analysis, the logarithm is the inverse of the exponential function,i.e., foranyx R+, thereexistsauniquey Rsuchthatey=x. Thisyisthendenotedy=ln(x). Wecanapplythesamemethodincomplexanalysis: for any z C, we look for a w C such that ew= z. Let us writew = u +ivwith(u, v) R2. Then,onhas:|z| = |eueiv| = euarg z = {v + 2k, k Z}. (2.3)Hence,wehave:ew= z w = ln |z| +i, with arg z.Wethusdenethelogarithmofanyz Cvia:ln z = {ln |z| +i, arg z}.This clearly shows that the logarithm of a complex number z is not uniquelydened: itsrealpartisunique, equaltoln |z|, butitsimaginarypartcanbeanyrealnumberthatisanargumentofz. Thisistherstexampleofamulti-valued function: to a given complex number z, the complex logarithmassociatesaninnityofcomplexnumbers. Inthenextsection, wewill seehowtotreatsuchmultifunctions.2.2.5 ExercisesSolve the following equations in C, and represent the solutions in the complexplane:(i) 2z5+ 1 = 0;(ii) z2+z + 1 = 0;2.3. MULTIFUNCTIONS 63(iii) z4+ 2z2+ 4 = 0;(ii) sin(2z) = 2;(iv) z4+i = 0;(v) z1/3= 2i;(vi) ln(z2) = 2 +4i.Togofurther2.3 MultifunctionsAswehaveseenwiththeexampleofthelogarithm, multifunctionsappearevery-time we try to invert a complex function (like the exponential) that isnotgloballyone-to-one. Inthissection, wearegoingtodevelopamethodto construct one-to-one function from multifunctions. We will come back tothesetechniqueslater,oncewehaveintroducedtheconceptofholomorphy.2.3.1 Example1: thelogarithmicfunctionLetusstartwiththelogarithmencounteredabove:z C, ln z = {ln |z| +i, arg z}.Hence, values of that dier fromeachother byaninteger multiple of2leadtothesamepointz=reiinthecomplexplaneplane, butgivedierentvalueforitslogarithmln |z| + i. Nevertheless, if werestrict [0, 2[ or] , ], weobtainasingle-valuedfunction. Wewill call sucharestrictionaprincipal-valuedeterminationof theargumentof z. Usually,64 CHAPTER2. COMPLEXFUNCTIONSFigure2.2: Cutofthecomplexplanefortheprincipaldeterminationofthelogarithm.suchaprincipal-valuedeterminationoftheargumentisnoted=Arg(z).One shouldnote that, despite its name, this prescriptiondoes not haveanythingparticular(despitebeingthemostwidelyused): anyinterval oflength2thatisclosedatoneendandopenattheotheronewoulddothetrick. Letuschoosetheprincipaldeterminationbyrestricting ] , ].Thisintroducesabranchcut(orcutforshort)inthecomplexplane,i.e.,thesemi-axisof negativereal numbers] , 0] iscutoutof thecomplexplane and zcannot cross this cut while roaming in the complex plane. Thiscuthastwoedges: theupperedgeisidentiedwiththeargument=,andtheloweredgeto= . Ourprincipal determinationimpliesthattheloweredgeofthecutisexcludedfromthecomplexplane,butitsupperedgeincluded. Thepoint0,thatisinthecutiscalledabranchpoint.Inthecutplane,wecannowdeneafamilyofsingle-valuedfunctions:k Z, ] , ], z= rei C, fk(z) = ln r +i( + 2k).2.3. MULTIFUNCTIONS 65Clearly,wehavethat:z C, ln z = {fk(z), k Z}.Inthecutplane, eachfkiscontinuous, bycontinuityofitsreal andimag-inaryparts. Butonthecut, theyarenotcontinuous, andincrossingthecutfromtheupperhalf-planetothelowerhalf-plane, onehastotransferfromfktofk+1: thenumberkcountsthenumberoftimewehaveaccom-plishedacomplete(anti-clockwise) tour aroundthebranchpoint 0(neg-ativevaluesindicateclockwisetours). Thistransferiscontinuousbecauselimh0+ fk(ih) = limh0 fk+1(ih).2.3.2 BranchpointsandmultibranchesNowthatwehaveseenhowtoproceedtoextractsingle-valuedfunctionsfrom the logarithm,we can tryandgeneralize ourintuition to more generalmultifunctions.Letusrstproveasimpleresult:Proposition 18. There is no restriction which selects a real function (z) arg zforall z C,sothat : z (z)isacontinuousfunction.Proof. Letusassumeforacontradictionthatsuchacontinuousargumentfunctionexists. Then,consider:v(t) = (eit)fort R.Bycompositionof continuous functions, v : R Ris thus continuous.Moreover,v(t + 2) = v(t),sovisperiodic,withaperiodof2. Now,v(t)andtarebothargumentsofeitbyconstruction,so,thereexistsafunctionn : R Zsuchthat:v(t) t = 2n(t).vbeingcontinuous,nisalsocontinuous. So,bythetheoremoncontinuousinteger-valued function from chapter1,it is constant: t R, n(t) = n,and66 CHAPTER2. COMPLEXFUNCTIONSv(t) =t + 2nfor anyt R. But, v(t) =v(t + 2); this implies that:t + 2n = t + 2 + 2n,hencethecontradiction.This results means that anyprincipal valueof theargument functionof z =reiobtainedbyrestrictingthedomainof has tohaveajumpdiscontinuitysomewhere. Thisisparticularlyobviousif [0, 2[: then,when z= eidescribes a complete circuit anticlockwise, starting from z= 1, starts at 0, and increases continuously towards 2, where z reaches 1 again.Thisresulthasimplicationsforothermultifunctions. Forexample, theimaginarypartof acomplexexponential isanargumentfunctionbycon-struction,sothereisnocontinuouslogarithmon C.BranchpointsConsidernowa multifunction w(z) thatis anon-emptysetof C foranyzinthedomainofdenitionofw.Denition 24. A branch point of w(z) is a point a C such that, for allr > 0,itisnotpossibletochoosef(z) w(z)suchthatfisacontinuousfunctiononthecirclecentredonaandofradiusr.Namely, thatmeansthatthedenitionof w(z)implicitlyorexplicitlyinvolvestheargument,wherez a = |z a|ei,inanothercongurationthanei(i.e. apurephase). Oneclearlyseesthatthemotivationforsuchadenitioncomesfromthepropositionaboveregardingtheargumentfunc-tion. Thefactthat0isabranchpointforthelogarithmthuscomesfromthe fact that the imaginary part of the logarithm is exactly the argument ofthecomplexnumberintothelogarithm,i.etheangleonacirclecentredon0.Lets look at another example: ln((z 1)/(z +1)). If one writes z 1 =|z 1|eiand z +1 = |z +1|ei, one has ln((z 1)/(z +1)) = ln(|(z 1)/(z +1)|ei()). Hence: ln((z 1)/(z +1)) = {ln |(z 1)/(z +1)| +i( ), arg(z 1), arg(z + 1)}. So, weseethatboththeanglesaround12.3. MULTIFUNCTIONS 67and 1appearinthedenitionofthemultifunction: 1and 1arebranchpointsforthismultifunction.MultibranchesAs we have seen previously, there is a way to construct continuous selectionsfrommultifunctions. The keytothis procedure is tomake achange ofvariableandreplacezby(r, )aroundeachbranchpointa, suchthatz=a + rei. Considerrstamultifunction w(z)withonlyonebranchpoint(suchasthelogarithm). Thatmeansthat,forz = a:w(z) = {w(z) = w(r, ), arg(z a)}.Henceisdeterminedonlyuptoaintegermultipleof 2. Therefore, byrestricting ]c, c + 2]withc R,wehavethemultibranches:k Z, Fk(r, ) = w(r, + 2k).Thatarecontinuousfunctionsofrand. Hence,w(z) = {Fk(r, ), k Z, ]c, c + 2]}.Theset(Fk)kZiscalledacompleteset of multibranchesfor w(z).Now,observewhathappenswhenzdescribesthecirclecentredonaandofradius r. Then z= reit, with t going from 0 to 2. Consider the kth branch:Fk(r, t = 2) = w(r, 2 + 2k) = w(r, 0 + 2(k + 1)) = Fk+1(r, t = 0). Thismeans that, whenz travels anticlockwisearoundthebranchpoint, thereis anatural, continuous transfer fromFktoFk+1. This technique onlyworks withone branchpoint. The theorywithseveral branchpoints ismoredelicateandwill not betreatedhere, but wewill seeonaspecicexamplehowtoaddresstheproblempractically.2.3.3 Example2: FractionalpowersWe have seen how to treat the logarithm multifunction in order to constructacompletesetofmultibranchesforit. Wewill seeinthissubsection, how68 CHAPTER2. COMPLEXFUNCTIONSto apply this method to another important class of functions: the fractionalpowers. Considern Z\{1, 1}. Considertheequationw=z1/n. Then,if w=reiandz=ei, wehave: r=1/nand=n+2kn, fork Z.Hence,wehaveamultifunction:z1/n = {|z|1/nei/n, z C, arg z},withabranchpointat0. Wechoose [0, 2[, sothatwecuttheplanealongthepositiverealaxis. Wedenethebranches:z= rei, z = 0, k {0, 1, ..., n 1}, gk(z) = r1/nei(+2kn).Then,wehave:z1/n = {gk(z), k {0, 1, ..., n 1}}.Thistimethecompletesetofmultibranchesisnite.2.3.4 Example3: AnexamplewithtwobranchpointsWhathappensifthemultifunctionhastwobranchpoints?Theideainthepreviouscases, withonebranchpoint, wastocutthecomplexplane ina waythat prevent the possibilityto construct closedcontour paths that include the branchpoint intheir interior. The idearemainsthesamewhentherearemorethanonebranchpoint. Considerf(z) =z2+ 1. This canberewrittenas: f(z) =_(z i)(z +i), andintroducing z i = reiand z +i = ei, so that and are the argumentsaround iandirespectively, wehave: f(z)=f(r, , , )=rei(+)/2.Hence, if zdescribes a circle around i (resp. i), varies by 2and variesabitbutcomesbacktoitsoriginalvalue(resp. variesby2andvariesabitbutcomesbacktoitsoriginalvalue),so,attheendoftheloop,f(z)hasanewvalue: fnew(z)= fold(z), whereaszhasreturnedtothesamevalue. Thisprovesthatiand iarebranchpointsforthemultifunctionf.Now, toconstruct well denedbranches, we needtwocuts, inorder to2.3. MULTIFUNCTIONS 69Figure2.3: Cutofthecomplexplaneforf(z) =z2+ 1.prevent the twoangles andtovarybyinteger multiples of 2. Forexample, wecancutparallel tothereal axis, alongthenegativereal partsforbothpoints, restricting(, ) ] , ]2. Wecouldalsocutalongtheimaginary axis, from i to innity and fromi to . Finally, let us mentionanotherinterestingcut. Ifwechooseabranchcutthatisthelinesegment[i, i], weclearlypreventanyclosedpatharoundior iseparately. But,weallowclosedpathsthatencirclebothbranchpointsatthesametime.Actually,itisnotaproblem: alongsuchapath,bothandvaryby2,andf(z)returnstoitsinitial value, sothatthesepathsdonotintroducetheneedforanynewbranchcut. Thisisageneralresult: toproducewell-behavedbranches of amultifunction, it is enoughtointroducecuts thatpreventtheexistenceofclosedpathsaroundeachisolatedbranchpoint.70 CHAPTER2. COMPLEXFUNCTIONSChapter3Complexdierentiation7172 CHAPTER3. DIFFERENTIATIONDierentiationinthecomplexplaneiscentral tocomplexanalysis. Itissimilar,butnotidenticaltodierentiationinrealanalysis. Itallowsonetointroduceaclassofcomplexfunctioncalledholomorphicfunctions, andcomplex analysis can be viewed as the study of these holomorphic functions.3.1 Holomorphicfunctions3.1.1 DierentiationandtheCauchy-RiemannequationsWewill rstdenedierentiabilityinthecomplexplane, andintroduceanecessary condition for a function to be dierentiable: the Cauchy-Riemannequations.Denition25. Let f: S C a complex valued function dened on S C.LetG SanopensubsetofS. Then,fisdierentiableatz Gi:limh0f(z +h) f(z)hforanyhsuchthatz +h Gexists. Whenthis limit exists, it is denotedbyf

