Complex Analysis
Fall 2007
Homework 11: Solutions
3.3.16
If
f
has a zero of multiplicity
k
at
z
0
then we can write
f
(
z
) = (
z

z
0
)
k
φ
(
z
), where
φ
is analytic and
φ
(
z
0
)
6
= 0. Diﬀerentiating this expression yields
f
0
(
z
) =
k
(
z

z
0
)
k

1
φ
(
z
) +
(
z

z
0
)
k
φ
0
(
z
). Therefore
f
0
(
z
)
f
(
z
)
=
k
(
z

z
0
)
k

1
φ
(
z
) + (
z

z
0
)
k
φ
0
(
z
)
(
z

z
0
)
k
φ
(
z
)
=
k
z

z
0
+
φ
0
(
z
)
φ
(
z
)
.
Since
φ
(
z
0
)
6
= 0, the function
φ
0
(
z
)
/φ
(
z
) is analytic at
z
0
and therefore has a convergent
Taylor series in a neighborhood of
z
0
. Thus
f
0
(
z
)
f
(
z
)
=
k
z

z
0
+
φ
0
(
z
)
φ
(
z
)
=
k
z

z
0
+
∞
X
n
=0
a
n
(
z

z
0
)
n
.
for all
z
in a deleted neighborhood of
z
0
. Uniqueness of such expressions implies that this is
the Laurent series for
f
0
/f
in a deleted neighborhood of
z
0
and therefore
f
0
/f
has a simple
pole at
z
0
with residue
k
.
3.3.20(a)
The closure of a set
A
is the intersection of all the closed sets that contain
A
.
It is not hard to show that an element belongs to the closure of
A
if and only if every
neighborhood of that point intersects
A
. Therefore, what we need to prove the following:
given any
w
∈
C
and any
± >
0, there exists
z
∈
U
so that
f
(
z
)
∈
D
(
w
;
±
).
So, let
w
∈
C
and
± >
0. According to the CasoratiWeierstrass Theorem, there is a
sequence
z
1
, z
2
, z
3
,
··· ∈
C
so that
z
n
→
z
0
and
f
(
z
n
)
→
w
. Since
z
0
∈
U
and
U
is open,
there is a
δ >
0 so that
D
(
z
0
;
δ
)
⊂
U
. Choose
N
1
∈
Z
+
so that

z
n

z
0

< δ
for
n
≥
N
1
and
choose
N
2
∈
Z
+
so that

f
(
z
n
)

w

< ±
. Let
N
= max
{
N
1
, N
2
}
. Then
z
N
∈
D
(
z
0
;
δ
)
⊂
U
and
f
(
z
N
)
∈
D
(
w
;
±
). That is,
z
N
satisﬁes the required conditions. Since
w
∈
C
and
± >
0
were arbitrary, we conclude that the necessary condition holds for all
w
∈
C
and
± >
0.
Thus, the closure of
f
(
U
) is
C
.
3.R.2
The function 1
/
cos
z
fails to be analytic precisely where cos
z
= 0. The latter occurs
if and only if
z
=
nπ
+
π/
2 for some
n
∈
Z
. Since the derivative of cos
z
is

sin
z
and
sin(
nπ
+
π/
2) = (

1)
n
6
= 0, we see that cos
z
has simple zeros at the points
z
=
nπ
+
π/
2,
n
∈
Z
. Consequently, 1
/
cos
z
has simple poles at these points.
3.R.6(a)
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 Fall '11
 RyanDaileda

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