8/3/2019 Complex Analysis 1990

1/38

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".

8/3/2019 Complex Analysis 1990

2/38

UPSCCivilServicesMain1990-Mathematics

8/3/2019 Complex Analysis 1990

3/38

ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University

ChandigarhMarch5,2010Question1(a)Letfberegularfor|z|

8/3/2019 Complex Analysis 1990

4/38

f(0)=2i12

0f(Rei)R2e2iRieid=2R12

f(Rei)eid0(1)

Wenowconsidertheintegral12iCR

f()t1(RRz

d

ByCauchys

residue

theorem,

the

above

integral

is

equal

to

2i(sum

of

residues

of

theintegrandwithinC

R

)t+1).Ift1,theonlypossibilityofapolecouldbeatthepoint=R2z

,1

8/3/2019 Complex Analysis 1990

5/38

=but|z|=|z|

8/3/2019 Complex Analysis 1990

6/38

R

,so|R2z

>R2

RR,soR2z

liesoutsideCR

andhencetheintegrand12iCR

f()t1(RRz

d=0fort1Inparticular,takingt=1,z=0,12i)t+1C

Rf()R2d=0Thusweget0=12R20

0=

12Rf(Rei)eid2

8/3/2019 Complex Analysis 1990

7/38

f(Rei)eid0(2)Adding(1),(2),wegetf(0)=2R

12

(f(Rei)+f(Rei))eid=0

1R2

u()exp(i)d

0asrequired.Note1:Togetthedesiredform,wecouldhaveconsideredtheintegralover{Cr

:|z|=r

8/3/2019 Complex Analysis 1990

8/38

isatleastr|ao

|

n=0

an

znseries,andMM+=|aM(r)0|whererisanynumbernotexceedingtheradiusofconvergenceofthe=sup|z|=r

|f(z)|.Solution.ByCauchysintegralformula,f(z)f(0)=||=r||=r

f()dwhere|z|

8/3/2019 Complex Analysis 1990

9/38

||=r

(1

z1

M

)d

M2|z|

02

rieidrei(r|z|)=rM|z||z|

because|z||||z|=r|z|on||=r.Thusr|f(0)||z|(M+|f(0)|)|z|M+|f(0)||f(0)|r

.Heref(0)=a0

8/3/2019 Complex Analysis 1990

10/38

,andtheresultfollows.2

8/3/2019 Complex Analysis 1990

11/38

Question1(c)Iff=u+ivisregularthroughoutthecomplexplane,andau+bvc0

8/3/2019 Complex Analysis 1990

12/38

forsuitableconstantsa,b,cthenfisconstant.Solution.Theorem:Iff(z)=u+ivisentire,andu0,thenfisconstant.Proof:ConsiderF(z)=ef(z),thenF(z)isalsoentire.Moreover

|F(z)|=|eu+iv|=|eu|1u0

ThusF(z)isentireandbounded,henceisaconstantbyLiouvillestheorem.NowF(z)=

f(z)ef(z)=0f(z)=0becauseef(z)=0,sof(z)isconstant.

Corollary:Iff(z)=u+ivisentire,andu0,thenfisconstant.Proof:Considerf(z)=uiv,thenu0andf(z)isconstant.NowconsiderF(z)=(aib)f(z)c=(au+bvc)+i(avbu).NowF(z)isentire,andReF(z)=au+bvc0,soF(z)isconstant,hencef(z)isconstant.Question2(a)Provethat

1x4+dxx8

=2sin8usingresiduecalculus.

Solution.Wetakef(z)=1+z8z4

(R,0)(0,0)(R,0)ByCauchysresiduetheoremR

limandthecontourCconsistingofasemicircleofradiusRwith

center(0,0)lyingintheupperhalfplane,andthelinejoining(R,0)and(R,0).FinallywewillletR.C

z4dz1+z8

8/3/2019 Complex Analysis 1990

13/38

=

1x4+dx

x8+Rlimz4dz

1+z8=2i(sumofresiduesatpolesoff(z)intheupperhalfplane)

Now

1z4+dzz8

0

R4e4iRieiR81

d

8/3/2019 Complex Analysis 1990

14/38

R5R81because|z8+1||z8|1=R81on|z|=R.ThereforeR

limz4dz

1+z8=0

f(z)haspolesatzerosofz8+1=0z8=1z8=e(2n+1)iz=e(2n+1)i8

,nZ.

Clearlyz=ei8

,e3i8

,e5i8

,e7i8

aretheonlypolesoff(z)intheupperhalfplaneandallthese3

8/3/2019 Complex Analysis 1990

15/38

z4

8/3/2019 Complex Analysis 1990

16/38

aresimplepoles.Theresidueatanysimplepolez0

is8z700

=18z30

,sumofresiduesatpolesoff(z)intheupperhalfplane=)=18

(e3i/8+e9i/8+e15i/8+e21i/8)=18(e3i/8ei/8+ei/8)e3i/8

=18(2isin82isin)3

==4ii(

8/3/2019 Complex Analysis 1990

17/38

42sin2

i(2cos8sin4

8cossin888cos

8sin4)Thus

x4dx

1+x88asrequired.Question2(b)Deriveaseriesexpansionoflog(1+ez)inpowersofz.Solution.Letf(z)=log(1+ez),thenf(z)=

8/3/2019 Complex Analysis 1990

18/38

=2i(2i2

sin8)=2sinez

1+ez1coshz2

Letg(z)=coshz2

=12ez

2

e2z2

+ez2

=12ez2

,theng(n)(z)={1

8/3/2019 Complex Analysis 1990

19/38

2n

sinhz2

,nodd12n

coshz2

,nevenInparticular,g(n)(0)=0whennisodd,andg(n)(0)=2n1

whenniseven.Moreoverf(z)coshz21z22

UsingLeibnitzruleforthederivativeoftheproductoftwofunctions,wegetdndzn=f(z)g(z)=e)g(np)(z)f(p+1)(z)

