Complex Analysis
Fall 2007
Homework 9: Solutions
3.1.4
(a) Let
z
∈
C
\ {
ni
:
n
∈
Z
}
. Then
lim
n
→∞
1
/
(
n
2
+
z
2
)
1
/n
2
= lim
n
→∞
n
2
n
2
+
z
2
= 1
.
According to the limit comparison test from calculus, the series
∞
n
=0
1
n
2
+
z
2
converges if and only if
∞
n
=1
1
n
2
converges. Since the latter series is known to converge, the former must as well. That
is,
∞
n
=0
1
n
2
+
z
2
converges absolutely for
z
∈
C
\ {
ni
:
n
∈
Z
}
.
(b) Let
D
be any bounded subset of
C
\ {
ni
:
n
∈
Z
}
. Then there is an
C >
0 so that

z

< C
for all
z
∈
D
. For
z
∈
D
we then have

n
2
+
z
2
 ≥ 
n
2
  
z
2

=
n
2
 
z

2
> n
2

C
2
.
If
n > C
then the above implies that
1
n
2
+
z
2
≤
1
n
2

C
2
=
M
n
for all
z
∈
D
.
Appealing to the limit comparison test as above, we conclude that
∑
n>C
M
n
converges. Hence, the Weierstrass
M
test implies that
n>C
1
n
2
+
z
2
converges uniformly and absolutely on
D
. Since neither of these modes of convergence
is altered by adding finitely many terms (Why? Convince yourself of this.) we conclude
that
∞
n
=1
1
n
2
+
z
2
1
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converges uniformly and absolutely on
D
. Since the functions 1
/
(
n
2
z
2
) are all analytic
on
D
, we conclude that the series is an analytic function on
D
.
Finally, since every
point
z
∈
C
\ {
ni
:
n
∈
Z
}
belongs to some bounded subset, we see that the series
represents an analytic function on
z
∈
C
\ {
ni
:
n
∈
Z
}
.
3.1.12
Let
>
0 and
A
=
{
z
∈
C
:

z
 ≥
}
. I will prove that
∞
n
=1
1
n
!
z
n
converges uniformly on
A
. Let
z
∈
A
. Then

z
 ≥
so that
1
n
!
z
n
≤
1
n
!
n
=
M
n
.
Since
lim
n
→∞
1
/
(
n
+ 1)!
n
+1
1
/n
!
n
= lim
n
→∞
1
n
= 0
the ratio test implies that
∑
M
n
converges. Since 1
/n
!
z
n
is analytic on
A
for each
n
, the
Weierstrass
M
test implies that the series in question converges uniformly and absolutely
to an analytic function on
A
. Since every point
z
∈
C
\ {
0
}
is contained in some
A
, this
proves that the series gives us a function that is analytic on all of
z
∈
C
\ {
0
}
.
As to the integral, since the series converges uniformly on
A
1
/
2
and the unit circle
γ
is
contained in this set, we have
γ
∞
n
=1
1
n
!
z
n
dz
=
∞
n
=1
1
n
!
γ
1
z
n
dz
= 2
πi
since
γ
1
z
n
dz
is zero unless
n
= 1, in which case we know it’s 2
πi
.
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 Fall '11
 RyanDaileda
 Calculus, Power Series, Taylor Series, lim

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