complex analysis 1994

8/3/2019 Complex Analysis 1994

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UPSCCivilServicesMain1994-Mathematics


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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhMarch1,2010Question1(a)SupposethatzisthepositionvectorofaparticlemovingontheellipseC:z=acost+ibsintwhere,a,barepositiveconstants,a>bandtistime.Determinewhere1.thevelocityhasthegreatestmagnitude.2.theaccelerationhastheleastmagnitude.Solution.See1996,question1(a).Question1(b)Howmanyzeroesdoesthepolynomialp(z)=z4+2z3+3z+4possess(i)inthefirstquadrant,(ii)inthefourthquadrant.Solution.1.p(1)=0.p(2)=20,thereforetheintermediatevaluetheoremshowsthatthereexistsx,3


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B(0,R)


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WenowconsiderthecontourOABOwhereOAisstraightlinejoining(0,0)and(R,0),ABisthearcofthecirclex2+y2=R2inthefirstquadrant,andBOisthelinejoin-ing(0,R)to(0,0).O(0,0)

A(R,0)BytheArgumentPrinciple,thenumberofzerosof(thechangeintheargumentofp(z)whenzmovesalongp(z)inthecontourthefirstOABOquadrantoriented=anti-21

clockwiseasR).

ChangeintheargumentalongOA:OnOA,p(z)=x4+2x3+3x+4>0argp(z)=0

foreveryxonOA.ThereforeaszmovesfromOtoA,thechangeintheargumentofp(z)i.e.OA

argp(z)=0.ChangeintheargumentalongBO:OnBO,z=iyandp(z)=y4+4+i(3y2y3).Thereforeargp(z)=tan1).BO

(3y2y3

y4+4argp(z)=tan1(3yy4+2y34)]0

=00=0

ChangeinargumentalongAB:OnarcAB,z=Rei,02

,sothat]p(z)=R4e4i+2R3e3i+3Rei+4=R4e4iR4e4iasR.Thus


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AB

[1+Rei2

+R3e3i3+R4e4i4argp(z)]HenceOABO

argp(z)=2=as4

R0

2

=2.1,sop(z)hasexactlyonezerointhefirstquadrant.

Sincep(z)isapolynomialwithrealcoefficients,itfollowsthatifisazeroofp(z)and

itliesinthefirstquadrant,thenisalsoazeroofp(z)anditliesinthefourthquadrant.Thusp(z)hasonezeroineachofthefirstandthefourthquadrants.Question1(c)Testforuniformconvergenceintheregion|z|1theseriesn=1

cosnzn3Solution.Bydefinitioncosnz=enyeinx+enyeinx21Alternately,p(z)=z4

einz+2einz=()

1+z2

+z33


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andtherefore


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n=1

cosnzn3=

n=1

enyeinx2n3+n=1

enyeinx2n3Case1:y>0.

n=1

enyeinx2n3

n=1

2n31showingthatthefirsttermisabsolutelyconvergent.Butthesecondtermisnotconvergent,becauseitsn-thterm

enyeinx


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2n3

0asninfact

enyeinx

2n3

asnwheny>0.Thereforen=1

cosnz

n3isnotevenconvergentwheny>0.Case2:y


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M-test,that|fn

which(z)|


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1.Thefunctione2zisanalyticeverywhereinthecomplexplane.TheTaylorseriesofe2zwithcenterz=1isgivenbye2z=dne2z

dznn=0

atz=1n!n=0

2ne2n!(z1)n3

(z1)n=


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(z1)n3


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=(ze21)3+(z2e2

1)2+2!(z4e21)+n=0

(n2n+3e2+3)!(z1)nwhichistherequiredLaurentseriesof(ze2z1)3withcenterz=1.Itisvalidinthering1


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,f(3)=332,...,f(n)(3)=n=0

(1)n(n3n+2n!+1)!(z3)n=n=0

(1)n(n3n+2+1)(z3)nThus

1z2(z3)2==n=0

(1)n(n3n+2+1)(z3)n2=32(z1

3)233(z23)+m=0

(1)m(m3m+4+3)(z3)mistherequiredLaurentseriesof1z2(z3)2

withcenterz=3validin0


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1!1dzdpoles.

((zsin2n)2ezz).Nowz=n

ddz((zn)2ezsin2z)=sin2z[(zn)2ez+2(zn)ez](zn)2ez2sinzcoszsin4z=ez(zsin3zn)(

(zn)sinz+2sinz2(zn)cosz)4

because

dne2zdzn=2ne2z.Thus

e2z(z1)3=e2(z1)3

+


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2e2(z1)2

+

4e22!(z1)

+

n=3

2ne2n!


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1


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Usingd((zznlimn)2ez

zsinzn)cosn=(1)n,wegetdz=sin2zz=n

=enznlim(zn)sin3z((zn)sinz+2sinz2(zn)cosz)=en(1)nznlim(zn)(sinz2cosz)sin2z

+2sinz=en(1)nznlimsinz2cosz+(z2sinzn)(coszcosz+2sinz)+2cosz=enznlimsinz+(zn)(cosz2sinz+2sinz)==enenznlimcosz+cosz+2sinz+2cosz(zn)(sinz+2cosz)Thustheresidueatz=nofezcsc2zisen.Question2(c)Bymeansofcontourintegrationevaluate


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0

(loge

u)2

u2

+1du.

Solution.CconsistingWetakef(z)ofthe=line(logz2+1z)2

joiningandthe(R,0)contourto(r,0),thesemicircleofradiusrwithcen-ter(0,0),thelinejoining(r,0)to(R,0)and

asemicircleofradiusRwithcenter(0,0).Thecontourliesintheupperhalfplaneandisorientedanticlockwise.Wehaveavoidedthebranchpointz=0ofthemultiplevaluedfunctionlogz.A(R,0)B(r,0)C(r,0)D(R,0)(EventuallyweshallletR,r0).(1)On,z=Reiand|1+z2||z|21=R21.Thus


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(log(Rei))2

R21

f(z)dz

iReid00

|logRR2+1i|2Rd=

0((logR)2+)ButR21R

RR21((logR)2+2)d=R2R

133((logR)2+33

)


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0asR,thereforeR

lim

f(z)dz=05


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(2)On,z=rei,|z|2+11|z|2=1r2.Thus


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0

(logr)2+2

1r2((logr)2+33f(z)dzrd=1r

r2)Buttherightside0asr0,itfollowsthatr0lim

f(z)dz=0.(3)f(z)hasasimpleatz=ioff(z)is(logi)22i

pole=1at(zi=iinthetheupperhalfplane(insideC)andtheresidue2i2


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)2

=2i8

.ThusR,r0

limC

R

f(z)dz=R,r0limr

r

f(x)dx+f(xei)dxeiR=2i2i8becauseonthelineCD,z=x,andonthelineAB,z=xei.Hence0

(log(xei))2

1+x2e2i0

(logx)21+x234

Now(log(xei))2=(logx)22+2ilogx,so2dx+dx=0


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(logx)21+x2dx20

dx1+x20

34Equatingrealparts,andnotingthat+2i

1logx+x2dx=0

dx1+x2=tan1x]

0

=2,weget20

(logx)2

1+x2323434


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sothatdx==0

(logx)21+x2dx=38.Notethatbyequatingimaginaryparts,weget0

1logx+x2dx=0.6

complex analysis 1994

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