complex analysis 1990
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UPSCCivilServicesMain1990-Mathematics
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ComplexAnalysisSunderLalRetired Professor of MathematicsPanjab University
ChandigarhMarch5,2010Question1(a)Letfberegularfor|z|
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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
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=but|z|=|z|
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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
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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
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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|
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||=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
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,andtheresultfollows.2
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Question1(c)Iff=u+ivisregularthroughoutthecomplexplane,andau+bvc0
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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
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=
1x4+dx
x8+Rlimz4dz
1+z8=2i(sumofresiduesatpolesoff(z)intheupperhalfplane)
Now
1z4+dzz8
0
R4e4iRieiR81
d
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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
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z4
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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(
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42sin2
i(2cos8sin4
8cossin888cos
8sin4)Thus
x4dx
1+x88asrequired.Question2(b)Deriveaseriesexpansionoflog(1+ez)inpowersofz.Solution.Letf(z)=log(1+ez),thenf(z)=
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=2i(2i2
sin8)=2sinez
1+ez1coshz2
Letg(z)=coshz2
=12ez
2
e2z2
+ez2
=12ez2
,theng(n)(z)={1
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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
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(np(n
)np
{p2npf(p+1)(0)=2n+110,nodd1,neven4wheren
=
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andtherefore
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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
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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
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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
...(
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21
...v)v1
0...u
50...u1
v...u2
.........vn
(
n)1
)vn1
(n2
vn2
...un
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dny
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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
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coefficientmatrixisvn+1.ThusbyCramersrule
yn
,...,yn
v0
=vn+11v1
v2
...(21
...v)v1
vn
00v......u...u1
...u2
(n
)......1
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)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
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F(1)(0)=f(2)(0)=
200...0112n+111...220...012...011(n
)1(n
)2
......(n
)n1
...
1
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1
4F(2)(0)=f(3)(0)=1421
11=
18
21102211
1
=0F(3)(0)=f(4)(0)=
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21110
22300231111
2111022300231000
18F(4)(0)=f(5)(0)=
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116=116
=216=
132
21110223002300021111
=014641
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Thuslog(1+ez)hastheexpansionasgivenabove.6
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(
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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(
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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
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.7