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UPSCCivilServicesMain1986-Mathematics

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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhJuly19,2010Question1(a)Letf(z)besinglevaluedandanalyticwithinandonasimpleclosedcurveC.Ifz0

isanypointintheinteriorofC,thenshowthatf(z0

1

2iC

f(z)dzzz0

wheretheintegralintakeninthepositivesensearoundC.Solution.ThisisknownastheCauchyintegralformula.Weshallshowthatgiven>0

)=1

2iC

f(z)dzzz

0f(z0

)

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0asabove,thereexistsa>0suchthat|zz0

|=|f(z)f(z0

)|0sosmallthatthedisc|zz0

|

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02

ieieid

=2i1

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Therefore

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1f(z)dz

2iC

zz0

1[f(z)dz2iC

zz0

f(z0

)=f(z0

)

zdzz0

]

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=

1

2i

f(z)f(z0

)zz0

dz

Thus2i1C

12|dz|=

2lengthof=f(z)dzzz0

f(z

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0

)=0andtheproofiscomplete.Question1(b)Bythecontourintegrationmethodshowthat1.

dx0

x4+a4=4a22wherea>0.

2.0

sinxxdx=2Solution.

1.Wetakef(z)consistingof=az4+a4semicircle1andthecontourCofradiusRwithcenter(0,0)lyingintheupperhalfplane,andthelineABjoining(R,0)and(R,0).Cispositivelyoriented.A(R,0)(0,0)B(R,0)(a)Polesoff(z)aregivenbyz=aei4

=aei

4

,z=aei

4

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a[cos4

areintheupperhalfplane.isin4

],outofwhichz=Residueatz=aei

4

is14a3e3i

4

[

]1

.Residueatz=aei4

=4a31

1[2+i2]1

.Sumofresiduesis2is14a3e3i

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4

=4a31

12i24a3[i1

1+1+1i]=i4a32

.ThusR

limdzC

z4+a42ii2

224a34a3

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2a3(b)

===

z4dz+a4

0

RieiR4a4

RR4a4becauseon|z4+a4||z4|a4=R4a4.ThusRlim

z4dz

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+a4=0.2

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(c)RlimAB

z4dz

+a4=

x4dx+a4=20

x4dx+a4.Using(a),(b)and(c)wegetR

limdzC

z4+a40

dxx4+a42a3Thus

0=2=24a3asrequired.

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2.Wetakef(z)=eizz

x4dx+a4=

22a3=andthecontourCcon-sistingofthelineABjoining(R,0)to(r,0),thesemicircleofradiusrwithcenter(0,0),thelineCDjoining(r,0)to(R,0)andasemicircleofradiusRwithcenter(0,0).Thecontourliesintheup-perhalfplaneandisorientedanticlock-wise.Wetookaspartofthecontourtoavoidthepoleat(0,0).A(R,0)B(r,0)C(r,0)D(R,0)WewilleventuallymakeRandr0.

(a)Sincetheintegrandeizzhasnopolesintheupperhalfplane,itfollowsthatR,r0

limC

eizzdz=0(b)InordertoprovethatRlim0,weuseJordaninequality,eizzwhich

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dz=statesthatsin2

y=sinfor0comparethegraphsasshowninthe

2

y=2

figure.=2

eizzdz

eR0sin

RRd=20

2

eRsind(sin()=sin)

22

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0

eR2

2R

[1eR]showingthatRlimd=2

eizzdz=0asRlim

1ReR=0.3

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(c)Now0sucheiz

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z

that=1z

|z|+(z)

(lengthof)

limr0

(z)dz=0.

dzz=0

reiidrei=i(d)R,r0

limAB

eizz0

dz=

eix

xdx,R,r0limeizCD

z

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dz=0

eix

xdxUsingtheseresults,weget0=R,r0limC

eizz

0dz=

eixxdxi+0

eixx

dx()

Since0

eixxdx=0

eiyydyweget

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0

eixxdx

0

eixxdx=ior0

eixeix1

2ix20

sinxx2

Notethatin(*)wecannotwritedx=

dx=

eixxdx=iandconcludethat

sinx

xdx=,

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cosxx

dx=0,because

cosx

xdxhasconvergenceproblematx=0.4