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(math304)[2007](s)final~PPSpider^sol_10500.pdf
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1. (a) Note that a2n.1 =3,a2n =1,n =1,2, . By the Root Test, 1
R = . =1. lim n |an|
n
..

. zn . 1 . 1
(b) For |z|1, . The in.nite series is con-
n2(n +1) n2(n +1) n2(n +1)
n=1

. n
z
vergent. Hence, we have uniform convergence of for |z|1.
n2(n +1)
n=1 n
. 4
z
(c) Let Sn(z)= , then
4)k.1
(1+z
k=1
. 1 .
|Sn(z).(1+z 4)|= ...
. 4)n.1.
(1+z
. 1 .
For |1+z4|> 1, we take .> 0, then .. <. when
. 4)n.1.
(1+z1
(n .1)ln < ln.
|1+z4|
or

ln.
n> +1.
1
ln
|1+z4|


. 4
z
Hence, =1+z 4 for |1+z4|> 1.
(1+z4)n
n=1
To show the property of uniform convergence for |1+z4| r,r > 1, we take ln.
N to be 1 +1, which is independent of z.
ln
r
ln.
Whenever n> 1 +1, we would have
ln
r
|Rn(z)|<., for |1+z 4|> 1,
where Rn(z)=1+z4 .Sn(z).
1
2. (a) f(z)= + , valid for |z .|> 0.
(z .)2 z .
(b) (i) Inside the domain |z|< ||, the Laurent series reduces to the Taylor series as f(z)is analytic in the region. Hence,

...2 . n 23
z zz nzz 2z3z
f(z)= =1. = =+++ .
2 n+1 2 3 4
(z .)2
n=1
(ii) For ||< |z|< , we have
11 (n +1)n 12 32
f(z)= ..2 = =+ 2 + 3 + .
zn
z zzzz 1.
zn=0
(c) Res(f,)=1; Res(f,0)is not de.ned since z =0 is not an isolated singularity
of f.

z
(d) dz.
(z .)2
|z|=1
z
If ||> 1, then f(z)= is analytic inside |z|=1, thus
(z .)2
z
dz =0.
(z .)2
|z|=1
If ||< 1, then
z
dz =2iRes(f,)=2i.
(z .)2
|z|=1
p (z0)(z .z0)+p (z0)(z .z0)2/2+
3. (a) f(z)= .
(z0)(z .(z0)(z .
qz0)3/3!+qz0)4/4!+
6p (z0)

Consider lim (z .z0)2f(z)= , which is .nite and non-zero.
(z0)
zz0 q
Hence, z = z0 is a double pole of f(z).

(b) Consider
Res(f,z0) = lim d [(z .z0)2f]
zz0 dz
dp (z0)+p (z0)(z .z0)/2!+
= lim
(z0)(z .
zz0 dz q (z0)/3!+qz0)/4!+
(z0)/3!+ (z0)/4!+
[p (z0)/2!+ ][q ].[q ][p (z0)+ ] = lim [q (z0)/3!+q (z0)(z .
zz0 z0)/4!]2 (z0) (z0)
3p 3p (z0)q
= . .
q (z0)2 q (z0)2
1
4. The singularities of f(z)are at z = 0 and sin /z = 0, that is, z =0,1, , .
2
1
For z = , n is any integer, they are all isolated singularities. Consider
n
.. 1
1 z .
lim z . cot = lim n lim cos
z1 nz z1 sin z1 z
nn z n
1 1
= lim lim cos = . .
z1 .2 cos z1 zn2
n zz n
Hence, z = 1, n is any integer, represents a pole of order 1 of f(z)= cot .
n z
Also,
11
res f, = . .
2
nn
1
z = 0 contains points of the form z =, n is some integer, and these points are
n
singularities of f(z).
5. The principal value of the integral is given by
4.
dx
I = lim + .

x(x .4)
4+
,0+
Choosing the branch
f(z)= 1 , for z = re i , 0 < < 2,
e