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    UPSCCivilServicesMain2002-Mathematics

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    ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

    ChandigarhJanuary30,2010Question1(a)SupposethatfandgaretwoanalyticfunctionsonthesetCofallcomplex

    numberswithf(n1

    )=g(n1

    )forn=1,2,3,...,thenshowthatf(z)=g(z)forallzC.

    Solution.LetG(z)=G(z)LetG(z)0for=

    z

    Cwhichf(z)wouldg(z),provethentheresult.

    G(n1

    )=0forn=1,2,....WeshallshowthatR,clearlyR>0.n=0

    Weashallnznbenowtheprovepowerthatseriesan

    of=G(z)0forwitheverycentern.0andradiusofconvergenceIfan

    =0forsomen,letak

    bethefirstnon-zerocoefficient.Then

    G(z)=zk(ak

    z+...)=zkH(z)

    ClearlyH(z)isanalyticin|z|

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    H(0)|

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    )=0asz1

    K0

    K

    1,andK0

    K1

    containsasequenceofpointsyn

    suchthatyn

    z1

    andG(yn)=0,wecanproveasbeforethatG(z)0inK1

    .Proceedinginthisway,innstepswegetG(z)0inKn

    ,orG(z)=0.SincezisanarbitrarypointofC,weget

    G(z)0inC.

    1

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    z

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    Question2(a)Showthatwhen0

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    whichisthedesiredLaurentseriesexpansion.Question2(b)Establishbycontourintegration0

    3cos(ax)x2+12ea,wherea0Solution.LetIbethegivenintegral.Putax=t,sothatI=dx=

    costdt0

    t2

    a2

    a0

    t2cost+a2dtWeshallnowprovethat+1=a0

    cost

    t2+a22aea,whichwillshowthatI=2

    eaasrequired.Clearly

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    dt=Weconsiderx2cosx+thea2integralisthereal

    partf(z)dzofx2whereeix+a2.f(z)ingof=thez2eizline+a2

    joiningand(R,0)isthecontourand(R,0)consist-and,whichisthearcofthecircleofradiusRandcenter(0,0)lyingintheupperhalfplane.(R,0)(0,0)(R,0)2

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    Clearlyon,ifweputz=Rei,then0and

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    z2eiz+dza2

    =

    0

    RieieiReiR2e2i+a2d

    0

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    R2e2iRieieiRei+a2dBut|eiRei|=|eiRcoseRsin|=eRsin1assin0for0.|z2+a2||z|2a2=

    R2a2.Therefore

    z2eiz+dza2

    0

    RR2a2RR2a2

    Henced=NowButthez2eizz2only+eizdz+a2

    dza2pole=02i(suminastheRupper.ofresidueshalfplaneatpolesisz=insideia,(a).>0)andtheresidueatz=iaisei(ia)2ia

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    =ea2ia.ThusR

    lim

    eizdz

    z2+a2

    eixdxx2+a2

    ea2iaeaa

    =

    ==2i=

    cosxdxx2+a2=eaa

    ,

    sinxdxx2+a2=0

    =

    0

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    cosxdxx2+a2=ea2a

    cosx=cos(x)

    Thiscompletestheproof.3