Complex Analysis 1992

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UPSCCivilServicesMain1992-Mathematics


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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhMarch1,2010Question1(a)Ifu=ex(xsinyycosy),findvsuchthatf(z)=u+ivisanalytic.Alsofindf(z)explicitlyasafunctionofz.Solution.See1993,question2(b).Question1(b)Letf(z)beanalyticinsideandonthecircleCdefinedby|z|=RandletreibeanypointinsideC.Provethatf(rei)=2102

R2(R22Rrcos(r2)f(Rei))+r2dSolution.ByCauchysintegralformulaf(z)=f(rei)=2i1CR

f()zd(1)Wenotethatthefunction:||=r

f()R2

zhasnosingularitywithinandonCR

,becausef()isanalyticwithinandonCR

and(R2z


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)1isalsoanalyticwithinandonCR

asR2z

lies

outsideCRandthus|R2z

|>R.thereforeR2z

ThusbyCauchys=0(NotethatR2=RR>R|z|,because|z|=r


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Using(1),(2)weget


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f(z)=[1z]

d=12if()||=r

[1R2z

zR2z

(z)(R2z

]d=1

2i||=r

f())]d

f(rei)=

12i

||=r

f()[([zzz)(zR2R2)


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r2R2(Reirei)(rRei()R2)]Reiid=

12

2i0

f(Rei)[r2R2(Rrei())(rei()R)]

d=12

20

f(Rei)[r2R2

R2r2+rR(ei()+ei())]d=12

20

f(Rei)

12

[20

R2r2


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R2+r2+2rRcos(]f(Rei))dasrequired.

Question1(c)Provethatalltherootsofz75z3+12=0liebetweenthecircles|z|=1and|z|=2.Solution.See2006question2(b).Question2(a)Findtheregionofconvergenceoftheserieswhosen-thtermis(1)n1z2n1(2n1)!.Solution.Clearly

CoefficientCoefficientoftheofthe(n+n-th1)-thterm

term

=(2n(2n+1)!1)!0asnThusseriesn

lim

|Coefficientofthen-thterm|

n1

=0.Sotheradiusofconvergenceofthepowern=1

(1)n1z2n1(2n1)!is,i.e.theregionofconvergenceistheentirecomplexplane.2


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1


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Question2(b)Expandf(z)=(z+1)(z+3)inaLaurentseriesvalidfor(i)|z|>3,(ii)1

Complex Analysis 1992

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