(z) andit is calledthederivativeoffatz.This denition relies strongly on the existence of the open set G. Indeed,sinceGisopen,foranyz G,thereisanr> 0suchthatD(z, r) G. Inother words, for any h C such that |h| < r, z+h G. This ensures that, inthe limit written above to dene the derivative, the point z+h can approachzfromanydirectionashtendstozero. Inotherwords, thederivativeexistsonlyif thevalueof thelimitdoesnotdependonthewayz + happroachesz. Thisissimilartowhathappensinreal analysis: areal functiondenedonanopeninterval of Risnotdierentiableatxifthederivativesfromtheleftandfromtherightofxarenotthesame. Wecanusethisideatoillustratethenon-dierentiabilityofasimplecomplex3.1. HOLOMORPHICFUNCTIONS 73function. Considerf(z) = Imzdenedin C. Letusconstruct:f(z +h) f(z)h=Im(z +h) Im(z)h=Im(h)h. (3.1)Chooseh R, sothatIm(h)=0, thenf(z+h)f(z)h0whenh 0. But,now, choose h such that Re(h) = 0; thenf(z+h)f(z)hi when h 0. So,inthatcase,wehavefoundtwowaysofapproachingzthatdonotgivethesamelimit; thatimpliesthatthefunctionf:z Imzisnotdierentiablein C. Selectingtwowaysofapproachingthepointzthatdonotleadtothesame limit, as we just did, is quite a general method to prove that a functionisnotdierentiable.Theorem9. Let f : G CwhereGis anopensubset of C. Let f bedierentiableatz G. Letz= x + iyandf(z) = u(x, y) + iv(x, y). Thenuandv, asreal functions, havepartial derivativesat (x, y) R2. Thesepartial derivativessatisfytheCauchy-Riemannequations:_ux=vyuy= vx.(3.2)Inthefollowing, wewill oftendenoteux=ux(andthesameforthederivativewithrespecttoy).Proof. Sincefisdierentiableatz,wehavethat,foranyh Csuchthatz +h G:f

