complex-analysis-1983.pdf

8/3/2019 complex-analysis-1983.pdf

1/16

This document contains text automatically extracted from a PDF or image file. Formatting may havebeen lost and not all text may have been recognized.

To remove this note, right-click and select "Delete table".


2/16

UPSCCivilServicesMain1983-Mathematics


3/16

ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhJuly19,2010Question1(a)ObtaintheTaylorandLaurentseriesexpansionswhichrepresentthefunc-tion(z+z22)(z1+3)intheregions(i)|z|


4/16

z(z32

)n

(z3)n

=3

2n=0

(1)n83n=0

(1)n1

6+n=1

(1)n[32n+13n+1

8]znistherequiredTaylorseriesvalidin|z|


5/16

2.2


6/16

f(z)=1+)1

=1+3

z(1+2z)1

83

(1+z3)n

=3n=1

2n

zn+1n=1

zn3nThisisvalidin2


7/16

n

83

n=0(1)n(z3(1)n53

83(1)n3z(1+2z

)1z8(1+z3)1

=1+n=0

(1)nzn+1(32n83n)Question1(b)Usethemethodofcontourintegrationtoevaluate


8/16

0

1xa1+x2

dx,0


9/16

1)+isin(a21))

.

1za1+z2dz

0

R2Ra11RdR2Ra1

0asR0


10/16

0

1ra1r2

rdr2ra1

0asr0a>0

2


11/16

Hereweuse|1+z2|1|z|2=1r2.Thus


12/16

Rr0

limC

0

f(xei)(dx)+0

f(x)dx=0

xa1ei(a1)1+x20

1xa1+x2dx=dx+

0

f(z)dz==2i12ixa11+(x2

cos(1(a+2cos(a1)1)+isin(a1))dx


13/16

(a1))2Equatingtherealpartsonbothsides,(

1+cos(a1)+isin)

xa10

1+x2(a1)2

or0

dx=cosxa11+x2(a1)2Equatingtheimaginarypartsalsogivesusthesameanswer.Alternatesolution:In1984,question1(b),weobtained

2sin2dx=seca

ta1logt

ta12

20

1+t20

1+t22a


14/16

2sinadt+sinadt=cos

0

ta1logt

1+t20

ta11+t2

22a2Multiplyingthefirstbycosa2

dtcosadt=sinandthesecondbysina

2andaddinggivesus(sinacosa2

=cosasin

sin(a

aa22)


15/16

ta10

1+t2(

dt=22cos2a2+sin2a2))

ta10

1+t222

=

dt=0

1ta1+t2dt=12sina2

2ascalculatedbefore.Note:Inthissolutiontheadvantageisthatweavoidtheuseofthemultiplevaluedfunctionlogz,howeveritismuchlonger.3=


16/16

2cos(

2

a2)=2sec(a1)

complex-analysis-1983.pdf

Documents