complex analysis 1979

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UPSCCivilServicesMain1979-Mathematics


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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhJuly19,2010Question1(a)Ifafunctionf(z)isanalyticandboundedinthewholeplane,showthatf(z)reducestoaconstant.Henceshowthateverypolynomialhasaroot.Solution.See1989,question2(b)forthefirstpart.See1996question2(a)forthesecondpart.Question1(b)Evaluatethefollowingintegralsbythemethodofresidues.1.20

sin2a+bcosd(a>b>0)2.0

(1x

16

+logxx)2dxSolution.1.LetI=20

sin2a+bcosd=120


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2

a1+bcoscos2dLetI

1=12

120

a+bcosd.Putz=eisothat

I1=12|z|=1

dziz(a+b2

=1i|z|=1

dz(z+1z

bz2+2az+b1))


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Theaintegranda2b21

bz2+2az+bhastwosimplepolesatz1

=a+ba2b2,z2

=

poleatbz=z1

.Sincea>b>0,|z2

|>1,but|z1

z2

|=1so|z1

|


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1=2a21

b2.ThusI1

=2i1i2a21b2=I2

+2a=1202

a+cos2bcosd=12Re02

e2id

a+bcos=12Re1i


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2z2dz|z|=1

bz2+2az+b=Re

1i2iResidueofz2bz2+2az+batz=z1

=21b

z1

z21

z2

ThusI1

I2

=b(z1

2z2

)b(z2z211z2

)=2


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a22b2(1z21

)=(1a2b2+(a2b2))b2=a2b2(a22a2a2b2b2)

ThusI=a2b2a2b2aa+2

a2b2.2.Letshown.f(z)r

=isz(1+z)2

a1


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6logcirclezandthecontourCasofradiusrorientedR

clockwise,andR

acircleofradiusRori-entedanticlockwise.ABisalongx-axisonwhichz=x,CDisthelineonwhichz=xe2i.Toavoidthebranchpointofthemultiplevaluedfunctionlogz,we

considerCpositivesideofthex-axis.

Wechoosethebranchoflogzforwhichlogz=log|z|+i,0


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(a)Clearlyf(z)hasadoublepoleatz=1.Residueoff(z)atz=1is


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11![(z+1)2z]atz=1=d1

6

dzlogz[z1

6

(z+1)2z16z56

]atz=1=eilog6z

z+6atz=ei=+logz=5

(6

cos)=logei+66e5i

6

(


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=i+6656

isin5632)=

3+i)(b)OnR

i+6612i

121(6+i)(,z=Rei,|z+1||z|1=R1and|logz|=|logRei|=|logR+i|logR+logR+2as02.Thus

R

(1z1

6

+logzz)2


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dz

2

0

R1

6

(logR(R1)2+2)

Rd=2(RR76

1)2(logR+2)ClearlyRlim[R

76

logR2R7

6

(R1)2(R1)2]+

=0,andthereforeRlimR

(1z1


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lim[r7

6

logr(1r)2+2r7

6

(1r)2]=0,andthereforer0

limr

z1

6

logz(1+z)2dz=0ByCauchysresiduetheorem,using1,2,3,wegetRr0

limC

f(z)dz=0

(1x1

6

+logxx)2dx+0

(xe2i)


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1

6

log(xe2i)

(1+x)2

dxbecauseonAB,z=xandonCD,z=xe2i.Therefore0

0

3+i)

(12x23i

)(101

6

+logxx)2dx

(1x1

6

+logxx)2dxx0


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1

6

ex

2i6

1

6

(1(1(log(12+++x)2x2

x)2+3i)2i2i)dxdx=2i12(6+i)(=

6[(6+3)+i(63)]3


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Equatingrealandimaginaryparts,weget


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120

0

3)(1)x1

6

logx(1+x)23dx+3(1+x1

6

x)2dx=6(6+2

00

3)(2)Multiplying(1)by(1x1

6

+logx


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x)2dx(1+x1

6

x)2dx=6(63andadding

0(1x1

6

+logxx)2dx=6

3+318[312]Thus

0[6+]=6


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21

6

2

3Inaddition,multiplying(2)by(1x+logxx)2dx=23andadding,weget20

x1

6

(1+x)2[3+3+6

3]givingus0

dx=66

x1

6

2(1+x)234


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dx=

complex analysis 1979

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