(z) =limh0f(z +h) f(z)hexists.Moreover,sincewecanchoosehfreely(i.e. wecanapproachzhoweverwewantinG),wecanrestrictittobepurelyrealontheonehand,andpurelyimaginaryontheother. Thus,wehave: forh R:f

(z) =limh0_u(x +h, y) u(x, y)h+iv(x +h, y) v(x, h)h_= ux+ivx;74 CHAPTER3. DIFFERENTIATION andforh = ik,k R:f

(z) =limh0_u(x, y +k) u(x, y)ik+v(x, y +k) v(x, y)k_=1iuy+vy.Thepartial derivatives exist becausethelimit deningf

(z) exists.Now,thelimitisuniquebyconstruction,so:ux +ivx=1iuy +vy,or vx + iux=uy+ ivy. So, equatingreal andimaginaryparts, werecovertheCauchy-Riemannequations.NotethattheCauchy-Riemannequationsareanecessaryconditionforafunctiontobedierentiable. Theyare not sucient. That meansthatthecontrapositiveof theprevioustheoremcanbeusedtoshowthatafunctionisnotdierentiableatapoint: if oneprovesthattheCauchy-Riemann equations do not hold, then the function is not dierentiable. But,if they hold, this is not sucient to prove that the function is dierentiable.Toseethis,letsgobacktoourpreviousexample: f(z) = Im(z)forz C.Wehaveseenthatthisfunctionisnotdierentiable. Wecanconrmthatbyprovingthat theCauchy-Riemannequations donot hold. Indeed, wehaveu(x, y) = 0andv(x, y) = y. So,ux= 0 = 1 = vy.Now, considerf(z) =_|Re(z)Im(z)| forz C. Atz =0, wehave, forh C:f(0 +h) f(0)h=_Re(h)Im(h)h.Hence, weseethatwhenweapproach0withh R, thelimitofthisratiois 0 (because the ratio is identically 0). But,if we approach 0 along the linemaking an angle /4 with the real axis, we can write h = t(1+i) with t R,theratioisconstantandequalto1/(1 + i) = 0. Thatprovesthatfisnotdierentiableat0. Ontheotherhand, theCauchy-Riemannequationsare3.1. HOLOMORPHICFUNCTIONS 75triviallysatisedatz=0. ThisshouldemphasizethatCauchy-Riemannequationsshouldbehandledwithcare.TogofurtherDespite this warning, a slight modication of our theorem provides a partialconversetotheprevioustheorem: itisenoughtoaddthecontinuityofthepartialderivatives.Theorem 10.Let f: G C with G an open subset of C. Let z= x+iy Candf(z) = u(x, y) +iv(x, y). IfuandvhavecontinuousrstorderpartialderivativesinGthatsatisfytheCauchy-Riemannequationsatz,thenf

