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DepartmentofMathematicsandStatisticsIndianInstituteofTechnologyKanpur

MSO202A/MSO202Assignment3SolutionsIntroductionToComplexAnalysis

The problems marked (T) need an explicit discussion in the tutorial class. Other problems areforenhancedpractice.1. (a) ,

C

z z dz whereCisthecounterclockwiseorientedsemicircularpartof thecircle 2z lyinginthesecondandthirdquadrants.(T)(b) ,

C

zz dzz where C is the clockwise oriented boundary of the part of the annulus

2 4z lyinginthethirdandfourthquadrants.(c) ( 2 ) ,n

C

z a dz whereCisthesemicircle 2 , 0 arg( 2 )z a R z a .Solution:(a) A parametric equation of C is 2 iz e , 3

2 2 . Therefore the given integral is

3 /2

/2

8 8C

z z dz i d i

.

(b) C

zz dzz AB BEC CD DFA

z z z zz dz z dz z dz z dzz z z z

2 2 4

4 2 0

2 42 2 4 42 4

i ii i

i ie ex dx ie d x dx ie de e

8 326 6 4

3 3

(c) ( 2 )nC

z a dz = 1 ( 1)0 0

( )n in i n i nR e iR e d iR e d i

11[( 1) 1], 1 0.

1

nnR for n

n

Therighthandintegralin(i)isobviously i forn+1=0.

F

A B C D

22. Evaluatetheintegral 1

C

dzz ,ineachofthefollowingcases:

(a) Cisthecounterclockwiseorientedsemicircularpartofthecircle 1z intheupperhalfplaneand z isdefinedsothat 1 1. (T)(b)Cisthecounterclockwiseorientedsemicircularpartofthecircle 1z inthelowerhalfplaneand z isdefinedsothat 1 1.

(c)Cistheclockwiseorientedcircle 1z and z isdefinedsothat 1 1. (d)Cisthecounterclockwiseorientedcircle 1z and z isdefinedsothat 1 .i Solution:(a) The two distinct values of z , iz r e , are given by ln ( 2 )exp( ), 0,1.

2 2z nz i n

1 1 0.n TheparametricequationofCis ,0itz e t / 2

0( / 2)

0

[ ]1 1 2( 1)/ 2

itit

i tC

edz i e dt i ie iz

.(b) 1 1 1.n TheparametricequationofCis , 2itz e t

2 /2 2

( /2)[ ]1 1 2(1 ).

/ 2

itit

i tC

edz i e dt i iiez

(c) 1 1 0.n TheparametricequationofCis , 2 0.itz e t

0 / 2 02

( / 2)2

[ ]1 1 4.( / 2)

itit

i tC

edz i e dt ie iz

(d) 1 0.i n TheparametricequationofCis , 0 2 .itz e t

2 /2 20

( /2)0

[ ]1 1 4.( / 2)

itit

i tC

edz i e dt iiez

3. Withoutactuallyevaluatingtheintegral,provethat

(T)(a) 21 ,1 3Cdz

z whereC is the arcof the circle 2z fromz=2 to z=2i lying in the first

quadrant. (b) 2 2( 1) ( 1),

C

z dz R R whereCisthesemicircleofradiusR>1withcenterattheorigin.Solution:

2 21 1( ) , ,

1 2 1a Length of C and on C

z 2 2

1 1 .1 2 1 3Cdz

z (byMLEstimate)

(b) Length ofC = R , and onC, 2 21 1z R . Consequently, the desired inequality follows byMLestimate

34. (T)DoesCauchyTheoremholdseparatelyfortherealorimaginarypartofananalyticfunctionf(z)?

Whyorwhynot?Solution:No,CauchyTheoremneednotholdseparatelyforrealorimaginarypartofananalyticfunction.Consider,forexamplef(z)=zandC: 1.z Then,

2

1 0

2

1 0

Re cos

Im sin

i

z

i

z

z dz i e d i

z dz i e d

5. Aboutthepointz=0,determinetheTaylorseriesforeachofthefollowingfunctions:

(T) 2( ) 2 ( ) ( 3 2)i z i ii Log z z Solution:(i) 2z i = 1/22 (1 )

2zii

2/ 2 1/ 2

1 1 11 2 22 ( ) [12 2 2 2

1 1 11 .... 12 2 2.... .....]

2

i

n

z zei i

nz

n i

12

1 1.3.5.....(2 3)(1 )[1 ( 1)2 2 22

nn

nn

z n zii in

.(ii)Log(z23z+2)=Log(1z)+Log(2z)+2m i ,forsomem. =Log(1z)+Log2+log(1

2z )+2m i ,forsomem

=Log2+2m i 1

n

n

zn

1

/ 2 n

n

zn

ThereforetheTaylorseriesofLog(z23z+2)isgivenby log(z23z+2)=log2+2m i

1

1(1 )2

n

nn

zn

46. Abouttheindicatedpointz=z0,determinetheTaylorseriesanditsregionofconvergenceforeachof

thefollowingfunctions.Ineachcase,doestheTaylorseriesnecessarilysumsuptothefunctionateverypointofitsregionofconvergence?

