Complex_Analysis-K_Houston.pdf

ComplexAnalysis2002-2003

c K.Houston2003

1ComplexFunctions

Inthissectionwewilldeﬁnewhatwemeanbyacomplexfunc-

tion.Wewillthengeneralisethedeﬁnitionsoftheexponential,

sineandcosinefunctionsusingcomplexpowerseries.Todeal

withcomplexpowerserieswedeﬁnethenotionsofconver-

gentandabsolutelyconvergent,andseehowtousetheratio

testfromrealanalysistodetermineconvergenceandradius

ofconvergenceforthesecomplexseries.

Westartbydeﬁningdomainsinthecomplexplane.This

requirestheprelimarydeﬁnition.

Deﬁnition1.1

The" -neighbourhood ofacomplexnumberzisthesetofcom-

plexnumbers{w2C:|z−w|<"}where"ispositivenumber.

Thusthe"-neigbourhoodofapointzisjustthesetofpoints

lyingwithinthecircleofradius"centredatz.Notethatit

doesn’tcontainthecircle.

Deﬁnition1.2

A domain isanon-emptysubsetDofCsuchthatforevery

pointinDthereexistsa"-neighbourhoodcontainedinD.

Examples1.3

Thefollowingaredomains.

(i)D=C.(Takec2C.Then,any">0willdoforan"-

neighbourhoodofc.)

(ii)D=C\{0}.(Takec2Dandlet"= 1 2 (|c|).Thisgivesa

"-neighbourhoodofcinD.)

(iii)D={z:|z−a|<R}forsomeR>0.(Takec2Candlet

"= 1 2 (R−|c−a|).Thisgivesa"-neighbourhoodofcinD.)

Example1.4

ThesetofrealnumbersRisnotadomain.Considerany

realnumber,thenany"-neighbourhoodmustcontainsome

complexnumbers,i.e.the"-neighbourhooddoesnotlieinthe

realnumbers.

Wecannowdeﬁnethebasicobjectofstudy.

Deﬁnition1.5

LetDbeadomaininC.A complexfunction ,denotedf:D!

C,isamapwhichassignstoeachzinDanelementofC,this

valueisdenotedf(z).

CommonError1.6

Notethatfisthefunctionandf(z)isthevalueofthefunction

atz.Itiswrongtosayf(z)isafunction,butsometimespeople

do.

Examples1.7

(i)Letf(z)=z 2 forallz2C.

(ii)Letf(z)=|z|forallz2C.Notethatherewehavea

complexfunctionforwhicheveryvalueisreal.

(iii)Letf(z)=3z 4 −(5−2i)z 2 +z−7forallz2C.Allcomplex

polynomialsgivecomplexfunctions.

(iv)Letf(z)=1/zforallz2C\{0}.Thisfunctioncannotbe

extendedtoallofC.

Remark1.8

Functionssuchassinxforxrealarenotcomplexfunctions

sincethereallineinCisnotadomain.Laterweseehowto

extendtheconceptofthesinesothatitiscomplexfunction

onthewholeofthecomplexplane.

Obviously,iffandgarecomplexfunctions,thenf+g,

f−g,andfgarefunctionsgivenby(f+g)(z)=f(z)+g(z),

(f−g)(z)=f(z)−g(z),and(fg)(z)=f(z)g(z),respectively.

Wecanalsodeﬁne(f/g)(z)=f(z)/g(z)providedthatg(z)6=0

onD.Thuswecanbuilduplotsofnewfunctionsbythese

elementaryoperations.

Theaimofcomplexanalysis

Wewishtostudycomplexfunctions.Canwedeﬁnedifferenti-

ation?Canweintegrate?WhichtheoremsfromRealAnalysis

canbeextendedtocomplexanalysis?Forexample,istherea

versionofthemeanvaluetheorem?Complexanalysisises-

sentiallytheattempttoanswerthesequestions.Thetheory

willbebuiltuponrealanalysisbutinmanywaysitiseasier

thanrealanalysis.Forexampleifacomplexfunctionisdif-

ferentiable(deﬁnedlater),thenitsderivativeisalsodifferen-

tiable.Thisisnottrueforrealfunctions.(Doyouknowanex-

ampleofadifferentiablerealfunctionwithnon-differentiable

derivative?)

Realandimaginarypartsoffunctions

Wewilloftenuseztodenoteacomplexnumberandwewill

havez=x+iywherexandyarebothreal.Thevaluef(z)

isacomplexnumberandsohasarealandimaginarypart.

Weoftenuseutodenotetherealpartandvtodenotethe

imaginarypart.Notethatuandvarefunctionsofz.

Weoftenwritef(x+iy)=u(x,y)+iv(x,y).Notethatuis

afunctionoftworealvariables,xandy.I.e.u:R

2 !R.