(z)exists.Proof. Let z G. Let r > 0 such that D(z, r) G. Consider h = p+iq Csuchthat |h| < r. Then,for(z, h) =f(z+h)f(z)h:(z, h) =ph_u(x +p, y +q) u(x, y +q)p+iv(x +p, y +q) v(x, y +q)p_+qh_u(x, y +q) u(x, y)q+iv(x, y +q) v(x, y)q_Now,sincethefunctionsuandvaredierentiable,theyarecontinuousonG, soonecanapplythemeanvaluetheoremandnd(, , , ) ]0, 1[4suchthat:f(z +h) f(z)h=pq_ux(x +p, y +q) +ivx(x +p, y +q)_+qp_uy(x, y +q) +ivy(x, y +q)_.Now,onecanusethecontinuityofthepartialderivativesonGtondthatfor> 0, there exists h > 0 small enough such that:f(z+h)f(z)hg(z) < where:g(z) =pq(ux(x, y)+ivx(x, y))+qh(uy(x, y)+ivy(x, y)) = ux(x, y)+1iuy(x, y).76 CHAPTER3. DIFFERENTIATIONThelast equalityholds becauseof theCauchy-Riemannequations. Thisshowsthatf

(z)existsandisequaltog(z).Inprinciple, thistheoremcanbeusedtotestthedierentiabilityof afunction. Butitisnotverypractical, andwewill shortlyseemuchmorepowerfulresultstoachievethisgoal.3.1.2 HolomorphicfunctionsUntil now, wehaveusedthedecompositionofcomplexnumbersandcom-plexfunctions intotheir real andimaginaryparts totalkabout complexdierentiation. Itistimetoforgetaboutallthisandtodealdirectlywiththecomplexvariable. Again, thenotionofopensetswillbecentraltothedevelopmentspresentedhere.Denition26. Acomplexfunctionfthatisdierentiableatanypointofan open set G included in its domain of denition is said to be holomorphicinG.Thismeansthat,foranyz G,irrespectiveofthewayhtendstozero,limh0(f(z+h)f(z))/h exists. We will denote by H(G) the set of functionsholomorphicinagivenopensetG.Denition27. A function fis said to be holomorphicatapoint a Cisthereexistsr > 0suchthatfisdenedandholomorphicinD(a, r).It is important to realize that being holomorphic at a point a is a strongerconditionthanbeingdierentiableata: inordertobeholomorphicata,fhastobedierentiableataandateverypointofadiskcentredona.Wecannowlistafewpropertiesofholomorphicfunctions. Theproofsare left tothe reader, as theyare identical totheir counterparts inrealanalysis. LetGbeanopensubsetof G. Thefollowingpropertiescanbe3.1. HOLOMORPHICFUNCTIONS 77derived by proving the appropriate dierentiability conditions at each pointz G. Let fand g be holomorphic in G and let C. Then, f, f +g and fgareholomorphicinGandthefollowingrulesfordierentiationapply,forallz G:(f)

(z) = f

(z)(f+g)

(z) = f

(z) +g

(z)(fg)

(z) = f

(z)g(z) +f(z)g

(z). LetfbeholomorphicinGandgbeholomorphicinanopensetcon-tainingf(G). Then,g fisholomorphicinGand,forallz G:(g f)

(z) = (g

f)(z)f

(z) = g

(f(z))f

(z).Thispropertyisoftenreferredtoasthechainrule. LetfbeholomorphicinGsuchthat z G, f(z) =0. Then,1/fisholomorphicinGand,foranyz G:_1f_

(z) = f

(z)(f(z))2.These rules can now be used to test the holomorphy of complicated functionsknowingtheholomorphyof simplefunctions. Itwill bemucheasierthanusingtheCauchy-Riemannequations.For example, the functionf(z) =z is triviallydierentiable for anyz C,asareanyconstantfunctions. Thisimpliesthatanypolynomial:P(z) =N

n=0cnznwhere N N and n {0, 1, .., N}, cn C, is holomorphic in C. Rememberthatapolynomialisthesumofnitelymanyterms. Powerseries,thatare78 CHAPTER3. DIFFERENTIATIONthesumsofinnitelymanyterms, will betreatedseparately. Inthesameway, any rational function P(z)/Q(z) where P(z) and Q(z) are polynomialsisholomorphicinanyopensetinwhichQ(z)isneverzero. Forexample,1/(1 +z2)isholomorphicin C\{i, i}.TogofurtherFinally, letusseewhathappensintheextendedcomplexplaneC. In C,holomorphyisstatedasabove, butwhathappensat ? Wehavealreadyseen a map z 1/zthat interchanges with a point of C, namely 0. Let fbe a function dened on a set {z C, |z| > r} for some r > 0. By deningfsuch thatf(z) = f(1/z), we have f() =f(0). Hence, any property offat0,suchascontinuity,limit,holomorphy,canbetransferredtofatinnity.Consider,forinstance,f(z) = z2. Then,f(w) = 1/w2,thatisnotholomor-phic at w = 0 (Check it); so we can say that fis not holomorphic at innity.Conversely, considerf(z)=1/(1 + z2)for |z| >1. Then, f()=0, andf(w) = w2/(1 +w2)for |w| < 1:fisholomorphicat0,sofisholomorphicat .3.1.3 SomeusefulresultsNow, wewill proveacertainnumberof resultsonholomorphicfunctionsthatwill beuseful intherestof thiscourse. Therstonestatesthatanholomorphic function is necessarily continuous. This is analogous to the realcase,wheredierentiabilityimpliescontinuity(butincomplexanalysis,weneedholomorphy,notmeredierentiability).Proposition19. Letf: S CwithS C. LetGbeanopensubsetofS.IffisholomorphicinG,thenitiscontinuousonG.Moreover,ifF GisacompactsubsetofG,thenfisboundedonF.3.1. HOLOMORPHICFUNCTIONS 79Proof. Supposef isholomorphiconG. Then, foranyz Gandforanyh Csuchthatz +h G:f

