Key to Homework 8, Thanks to Da Zheng for the
textfile
November 18, 2011
1. Prove that
csc
z
=
1
z
+ 2
z
+
∞
X
n
=1
(

1)
n
1
z
2

n
2
π
2
,
z
6
= 0
,
±
π,
±
2
π,
· · ·
Proof.
We consider the following auxiliary function, where
z
6
= 0
,
±
π,
±
2
π,
· · ·
f
(
ζ
) =
csc
ζ

1
ζ
ζ
(
ζ

z
)
Then, it is easy to check that csc
ζ

1
ζ
has a removable singularity at
ζ
= 0, so
we will regard it holomorphic at
ζ
= 0.
Thus, the function
f
has poles at
ζ
=
nπ
,
n
∈
Z
, each of order one.Here, notice
that
ζ
= 0 is also a removable singularity for
f
.
Next, choosing the following sequence of counterclockwise
square
contour
γ
n
,
centered at the origin with vertices (
n
+
1
2
)(
±
1
±
i
)
π
. Hence, integrating
f
(
ζ
)
over contour
γ
n
and applying the residue’s theorem, we obtain
1
2
πi
I
γ
n
f
(
ζ
)
dζ
=
csc
z

1
z
z
+
n
X
k
=1
(

1)
k
1
kπ
(
kπ

z
)
+
n
X
k
=1
(

1)
k
1
kπ
(
kπ
+
z
)
(
*
)
Then, for enough large
n
(i.e.
γ
n
contains
z
), we evaluate the integral on the
left hand side as follows: we first
claim
that csc
z
is bounded on the boundary
of the square
γ
n
(
please see this argument carefully
): Indeed, let
z
=
x
+
iy
,
for
y >
π
2
,

csc
z

=
2
i
e
iz

e

iz
≤
2
e
y

e

y
≤
2
e
π/
2

e

π/
2
≤
1
.
Similarly, for
y <

π
2
,

csc
z
 ≤
1. In other words,

csc
z
 ≤
1 on two horizontal
lines of the square
and also on the part of two vertical lines with

y

>
π
2
. Hence,
1
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it only remains to see the points on the part of two vertical lines with

y
 ≤
π
2
.
Notice that on the line segment with ends
π
2

π
2
i,
π
2
+
π
2
i
, the function sin
z
has no zeros, so

csc
z
 ≤
M
1
on the line segment. Since

csc(
z
+
π
)

=

csc
z

,
so

csc
z
 ≤
M
1
on the line segment with ends
n
+
π
2

π
2
i, n
+
π
2
+
π
2
i
. Hence

csc
z
 ≤
M
on
γ
n
. Thus
I
γ
n
f
(
ζ
)
dζ
=
I
γ
n
csc
ζ

1
ζ
ζ
(
ζ

z
)
dζ
≤
I
γ
n
csc
ζ

1
ζ
ζ
(
ζ

z
)

dζ

≤
I
γ
n

csc
ζ


ζ
(
ζ

z
)

+

1
ζ


ζ
(
ζ

z
)


dζ

≤
2
πM
(
n
+ 1
/
2)
π
 
z

+
2
π
(
n
+ 1
/
2)
π
((
n
+ 1
/
2)
π
 
z

)
→
0
,
as
n
→
+
∞
So, the expression (*) becomes
0 =
csc
z

1
z
z
+
∞
X
k
=1
(

1)
k
1
kπ
(
kπ

z
)
+
∞
X
k
=1
(

1)
k
1
kπ
(
kπ
+
z
)
Simply the above and we obtain our desired equality
csc
z
=
1
z
+ 2
z
+
∞
X
n
=1
(

1)
n
1
z
2

n
2
π
2
,
z
6
= 0
,
±
π,
±
2
π,
· · ·
2
2. Find the expression of
+
∞
X
∞
1
n
2
+
a
2
where
a >
0 is not a integer.
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 Uniform convergence, Conformal map, Normal family, (πz )

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