Complex Analysis 1998

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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhFebruary20,2010Question1(a)Showthatthefunctionf(z)=

x3(1+i)y3(1x2+y2i),z=00,z=0iscontinuousandC-Rconditionsaresatisfiedatz=0,butf(z)doesnotexistatz=0.Solution.Letf(z)=u+iv,thenu=x3x2+y2y3,v=x3+y3x2+y2forz=0,andu(0,0)=

v(0,0)=0.ux(0,0)=h0limu(h,0)hu(0,0)=h0limh3

h2h0=1uy(0,0)=k0


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limu(0,k)ku(0,0)=k0limk3k2

k0=1xv(0,0)=h0limv(h,0)hv(0,0)=h0limh3h2

h0=1v

y(0,0)=k0limv(0,k)kv(0,0)=k0limk3k2

k

0=1Thusux=v


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y,uy=

xvat(0,0),i.e.theCauchyRiemannequationsaresatisfiedat(0,0).f(z)isclearlycontinuousatz=0,because|u(x,y)u(0,0)|=

|v(x,y)v(0,0)|2x3x2x2+y2y3+y2=r3(cos3r2sin3)

2x2+y2


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1


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Thusu,varecontinuousat(0,0),sof(z)iscontinuousat(0,0).


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Iff(z)istobedifferentiableat0,thenz0

limf(z)z0

=x0,y0lim(x3y3)+i(x3(x2+y2)(x++iy)y3)=x0,y0lim(x3+iy3)(1+i)(xiy)(x2+y2)2shouldButexistifweandtakeittheshouldlimitbealongequaly=to

x,uxthen(0,0)+ixv(0,0)=1+i.z0

limf(z)z

0=x0lim(x3+ix3)(1(2x2)2+i)(xix)=1+2i

Thereforef(z)isnotdifferentiableatz=0.Question1(b)FindtheLaurentexpansionof

(z+1)(zz+2)aboutthesingularityz=2.Specifytheregionofconvergenceandnatureofsingularityatz=2.Solution.Clearlyf(z)=


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z(z+1)(z+2)2z+21

z+12z+211(z+2)===

+z+22+n=0

(z+2)nfor|z+2|


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n=0

(xnn!

)2

=122

e2xcosd0

Solution.ItiseasilydeduciblefromCauchysIntegralformulathatiff(z)isanalytic

withinandonasimpleclosedcontourCandz0isapointintheinteriorofC,thenf(n)(z0

)=n!2i

f(z)C

(zz0

,dz2)n+1


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Letf(z)=exz(HerexisnotRexbutaparameter),thenf(z)isanentirefunctionand


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thereforef(n)(0)=xn=2in!

C

zn+1exz,dzwhereCisanyclosedcontourcontaining0initsinterior.Hence(xnn!)2

=xn(n!)2n!

2iC

exzzn+1

,dz=2i1

xnexzC

n!zn+1dzasrequired.WetakeCtobetheunitcircleforconvenience.Then

n=0

(xnn!)2


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=12i

n=0C

xnexzn!zn+1dz=2i1

xnexzCn=0

n!zn+1dzInterchangeofsummationandintegralisjustified.Thusn=0

(xnn!

)2

=12i|z|=1

exzz

n=0(xz

n!)ndz=1


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2i|z|=1

exzz

exz

dzPutz=eisothatdz=ieidandn=0

(xnn!

)2=2i12

ex(ei+ei)0

eiieid=

212

e2xcosd0

asrequired.Question2(a)Provethatallrootsofz75z3+12=0liebetweenthecircles|z|=1and|z|=2.Solution.See2006question2(b).

Question2(b)Byintegratingaroundasuitablecontourshowthat0

xsinmxx4+a44b2


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embsinmbwhereb=a2

dx=.3


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Solution.


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Letf(z)=tegralz4zeimz+a4.sisting

off(z)dztheWeconsiderthein-whereisthecontourcon-linejoining(R,0)and(R,0)and,thearcofthecircleofradiusRandcenter(0,0)lyingintheupperhalfplane.(R,0)(0,0)

(R,0)

=

0

ReieimR(cosz4+a4+isin)Rieid

R4R2a4because|z4+a4||z|4|a4|=R4a4on,andemRsin1assin>0for0


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Thusf(z)dz

f(z)dz0asRandR

lim

x4xeimx+a4dxButbyCauchysresiduetheoremf(z)dz=

f(z)dz=2i(thesumoftheresiduesofpolesoff(z)inside).Thepolesoff(z)aresimplepolesatae3i4

areinside.Residueatz=aei4

,ae3i4

,outofwhichaei

4

,aei

4

isaei

4

eimae4a3e3i


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4i

4

.Residueatz=ae3i

4isae3i4

4a3eeimae9i43i4

.]Sumofresidues==i4a2[eima(cos4+isin4

)+eima(cos34+isin34)

]=i4a2

[ema

2(i1)

+ema


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2(i1)

4a2ma

2

(2isin)=ma

2

2a2ma

2Thus

iema2

esinxeimxma

2

x4+a42a2ma

2Takingimaginarypartsofbothsides,weget

dx=2ie


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sinxsinmx

xsinmx

ema

2

x4+a40

x4+a4a2ma

2emb2b2sinmbwhereb=a2

dx=2dx=sin

=.Thus0

xsinmxx4+a4dx=emb4b2sinmbasrequired.4


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20

d32cos+sin.

Solution.Weputz=ei,sothatd=dziz,cos=12

(z+1z

),sin=2i1

(z1z

).Thus

I=20

d32cos+sin=

|z|=1

dz

iz[3(z+1z)+2i1

(z1z

)]=2

|z|=1

dz

6iz2iz22i+z21=2Question2(c)Usingtheresiduetheoremevaluate

dz|z|=1

(12i)(z+i


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12i

)(z+12i5i

)Clearly(6iz2iz22i+z21)1hastwosimplepoles12i

iliesinside|z|=1.Theresidueatthispoleislimandz+i12i5i

ofwhichonly12ii12iz

byCauchysresiduetheorem(12i)(z+i12i

)(z+12i5i

)=4i1.ThusI=i

12i

20

d32cos+sin=22i14i=5

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Complex Analysis 1998