(z) =limh0f(z +h) f(z)h.So,wecandene,forh = 0,thefunction:(h) =f(z +h) f(z)hf

(z).So,bydenition,(h) 0whenh 0. Hence,wehave:f(z +h) f(z) = h(f

(z) +(h)).Thisimpliesthat:limh0|f(z +h) f(z)| = 0,whichisexactlytherequirementforftobecontinuousonG.Now,rememberthatwehavestated,intherstchapterthatacontinuousfunctiononacompactsubsetof Cisboundedonthisdomain. So, sincefiscontinuousonG,itiscontinuousonF G,andsinceFiscompact,fisboundedonF.Proposition20. Let f:S C CbeholomorphiconaregionG S.Then,fisconstantonGifanyofthefollowingconditionistrue:(i) z G, f

(z) = 0;(ii) |f|isconstantinG;(iii) z G, Im(f(z)) = 0.Proof. Remember that a region is a non-empty open connected subset of C.LetussupposethatG = D(0, 1). Foranyz= x +iy D(0, 1),letuswritef(z) = u(x, y)+iv(x, y). Since fis holomorphic on G, the Cauchy-Riemannequationshold,and:f

(z) = ux +ivx= vyiuy.80 CHAPTER3. DIFFERENTIATION(i) Suppose z G, f

(z)=0. Then, ux=vx=uy=vy=0identicallyon G. Let p = a+ib and q= c+id be two arbitrary point of G. Then,constructr=c + ibands=a + id. Itisobviousthat |r|2+ |s|2=|p|2+ |q|2< 2,so,atleastoneofrorsisinD(0, 1). Letussupposer D(0, 1),withoutlossofgenerality. Thefunctionsx u(x, b)andy u(c, y)arerealfunctionswithvanishingderivatives,so,byvirtueofthemeanvaluetheorem,theyareconstant. Hence:u(a, b) = u(c, b)andu(c, b) = u(c, d).So,u(a, b) = u(c, d). Withthesameargument,wecanprovethat:v(a, b) = v(c, b)andv(c, b) = v(c, d),sothatv(a, b) = v(c, d). Thisshowsthatf(p) = f(q)foranypandqinG.(ii) Now, letussupposethat z D(0, 1), |f(z)| =c, withc R+con-stant. Then: u2+v2= c2,sothat:uux +vvx= 0anduuy +vvy= 0.ByusingtheCauchy-Riemannequations, thisleadstouux vuy=uuy+ vux=0, so: (u2+ v2)ux=0. If u2+ v2=0, thenf =0onD(0, 1), sofisconstant. Ifu2+ v2=0, then, ux=0onD(0, 1).Similarly, uy= vx= vy= 0 on D(0, 1), so that fis constant on D(0, 1)(accordingtotherstpoint).(iii) Finally, takef real valuedonD(0, 1). Then, v =0, sothat vx=vy=0, whichimplies, throughtheCauchy-Riemannequations, thatux=uy=0. Hence, fisconstantonD(0, 1)(accordingtotherstpoint).Theproof presentedheregeneralizes toanyregionofCbyreplacingthesimpleroute(p, r, q)oftherstpointtoanarbitrarypolygonal routecon-sistingofhorizontal andvertical linesegments. Itismoremessy, butdoesnotinvolveanythingnew.3.2. SOMEHOLOMORPHICFUNCTIONS 813.1.4 Exercises1. Whichofthesefunctionsaredierentiableatthegivenpointa C:(i) f(z) = z|z|,a = 0;(ii) f(z) = |z|2,a = 0;(iii) f(z) = arg(h),a = 0(whereargisrestrictedto[0, 2[);(iv) f(z) = z,foranya C.2. Give the domain of holomorphy of the following function, and calculateitsderivativeinthisdomain:f(z) =zz2+ 1.3. Let f: G C C be holomorphic in the open set G. For z= x+iy C,dene:f z=12_fx+ify_fz=12_fx ify_,wherefisregardedasafunctionof(x, y), f(z)=f(z(x, y)), ontheright-handside.Checkthatthepartialderivativesoff(z(x, y))withrespecttoxandyexist,andshowthat:f z= 0andfz= f

(z).Conversely,prove that a dierentiable function fthat satisesf z= 0isholomorphicinG.3.2 SomeholomorphicfunctionsWe will now come back to the functions we have dened previously, and lookattheirholomorphy. Beforethat, wehavetoproveaverypowerful result82 CHAPTER3. DIFFERENTIATIONonthedierentiationofpowerseries: anyfunctionthatcanbewrittenasapowerseriesisholomorphicinthediscof convergenceof thepowerseries.Later in the course, we will prove another very powerful result: the fact thatthe contrapositive is also true, namely that any holomorphic function can bewritten as a power series. These two results together mean that holomorphicfunctionsandpowerseriesareonesamenotion!3.2.1 AresultonthedierentiationofpowerseriesLetusseewhatthederivativeofapowerseriescouldbe.Lemma2. Let(cn)nN. Thepowerseries

cnznand

ncnzn1havethesameradiusofconvergence.Proof. Letussupposerstthat

cnznconvergesfor |z| R. Conversely,if

|ncnzn1|convergesfor |z| < R,then:n N, |cnzn| |z||ncnzn1|,3.2. SOMEHOLOMORPHICFUNCTIONS 83so

|cnzn|convergesbythecomparisontest,andthus,

cnznconverges.Theorem11. Letf(z)=

+n=0cnznwitharadiusof convergenceR>0.Then,f H(D(0, R)),and:z D(0, R), f

(z) =+

n=1ncnzn1.Proof. Thepreviouslemmaallowstodene,forallz D(0, R),afunctiongsuchthat:g(z) =+

n=1ncnzn1.Forz D(0, R)andh Csuchthatz +h D(0, R):f(z +h) f(z)hg(z) =+

n=1_(z +h)nznhnzn1_cn.Wewillneedthebinomialexpansion:(z +h)n=n

k=0CknznkhkwithCkn=n!k!(n k)!.Then:(z +h)nznhnzn1=nhzn1+... +Cknhkznk+... +hnhnzn1= h_C2nzn2+... +Cknhk2znk+... +hn2_= hn