01( ) , 11i zz (T) 0( ) cosh ,ii z z i (T) 0( ) , 1iii Log z z i Solution:(i)

11 1 111 2 2

zz

= 01 1( 1)2 2

nn

n

z

.TheregionofconvergenceofitsTaylorseriesis1 2.z TheTaylorseriessumsupto 1

1 z ateverypointof 1 2z ,since1

1 z isanalyticateverypointof 1 2z .

(ii)coshz=cosh(z i+ i)=cos(i(z i) )=cos(i(z i)).Therefore,

2

0

( )cosh2

n

n

z izn

.TheregionofconvergenceofTaylorseriesiswholecomplexplane.The

Taylorseriessumsuptocoshzateverypointofthecomplexplane,sincecoshzisanalyticateverypointofthecomplexplane.(iii) 1

1

1( 1) , 1, 2,......( 1 )

nn

n nz i

d nLog z for ndz i

1

(1 )( 1 ) ( 1 )2

nn

nn

iLog z Log i z in

.

TheregionofconvergenceoftheTaylorseriesonRHSis 1 2z i .However,theTaylorseriesdoesnotsumuptoLogzateverypointof 1 2z i , sinceLogzisnotanalyticonnegativerealaxis,apartofwhich iscontained in thedisk 1 2z i ,while thesumfunctionrepresentedby theaboveTaylorseriesisanalyticateachpointofthisdisk.

57. (T)Evaluate the integral 2( 1)C

dzz z , for all possible choices of the contour C that does not pass

throughanyofthepointsz=0, i .Solution:LetcurveC beorientedcounterclockwiseinthefollowingcases.ForclockwiseorientedC thevalueoftheintegralwillbenegativeofthevalueobtainedinthesecases.Case1(Cdoesnotencloseanyofthepoints0, i ):Inthiscase,I= 2( 1)C

dzz z =0,byCauchyTheorem.

Case2WhenCenclosesonlythepoint0,I= 2(1/ 1 )

C

zdz

z

=2 i ,byCauchyIntegralFormula.Similarly,whenCenclosesonlythepointi,I= (1/ )

C

z z idz

z i

= 12 2i ii i andwhenCenclosesonlythepointi,I= (1/ )

C

z z idz

z i

= 12 2i ii i .Case3:WhenCenclosesonlythepoints0,i,

1 2

2

1 2(1/( 1)) (1/ ( )) ,

0 .C C

z z z iI dz dz where C and C are sufficientlyz z i

small circles around and i respectively

12 1 2

2i i i

i i ,byCauchyIntegralFormula.

Othercases,i.e.whenCenclosesonlythepoints0,iori,iaretreatedsimilarly.Case4(Cenclosesallofthepoints0,i,i):Inthiscase

1 2 3

2

1 2 3

(1/( 1)) (1/ ( )) (1/ ( )) ,

,0 , .

C C C

z z z i z z iI dz dz dzz z i z i

where C C and C are sufficiently small circles aroundi and i respectively

1 12 1 2 2 02 2

i i ii i i i

.

6

8. (T)UseCauchyTheorem formultiplyconnecteddomainsandCauchy IntegralFormula toevaluatetheintegral

2cos

( 1) 2C

z dzz z

,C:thecircle 3z orientedcounterclockwise.

Solution:LetC1,C2bethecounterclockwiseorientedcirclesofsufficientlysmallradiuscenteredat1and2respectively.TheintegrandisananalyticfunctionintheregionlyingbetweenthecirclesCandC1,C2Therefore,byCauchyTheoremformultiplyconnecteddomains,

2cos

( 1) 2C

z dzz z

=

1 2

2 2(cos / ( 2)) (cos / ( 1))( 1) ( 2)C C

z z z zdz dzz z

2 2

1 2cos cos2 ( ) 2 ( ) 4

2 1z zz zi i i

z z .(byCauchyIntegralFormula)

9. Evaluate

(T)(a)2

4( 1)

z

C

e dzz z ,C:thecircle 2z orientedclockwise

(b) 2sin

( ) nC

z dzz ,C:thecircle 1z orientedcounterclockwise

Solution:(a) UsingCauchyTheoremformultiplyconnecteddomains,

1 2

2 4 2

41

1 2

( / ( 1) ) ( / ) ,( 1)

, 1 .

z z

z z

e z e zGiven Integral dz dzz z

where z z are sufficiently small clockwise oriented circles

2 3 2

0 14 3

2[2 ( ) { ( )} ]( 1) 3

z z

z ze i d eiz dz z

(byCauchyIntegralFormulafornthderivatives)

(b) GivenIntegral=2 1

2 12 2( sin ) ( 1)

2 1 2 1

nn

zni d iz

n ndz

10. Ifuisaharmonicfunctionin 0 ,z R and r R showthat

2

0

1(0) ( ) .2

iu u re d

Solution:Let vbe theharmonic conjugateofu in z R so that f(z)=u+ i v is analytic in z R . ByCauchyIntegralFormula

2

0

1 ( ) 1(0) ( ) .2 2

i

z r

f zf dz f re i di z i

Nowtakingtherealpartonboththesidesgivesthedesiredresult.