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Mathematics 185 – Intro to Complex Analysis
Fall 2009 – M. Christ
Solutions to Selecta from Problem Set 12
4.4.4
Show that
π
2
sin
2
(
πz
)
=
∑
∞
n
=
∞
(
z

n
)

2
for all
z
∈
C
\
Z
.
Solution.
We deduce this from the formula
π
cot(
πz
) =
z

1
+
X
n
6
=0
±
(
z

n
)

1
+
n

1
²
,
(1)
where the sum is over all integers
n
6
= 0, positive and negative. We know the latter formula
to be valid for all
z
∈
C
\
Z
.
The connection between the two formulas is that
π
2
sin
2
(
πz
)
is the derivative of
π
cot(
πz
),
while
∑
∞
n
=
∞
(
z

n
)

2
is obtained, at least formally, by diﬀerentiating the righthand side
∑
∞
n
=
∞
(
z

n
)

1
term by term. Thus our task is to justify this term by term diﬀerentiation.
Let
F
N
(
z
) =
∑
N
n
=

N
(
z

n
)

1
. It was implicitly shown in the course of the proof of (1)
that
F
N
(
z
)
→
π
cot(
πz
)
uniformly on
D
,
as
N
→ ∞
. (The diﬀerence
F
N
(
z
)

π
cot(
πz
) is
expressed as an integral over
C
N
, and the bound established for this integral tends to zero
as
N
→ ∞
, uniformly for
z
∈
D
.)
Therefore the sequence of derivatives also converges uniformly on any disk centered at
w
of strictly smaller radius;
F
0
N
→
d
dz
(
π
cot(
πz
)) uniformly for all
z
∈
B
(
w,r/
2),
by one of our general theorems. Since
F
0
N
(
z
) =
∑
N
n
=

N
(
z

n
)

2
, this is the desired
conclusion.
4.4.8
Prove that
∑
∞
n
=0
(

1)
n

1
n

2
=
π
2
12
.
Solution.
Apply the residue theorem to
f
(
z
) =
π
z
2
sin(
πz
)
=
z

2
π
csc(
πz
), using the familiar
contours
C
N
. Calculate
Res(
f,n
) =
n

2
1
cos(
πn
)
= (

1)
n
n

2
for 0
6
=
n
∈
Z
;
using csc instead of cot is natural, because it gives the alternating
±
signs which we’re
aiming for.
Use the expansion sin(
πz
) =
πz
(1

π
2
z
2
/
6 +
O
(
z
4
)) to calculate
π
π
sin(
πz
)
=
π
πz
(1

π
2
z
2
/
6 +
O
(
z
4
))

1
=
z

1
(1 +
π
2
z
2
/
6 +
O
(
z
4
))
to obtain
z

2
π
π
sin(
πz
)
=
z

3
(1 +
π
2
z
2
/
6 +
O
(
z
4
)) =
z

3
+
π
2
6
z

1
+
O
(
z
)
1
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View Full Documentand therefore
Res(
f,
0) =
π
2
6
.
By the Residue Theorem, for each integer
N
≥
1,
(2
πi
)

1
Z
C
N
f
(
z
)
dz
=
π
2
6
+

1
X
n
=

N
(

1)
n
n

2
+
N
X
n
=1
(

1)
n
n

2
.
If we can show that
R
C
N
f
(
z
)
dz
→
0 as
N
→ ∞
then we obtain the desired formula.
Now

Z
C
N
f
(
z
)
dz
 ≤
‘
(
C
N
) max
z
∈
C
N
h

z


2
π

sin(
πz
)


1
i
.
Since
‘
(
C
N
) = 8
N
+ 4 and

z


2
≤
N

2
for every
z
∈
C
N
, it suﬃces to prove that

sin(
πz
)


1
≤
1000 for every
z
∈
C
N
, for every integer
N
≥
1.
This is very similar to a corresponding bound established in class and in our text for
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 Fall '07
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