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(math304)[2006](s)fianl~PPSpider^_10497.pdf
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MATH304, Spring 2006
Solution to Final Examination
1. Since |an||Re an|so that |an|1/n |Re an|1/n. By the root test, the radius of

convergence R1 of anz n is given by
n=0

1 = lim |an|1/n
R1 n

while the radius of convergence R2 of (Rean)z n is
n=0

11
= lim |Re an|1/n lim |an|1/n = .
R2 nnR1
Hence, we obtain R1 R2. Equality holds when Re an = an, that is, an is real.
2. Let u = z +1 so that
..n
1 1 112
== .. = .
(z +1)(z +3) u(u +2) u2 1+ 2 u2 u
un=0
1248
= . + . +
u2 u3 u4 u5
1248
= . + . + ,
(z +1)2 (z +1)3 (z +1)4 (z +1)5
. 2.
which is valid when < 1 or |z +1|> 2. Since the Laurent series is not valid
u
inside a deleted neighborhood of the isolated singularity z = .1, so one cannot
infer any conclusion on the classi.cation of the isolated singularity using the above
Laurent series.
3. (a) Since zj is a simple pole, so
Res(f,zj) = lim [(z .zj)f(z)]
zzj
..
. k z .zj.
= lim . cj + ci .
zzj z .zi
i=1 i.
=j
k z .zj
= cj + ci lim = cj.
zzj z .zi
i=1 i.
=j
1
2.i/n,. =
(b) The poles of are z. = e0,1, ,n .1. They are all simple
zn .1
poles.

11 z. z.
Res ,z. = == ,. =0,1, ,n .1.
n.1 n
zn .1nz nzn
..
1
We then have
11 n.1 z.
= .
zn .1 nz .z.
.=0
4. The function has a pole of order 2 at z =0 and a pole of order 3 at z = .
1 dd cosz
Res(f,0) = lim [z 2f(z)]=lim
z0 1! dzz0 dz (z .)3
.(z .)sinz .3cosz .3
= lim = ,
z0 (z .)4 4
d2 d2 Res(f,) = lim [(z .)3f(z)]=lim
1 1 cosz
z 2! dz2 z 2dz2 z2
1 (6.z2)cosz +4z sinz2 .6
= lim = .
z 2 z4 24
5. (i) When f(z0)=0,theTaylorexpansionof f(z)inside some neighborhood of z0 is given by
(z0)
f(z0) f
f(z)= (z .z0)+ (z .z0)2 +
1! 2!
so that
f (z0)

g(z)= f (z0)+ (z .z0)+
2!
Though g(z)is not de.ned at z0, but it admits a Taylor expansion in a neigh-borhood of z0, so z = z0 is a removable singularity of g(z).
(ii) When f(z0)=.0, the Laurent expansion of g(z)in a deleted neighborhood of z0 takes the form
(z0)
f(z0) f(z0) f
g(z)= ++(z .z0)+
z .z0 1! 2!
Hence, z = z0 is a simple pole of g(z), and Res(g,z0)= f(z0).
6. (a) Suppose f(z) 0 as z , the Jordan Lemma states that
izf(z)dz
lim e =0,
R
CR
where is a positive real number and the contour CR is the upper semi-circle 1
with radius R. In our integral, we take = 2 and f(z) = . Obviously,
z2 . 2iz
1 e
0 as z . Therefore, dz 0 as R . Also,
z2 z2
CR
1 R 1
dz = O = O 0 as R .
z2 R2 R
CR
Hence,
. 2iz
1.e
lim dz =0.
R z2
CR
2

2iz
1.e
(b) Note that z = 0 is an isolated singularity of . Indeed, it is a simple
z2
pole since
2iz
1.e
lim z = .2i .=0.
z0 z2
We cannotchoose a contour whichpasses throughtheisolated singularity. This
2iz
1.e
is why we have to choose an indented con