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UPSCCivilServicesMain1985-Mathematics

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(|z|+|h|)n1+|z|(|z|+|h|)n2+...+|z|n1nn1(1)Sincetheseries

n=1nan

n1isconvergent,given>0N1

>0suchthat

r=n+1

r|ar

|r10suchthat

+1

(zh)nhznnan

zn1N

1

[an

]

n=1

(zh)nznhnan

zn1

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h=2i1

f(z)dzC

(zz0

)2

i.e.f(z)isdifferentiableatz0

andsincez0

isanarbitrarypointofD,itfollowsthat

f(z)=C

f()d(z)2whereCisanypositivelyorientedcirclecontainingzinitsinterior.Weshallnowprovethatf(z0

1

2i)=2i2!f(z)dzC

(zz0

)3

wherez0,Careaschosenabove.Lethbef(z0

+h)hf(z0

)

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2i2!alsochosenasabove.Then

f(z)dzC

(zz0

)3=]dz=1

2ih1C

f(z)[(zz10

h)2(z1z0

)2(z2hz0

)3(zz0

)3(zz0

h)2(zz0

)h)2

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2ihC

(zz0

2h(zz

0dzNow(zz0

f(z)h)2(zz0

)3h)2=(zz0

)3(zz0

h)2(zz0

)2h(zz0

)+h2]=(zz0

)[(zz0

)2(zz0

h)2]2h[(zz0

)22h(zz0

)h[2(zz0

)h]2h(zz0

)2+4h2(zz0

)2h3=h[2(zz0

)2h(zz0

)2(zz

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0

)2+4h(zz0

)2h2]=h2[3(zz

0)2h]Thusweget

f(z0

+h)f(z0

)

h2!f(z)dz2iC

(zz0

M|h|(3+2|h|2)l2d3(d|h|)2whereM,d,areasbefore.Sincetherighthandsideoftheaboveinequalitytendsto0ash0,itfollowsthat

f(z0

)3)=h0lim

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f(z0

+h)hf(z0

)=2i2!f(z)dzC

(zz0

)3

i.e.f(z)isalsoanalyticinD.3

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0

xsinaxx2b2dx.

Solution.Wetakef(z)=z2b2zeiazandthecontourCconsistingofthefollowing1.ThelineABjoiningA=(R,0)andB=(br1

,0).2.1

,thesemicircle(x+b)2+y2intheupperhalfplane.=r21

lying3.LineCDjoiningC=(b+r1

,0)andD=(br2

,0).4.

2,thesemicircle(xb)2+y2intheupperhalfplane.=r22

lying5.LineEFjoiningE=(b+r2

1

2

A(R,0)BCDEF(R,0)EventuallywewillletR,r1

,0)andF=(R,0).6.,thesemicirclex2+y2=R2lyingintheupperhalfplane.

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,r2

0.Nowtheintegrandhasnopoleintheupperhalfplane,thereforeR

rlim1

C

zeiaz(z2dzb2)=01.On,

r

2

00

Question1(c)Evaluatebythemethodofcontourintegrationzeiazdz

(z2b2)

becauseof,|z2b2||z|2b2=R2b2.

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ReieiaRei0

R2b2Rieid

zeiaz

(z2dzb2)

R2R2b20

eaRsind=2R2R2b2

2

eaRsind0

(Wecandoubletheintegralandhalvethelimit,becausesin()=sin).Using

Jordansinequalitysin2

for02

weget

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(z2zeiazdzb2)

2R2

2

2R2R2b20

R2b2(12aR/eaR)=R(1a(R2eaR)

b2)showingthatRlimeaR2

d=

zeiaz(z2dzb2)=0.2.4

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,

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Togetthevalueoftheintegralalong12

weobservethatiff(z)hasasimplepoleatz=aand

rr

isapartofacircleofradiusrwithcentera,thenr0

limr

21

Proof:Letf(z)=)wherea1

f(z)dz=ia1

(2

1

istheresidueoff(z)ata.za1a+a0

+a

1(za)+a2

(za)2+...=za1a

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+(z)

where(z)isanalyticinthe

circler

|za|

r.Thus

)whereM=sup|za|=r

(z)dzMr(2

1

r

(z)dz=0andr0lim|(z)|.Thusr0lima1

dz

2

rr

za

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0=Rr

lim1

Czeiazdz

(z2b2)

xeiaxdx(x2b2)i2

eiab

i2eiabor

=r

2

00

xeiax(x2dxb2)=icos(ab)Takingimaginaryparts,weget

xsinaxdx(x2b2)=cos(ab)or0

xsinaxdx(x2b2)cos(ab)

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25=