k=2Cknhk2zn2= hn2

i=0n!(n (i + 2))!(i + 2)!hizni284 CHAPTER3. DIFFERENTIATIONbythechangei = k 2. So,usingthetriangleinequalityrecursively:+

n=0(z +h)nznhnzn1|cn| |h|+

n=0n2

i=0|cn|n!(n 2 i)!(i + 2)!|h|i|z|n2i |h|+

n=0n(n 1)n2

i=0|cn|(n 2)!|h|i|z|n2i(n 2 i)!(i + 2)! |h|+

n=0n(n 1)|cn|(|z| +|h|)n2.Let > 0 such that |z| < < R and |z|+|h| < (always possible since h canbeassmallasdesired). Then,clearly, |h|

+n=0n(n 1)|cn|(|z| +|h|)n2 0, =] , ], fk(z) = fk(r, ) = ln r +i( + 2k).Then,wehavetheresult:86 CHAPTER3. DIFFERENTIATIONProposition21. Forall k Z, fkisholomorphicinC= C\] , 0],with:z C, f

k(z) =1z.Proof. Let z C, andlet h Csuchthat z+ h C. Denote=fk(z+h)fk(z). Then, the continuity of fk on C implies that limh0 = 0.Now,usingthefactthatefk(z)= zforanyz C,wehave:h = efk(z+h)efk(z)= efk(z)(e1) = z(e1),sothat:fk(z +h) fk(z)h=1ze1. (3.4)Considerg() =e1,for = 0. Then:1g()=1_+

n=0nn! 1_=1+

n=1nn!=+

k=0k(k + 1)!.Hence:0 1g() 1=+

k=1k(k + 1)!+

k=1||k(k + 1)! ||2when 0.So,whentendstozero,thedominantterminthissumis ||andittendstozero, so, wehave: 1/g() 1whentends tozero; or equivalently,g() 1. Thismeansthat, whenhtendstozero,fk(z+h)fk(z)htendsto1/z. Hencefkisholomorphicwithf

k(z) = 1/zon C.3.3. CONFORMALMAPPING 873.2.5 ExercisesFindthedomainof holomorphyof thefollowingfunctions, andcalculatetheirderivativesinthisdomain:(i) f(z) = exp_1 +iz3_;(ii) f(z) =sin(iz+3)z21;(iii) f(z) = tan(z)cos(z);(iv) f(z) = tanh(z)cosh(z);(v) f(z) = cosh (2 sin(z) +i).Togofurther3.3 ConformalmappingInthissection,wewillbrieystudymappingsbetweenregionsofthecom-plex plane that preserve angles. In particular, we will see that any holomor-phicfunctionwhosederivativeisnon-zerodenessuchamapping.3.3.1 ConformalmappingConsider apathwithparameter interval [0, 1], for convenience. Then,there is awell-denedtangent to at =(0): +t

(0), for t 0,provided

(0) =0. This tangent makes ananglearg

(0) withtherealaxis.Now,let 1and 2be two paths,both with parameter interval [0, 1],with acommonstartingpoint1(0) = 2(0) = . Assumethat

1(0)and

2(0)are88 CHAPTER3. DIFFERENTIATIONboth non-zero,so that each path has a well-dened tangent at . The anglebetween1and2atisthensimplytheanglebetweentheirtangentsatthatpoint: arg

1(0) arg

2(0).Theorem12. Conformalitytheorem.Letf: C CbeholomorphicinanopensetG,andlet1and2bepaths,withparameterinterval [0, 1], inGmeetingat=1(0)=2(0). Supposethatf

() =0. Then, f preservesanglesbetweenpathsinGmeetingat.Thismeansthattwopathsmeetingatwithananglesaretransformedbyfintotwopathsmeetingatf()withanangle.Proof. Let=arg

1(0) arg

2(0)betheanglebetween1and2at.The paths 1and 2are mapped by fto paths f 1and f 2, respectively(notethattheyare, indeed, paths, becausef isholomorphic). Thesetwopathsmeetatf(), withanangle=arg (f 1)

(0) arg (f 2)

(0).Wehave:(f 1)

(0)(f 2)

(0)=f

()

1(0)f

()

2(0)=

1(0)

2(0).Hence, taking the argument of both side, and remembering that arg(z/w) =arg(z) arg(w): = ,whichistheresultthatneededtobeproven.Denition28. Acomplex-valuedfunctionf isconformal inanopensetG C(orC), if f H(G)and z G, f

(z) =0. Itissaidtobeconformal atapoint Cifitisconformal inadiscD(, r)forsomer > 0.Hence, theconformalitytheoremshowsthataconformal mappingpre-servesboththemagnitudeandsenseofanglesbetweenpaths.Theconformalitytheoremadmitsapartialconverse:Theorem13. LetGbeanopensetof C. letf: C Cbeafunctionsuchthatitspartial derivativesfxandfyexistandarecontinuousinG. Iffisconformal inG,then,fisholomorphicinG.3.3. CONFORMALMAPPING 89Proof. ConsiderapathinG. Let = f . Then,byasimplemanipu-lation,onecanwrite:

(t) =12 (fxify)

(t) +12 (fx +ify)

(t),wherethepartialderivativesareevaluatedat(t). Now,uptoachangeoforigin and units, we can consider that the square[1, 1] [i, i] is in G, andconsidertwopaths: 1(t) = itfort [0, 1],i.e. thelinesegment[0, i]; 2(t) = tfort [0, 1],i.e. thelinesegment[0, 1].Theyintersectat0, withanangle/2: arg (