Similarlyforv.

Examples1.9

(i)Letf(z)=z 2 .Then,f(x+iy)=(x+iy) 2 =x 2 −y 2 +2ixy.So,

u(x,y)=x 2 −y 2 andv(x,y)=xy.

(ii)L etf(z) =|z|.Then,f(x+iy)= p x 2 +y 2 .So,u(x,y)=

Exercises1.10

Finduandvforthefollowing:

(i)f(z)=1/zforz2C\{0}.

(ii)f(z)=z 3 .

Visualisingcomplexfunctions

InRealAnalysiswecoulddrawthegraphofafunction.We

haveanaxisforthevariableandanaxisforthevalue,andso

wecandrawthegraphofthefunctiononapieceofpaper.

Forcomplexfunctionswehaveacomplexvariable(that’s

tworealvariables)andthevalue(anothertworealvariables),

soifwewanttodrawagraphwewillneed2+2=4real

variables,i.e.wewillhavetoworkin4-dimensionalspace.

Nowobviouslythisisabittrickybecauseweareusedto3

spacedimensionsandﬁndvisualising4dimensionalspace

veryhard.

Thus,itisverydifﬁculttovisualisecomplexfunctions.How-

ever,therearesomemethodsavailable:

(i)Wecandrawtwocomplexplanes,oneforthedomainand

onefortherange.

p x 2 +y 2 andv(x,y)=0.

(ii)Thetwo-variablefunctionsuandvcanbevisualisedsep-

arately.Thegraphofafunctionoftwovariablesisasur-

faceinthreespace.

u(x,y)=cosx+sinyandv(x,y)=x 2 −y 2

(iii)Makeoneofthevariablestimeandviewthegraphas

somethingthatevolvesovertime.Thisisnotveryhelpful.

Deﬁning e z , cosz and sinz

Firstwewilltryanddeﬁnesomeelementarycomplexfunc-

tionstoplaywith.Howshallwedeﬁnefunctionssuchase z ,

coszandsinz?Werequirethattheirdeﬁnitionshouldcoincide

withtherealversionwhenzisarealnumber,andwewould

likethemtohavepropertiessimilartotherealversionsofthe

functions,e.g.sin 2 z+cos 2 z=1wouldbenice.However,sine

andcosinearedeﬁnedusingtrigonometryandsoarehardto

generalise:forexample,whatdoesitmeanforatriangleto

haveanhypotenuseoflength2+3i?Theexponentialisde-

ﬁnedusingdifferentialcalculusandwehavenotyetdeﬁned

differentiationofcomplexfunctions.

However,weknowfromRealAnalysisthatthefunctions

canbedescribedusingapowerseries,e.g.,

sinx=x− x 3

(−1) n x 2n+1

5! −···=

(2n+1)! .

n=0

Thus,forz2C,weshalldeﬁnetheexponential,sineand

cosineofzasfollows:

z n

n! ,

e z :=

n=0

(−1) n z 2n+1

sinz:=

(2n+1)! ,

n=0

(−1) n z 2n

cosz:=

(2n)! .

n=0

3! + x 5

Thus,

n! =1+(3+2i)+ (3+2i) 2

2! + (3+2i) 3

e 3+2i =

3! +...

n=0

Thesedeﬁnitionsobviouslysatisfytherequirementthatthey

coincidewiththedeﬁnitionsweknowandloveforrealz,but

howcanwebesurethattheseriesconverges?I.e.whenwe

putinaz,suchas3+2i,intothedeﬁnition,doesacomplex

numbercomesout?

Toanswerthiswewillhavetostudycomplexseriesand

asthetheoryofrealserieswasbuiltonthetheoryofreal

sequenceswehadbetterstartwithcomplexsequences.

ComplexSequences

Thedeﬁnitionofconvergenceofacomplexsequenceisthe

sameasthatforconvergenceofarealsequence.

Deﬁnition1.11

Acomplexsequencehc n i converges toc2C,ifgivenany">0,

thenthereexistsNsuchthat|c n −c|<"forallnN.

Wewritec n !corlim n!1 c n =c.

Example1.12

Thesequencec n =

4−3i

convergestozero.

Consider

r 25

! n

4−3i

n =

4−3i

|c n −0|=|c n |=

|c n −0|<"()(5/7) n <"

()nlog(5/7)<log"

()n> log"

log(5/7) .

So,givenany"wecanchooseNtobeanynaturalnumber

greaterthanlog"/log(5/7).Thusthesequenceconvergesto

zero.

Remark1.13

Noticethata n =|c n −c|isarealsequence,andthatc n !cif

andonlyiftherealsequence|c n −c|!0.Hence,wearesaying

somethingaboutacomplexsequenceusingrealanalysis.

(3+2i) n

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