Thuswhenz=0,wegetnp=0

(12ez2

)

=ez2

2n+1=np=0

8/3/2019 Complex Analysis 1990

20/38

(np(n

)np

{p2npf(p+1)(0)=2n+110,nodd1,neven4wheren

=

8/3/2019 Complex Analysis 1990

21/38

andtherefore

8/3/2019 Complex Analysis 1990

22/38

2n+1f(n+1)(0)=1n1p=0

(n

p)2p+1np

f(p+1)(0)Case(1):Whenniseven2n+1f(n+1)(0)=1(n)

0)2f(0)n2(2p+1npp=1

np

f(p+1)(0)Notethatoddpdonotcontributeanythingtothesummation,asnp

=0foroddp.Nowwecanseebyinductionthatf(n)(0)=01then2letting12

=0.Assumebyinductionhypothesisn=2mintheaboveformula,whenevernisoddthatf(3)(0)andn>=f(5)(0)1.=f...(0)==f12

(2m1)(0).23f(3)(0)=

=0,22m+1f(2m+1)(0)=)22p+1f(2p+1)(0)=0Case(2):Whennisodd:Thetermswithevenpintheformulaabovedonotmakeanycontribution.Thuslettingn=2m+1,22m+2f(2m+2)(0)=1

8/3/2019 Complex Analysis 1990

23/38

m1(2mp=1

2p

m1)r=0

(2m2r++11)22r+2f(2r+2)(0)=1mr=1

(22rf(2r)(0)()

Wecannowseethatf(0)=14

2m+12r1,fThus,f(4)(0)=18

(6)(0)=1

4.log(1+ez)=log2+z21142!1

184!146!1

8/3/2019 Complex Analysis 1990

24/38

z6+...=log2++z2z4+n=1

f(2n)(0)z2n(2n)!

wheref(2n)(0)isgivenby()forn1.

Note:Wenowpresentanalternativesolution,whereweuseLeibnitzruleforthen-thderivativeofthequotientoftwofunctions.Itisagoodexerciseinitselfandisusuallymissingfromtextbooks.Theorem:Lety=uv

z2+,whereu,varefunctionswithderivativesuptoordern.Thenyn

v0

=vn+11v1

v2

...(

8/3/2019 Complex Analysis 1990

25/38

21

...v)v1

0...u

50...u1

v...u2

.........vn

(

n)1

)vn1

(n2

vn2

...un

8/3/2019 Complex Analysis 1990

26/38

dny

8/3/2019 Complex Analysis 1990

27/38

HerethedeterminantProof:vy=u,is(n+1)(n+1),andtherefore,bytakingsuccessiveyn

=derivativesdxn

.usingLeibnitzproductrulewegetvy=uv1

=u1

v2

y+vy1y+2v1

y1

+vy2

=u2

...v

ny+(n1

)vn1

y1

+...+vy

n...=un

Thesearen+1equationsinn+1unknownsy,y1

,andthedeterminantofthe

8/3/2019 Complex Analysis 1990

28/38

coefficientmatrixisvn+1.ThusbyCramersrule

yn

,...,yn

v0

=vn+11v1

v2

...(21

...v)v1

vn

00v......u...u1

...u2

(n

)......1

8/3/2019 Complex Analysis 1990

29/38

)vn1

(n

2vn2

asrequired.Nowf(z)=log(1+ez),f(0)=log2.f(z)=1+ezez

...un

Thenun

(0)=1foreveryn,andv(0)=2,vn(0)=1for,fn(0)1.=Let12

.LetF(z)u=ez,v=uv

,=1+ez.F(n)(0)=f(n+1)(0)=then

8/3/2019 Complex Analysis 1990

30/38

F(1)(0)=f(2)(0)=

200...0112n+111...220...012...011(n

)1(n

)2

......(n

)n1

...

1

8/3/2019 Complex Analysis 1990

31/38

1

4F(2)(0)=f(3)(0)=1421

11=

18

21102211

1

=0F(3)(0)=f(4)(0)=

8/3/2019 Complex Analysis 1990

32/38

21110

22300231111

2111022300231000

18F(4)(0)=f(5)(0)=

8/3/2019 Complex Analysis 1990

33/38

116=116

=216=

132

21110223002300021111

=014641

8/3/2019 Complex Analysis 1990

34/38

Thuslog(1+ez)hastheexpansionasgivenabove.6

8/3/2019 Complex Analysis 1990

35/38

(

8/3/2019 Complex Analysis 1990

36/38

1cos1z

)Question2(c)Determinethenatureofsingularpointsofsin

andinvestigateitsbehavioratz=.Solution.1.Let=1z

),and()=f(1

(1

cos.Therefore0lim()=sin1,showingthat()hasaremovablesingularityat=0.Infact()isanalyticat=0if(0)isdefinedtobesin1.Notethat0

lim)=sin)

sectan=0Thussin()(0)=0limsin(cos1

)sin1

=0limcos(1cos(

8/3/2019 Complex Analysis 1990

37/38

1)cos1z

isregularat.

2.Atallzerosofcos1z

i.e.laritiesbecauselimx

z=2(2n+1)thefunctionsin(cos1

z1

)hasessentialsingu-sinxdoesnotexistifitdid,thengiven>0,wewouldhaveNsuchthatx1

|N,x2

>N|sinx1

sinx2

=2n+2

>x2

=2n>N,then|sinx1

sinx2

|=1

8/3/2019 Complex Analysis 1990

38/38

.7