1(0)/

2(0))=/2. Iffpre-serves the angle and sense in G we must therefore have: arg (

1(0)/

2(0)) =/2. Thisimpliesthatarg (fx/fy)=/2, hence: fy=ifx, at0. Writingf(z) = u(x, y) +iv(x, y),thisgivestheCauchy-Riemannequations.NotethattheargumentwouldbevalidforanyperpendicularcurveatanypointofG,providedweusethecorrectreparametrizationofthepaths,so,we can say that fhas continuous partial derivatives that satisfy the Cauchy-RiemannequationseverywhereinG. ThisimpliesthatfisholomorphicinG.3.3.2 SomeexamplesWestartbyprovingthatMobiustransformationsareconformal.Theorem14. Letf: z (az + b)/(cz + d)withad bc = 0beageneralMobiustransformation. fisconformal in C\{d/c}forc = 0.Proof. ItisenoughtoprovethatMobiustransformationareholomorphicwithf

(z) =0on C\{d/c}. Thisiseasilydonebyrealizingthatfistheratiooftwoholomorphicfunctions. Moreover,wehave:z C\{d/c}, f

(z) =ad bc(cz +d)2 = 0.90 CHAPTER3. DIFFERENTIATIONWhen we studied Mobius transformation, we considered their behaviourinCratherthat C,Wewouldthereforeliketoextendournotionofconfor-malitytoC. Iffmaps Cto , wewill buildg:z 1/f(z), andsaythatf isconformal at if gisconformal at. Wewill alsosaythatf isconformal at iff suchthatf(z)=f(1/z)isconformal atz=0. Letc = 0. Then:f() =b +ad +c.Fromthe previous theorem, this is conformal at 0, sof is conformal atinnity.Now,considerthebehaviouratz= d/c. Thispointismappedinto byf. Let= 1/wwherew = 1/f(z). Then:=cz +daz +b,andthishasanon-zeroderivativeatz= d/c, sof isconformal atz=d/c. Geometrically, it means that f maps apair of circles tangent atz= d/cintoapairofparallellines.Finally, in the case c = 0, f() = . By considering = 1/was a functionof=1/z, onecanshowimmediatelythatthederivativeatz=0isnon-zero.Hence, we can conclude that Mobius transformations are conformal at everypointofC.Hence, wecannowlistalotof standardconformal mappingsthatareMobiustransformations: z ziz+imapstheopenupperhalf-planeontotheunitdisc; z z+izimapstheopenlowerhalf-planeontotheunitdisc; z z1z+1mapstheopenrighthalf-planeontotheunitdisc; z z+1z1mapstheopenlefthalf-planeontotheunitdisc.Otherexamplesofconformalmapsare:3.3. CONFORMALMAPPING 91 Theexponentialmap. Itisconformalin Cbecauseitisholomorphicin C,with z C,_eiz_

= eiz= 0. z znforn N\{0, 1}isholomorphicin C. Atz=0, anglesaremagniedbyafactorofn,sothemapisnotconformal. Anyholomorphicbranchofthelogarithm. Theconformalityalsofol-lowsfromtheholomorphyandthenon-zerovalueofthederivative. Anyholomorphicbranchofageneralpowerz z,for > 0.In the last two cases, it is important to choose the cut so that the region wewishtomapisnotaectedbytheintroductionofthecut.Then, conformal mappingscanbeconstructedatwill byanystandardoperationsonfunctionsappliedtothefewwehavelistedabove,aslongasthe resulting function remains holomorphic with a non-zero derivative in theregiononewishestotransform.92 CHAPTER3. DIFFERENTIATIONChapter4Complexintegration9394 CHAPTER4. INTEGRATIONThecontentof thischapteristhecoreof complexanalysis. Inpartic-ular, wewill proveCauchystheorem, Cauchysformulaeandtheresiduetheorem. We will see that all this machineryof complexintegrational-lowsustoproveveryimportanttheoremsthatareencounteredeverywhereinmathematics, as, forexample, thefundamental theoremofcalculus, thefundamental theoremof algebra, Liouvillestheorem, butalsomethodstointegraterealfunctions,todealwithLaplaceandFouriertransformsetc.4.1 IntegrationinthecomplexplaneTostart, let us remember that wehavedenedapathintherst chap-terof thiscourse, asthejoinof nitelymanysmooth(i.e. dierentiable)curves: Itisafunction:[a, b] R Cthatispiecewisecontinuousanddierentiable. Acontourwasjust aclosedpathmadeof bitsof circlines(circular arcs and line segments). These paths and contours will be essentialin the theory of complex integration. Let us recall what piecewise continuitymeans. Afunctionh : [a, b] R Cispiecewisecontinuouson[a, b]ithereexistsrealnumbers(ti)i{0,1,...,n}suchthata = t0< t1< ... < tn= bandcontinuous functions hk: [tk, tk+1] Csuchthat h(t) =hk(t) fort ]tk, tk+1[. Notethathneednotbedenedatthepointstk. Thatmeansthat h is continuous everywhere on [a, b] except possibly for a nite numberofdiscontinuities. Areal-valuedfunctionhthatispiecewisecontinuousisintegrablewith:_bah(t)dt =n1

n=0_tk+1tkhk(t)dt.Thiscomesfromthefactthatcontinuousrealfunctionsareintegrable.4.1.1 IntegrationalongpathsWhatisthemeaningofanobjectlike:_baf(z)dz,4.1. INTEGRATIONINTHECOMPLEXPLANE 95when a and b are complex numbers, and f: S C C a complex function?Inrealanalysis,Riemannsconstructionoftheintegral:_baf(x)dxfor[a, b] Randfarealfunction,reliesonthepartitionof[a, b]intosmallerinterval[xi, xi+1]suchthat i {0, 1, ..., n}, xi=x0 + ixnx0n, withx0=aandxn=b. Then, theintegralisdenedasthelimit, whenntendstoinnity, i.e. whenthesizeof thesubintervalstendstozero,ofthesum:n1

i=0f(xi)(xi+1xi).Then,itcanbeshownthatthelimitdoesnotdependonthewaytheinter-val[a, b]iscutintosubintervals. Onecouldhavetheideatogeneralizethattoacomplexintegral, andtocalculatetheintegral of afunctionbetweentwocomplexnumbersbyasuccessionof small incrementsthatstartataandconnectittob. But,inthecomplexplane,thereexistsinnitelymanycurvesthatjoinapointatoab. Whichoneshouldonechoose?Actually,wewilldenecomplexintegrationbyusingaparticularpathbe-tween a and b. Then,we will show,later that,undercertainconditions,the result of the integrationdoes not dependonthe path. Be-foretreatingthegeneral caseof complexfunction, letusseethecaseof acomplex-valuedfunctionsdenedonanintervalof R.Denition29. Letf:[a, b] R C. Let x [a, b], f(x)=Re(f(x)) +iIm(f(x)), where Re(f) andIm(f) are real functions. We saythat f isintegrableiRe(f) andIm(f) are bothintegrable. If this is the case,then,wedene:_baf(x)dx =_baRe(f(x))dx +i_baIm(f(x))dx.Theeasiestexampleofsuchanintegralis:_20eixdx =_20cos(x)dx +i_20sin(x)dx = [sin(x)]20i[cos(x)]20= 0.96 CHAPTER4. INTEGRATIONWearenowequippedtodeneproperlytheintegralofacomplexfunctionalongapath.Denition30. Let be a path such that : [a, b] R C. By denition,thereexists(ti)i{0,1,...,n}witha=t00suchthat z , D(z, m) G. Wehavetoprovethatcanbecoveredbyanitesetof suchdisks, eachoverlappingthenext. Suppose, for astart, that issmooth. Then,byapplyingtherealmeanvaluetheorem,wehave:(s, t) [a, b]2, c [a, b], (Re) (s) (Re) (t) = (s t) (Re)

(c),andthesamefortheimaginarypartof . Sincetheyarecontinuousonaclosedinterval,[a, b],(Re)

and(Im)

arebounded. Hence:(s, t) [a, b]2, > 0, |s t| < |(s) (t)| < m.This states the uniform continuity of . This result remains valid if is notsmooth, sincewecanapplythesameargumenttothesmoothpiecesthatconstitute. Now, wecanchoosea=t0 0. fis dierentiable at Zsince it is holomorphic in the triangle.So,thereisar > 0suchthat:z D(Z, r),f(z) f(Z) (z Z)f

(Z)< |z Z|.LetN NsuchthatN D(Z, r). Then,forallz N,|z Z| 2NL.110 CHAPTER4. INTEGRATIONMoreover:_N_f(Z) + (z Z)f

(Z)_dz= 0,accordingtothefundamental theoremof calculusforaclosedcontourap-pliedtothefunctionF(z) = f(Z)z + (z2/2 Zz)f

(Z). Now:_Nf(z)dz=_N_f(z) f(Z) (z Z)f

(Z)_dz,and,sincetheintegrandontheright-handsideisbounded:_Nf(z)dz 2NL length() = 22NL2.Sincewehave: 4N|I| _Nf(z)dz, wehave:|I| L2foranarbitrarysmall,soI= 0.Theorem22. IndeniteintegralTheorem.Let f: C CbeacontinuousfunctiononaconvexregionG Csuchthat_ f(z)dz=0foranytriangle G. Leta G. Then, thefunctionFdenedby:z G, F(z) =_[a,z]f(w)dw,isholomorphicinGwithF

= f.Proof. Let z G, andD(z, r) Gforr>0, suchthat |h| r (this means that b is the rst pointatwhich1meetsthecircle |z a|=r). Then2=[c, d] 1joinsctob. Now, letbethejoin: = 2 ((a, r)) (2). isnotacontourbecausewetrace2twice,inbothdirections. However,the proof of Cauchys theorem for contours can be generalized to suchpaths(exercise). So:_f(z)dz= 0 =_ f(z)dz _(a,r)f(z)dz.(ii) ThispointcanbeprovenbychoosingsomediskD(a, r) I( )andbyapplyingtherstpointtwicetoand , toobtain:_ f(z)dz=_(a,r)f(z)dz=_ f(z)dz.4.2. CAUCHYSTHEOREM 117(iii) Thelastpointisadirectresultof Cauchystheorem, togetherwiththedecompositionoftheintegralalongajoin.Thistheoremcangiveusageneralizationof thefundamental integral_(a,r)(z a)ndzforn Z.Proposition26. Letbeapositivelyorientedcontour,andleta Csuchthata . Then:_(z a)ndz=___0 ifa O()0 ifa I()andn = 12i ifa I()andn = 1Proof. Forn = 1,theresultscomefromthefundamentaltheoremofcal-culus. Forn= 1, thereisnoantiderivativefor(z a)1, sowecannotusethefundamental theoremof calculus. However, Cauchys theoremisapplicablewhena O(), becausethenf is holomorphicinsideandonthecontour. Whena I(), wecanusethedeformationtheoremandtheknownresultfor(a, r)toprovethestatement.Example 7. Consider f(z) =2/(4z21) andI =_(0,1)f(z)dz. Thefunctionf isholomorphiceverywhereexceptatthetwopointsz= 1/2,where its denominator cancels. Let us separate these poles and write f(z) =1/(2z 1) 1/(2z + 1). Then, wecanapplythedeformationtheoremtoeachpartandwrite:I=_(1/2,1/4)12(z 1/2)dz _(1/2,1/4)12(z + 1/2)dz=122i 122i = 0.118 CHAPTER4. INTEGRATIONTogofurther4.2.4 Thecomplexlogarithm... againWhen we introduced the complex logarithm, we dened it as the innite setof solutions to the equation ew= z, for z = 0. In order to get a well-denedlogarithm, wehaveseenhowtointr