MATH 185: COMPLEX ANALYSIS
FALL 2009/10
PROBLEM SET 10 SOLUTIONS
1.
(a) Show that if
f
has a pole or an essential singularity at
a
, then
e
f
has an essential singularity
at
a
.
Solution.
If
f
has a pole of order
m
at
a
, then there exists
ε >
0 and
g
:
D
(
a,ε
)
→
C
analytic,
g
(
a
)
6
= 0, such that
f
(
z
) =
g
(
z
)
(
z

a
)
m
for all
z
∈
D
*
(
a,ε
). Let the power series representation of
g
on
D
(
a,ε
) be
g
(
z
) =
∞
X
n
=0
a
n
(
z

a
)
n
.
Let
h
(
z
) :=
∞
X
n
=0
a
n
+
m
(
z

a
)
n
.
Then
1
f
(
z
) =
h
(
z
) +
m

1
X
n
=0
a
n
(
z

a
)
m

n
and so
e
f
(
z
)
=
e
h
(
z
)
m

1
Y
n
=0
e
an
(
z

a
)
m

n
=:
e
h
(
z
)
F
(
z
)
.
Note that
e
h
(
z
)
is analytic and nonzero. If
e
f
(
z
)
has a pole or a removable singularity at
a
,
then
F
(
z
) =
e
f
(
z
)
e

h
(
z
)
will have a pole or a removable singularity at
a
. So since
F
(
z
) has
an essential singularity at
a
,
e
f
(
z
)
must have an essential singularity at
a
.
If
f
has an essential singularity at
a
, then
f
(
D
*
(
a,ε
)) is dense in
C
for all
ε >
0, ie.
f
(
D
*
(
a,ε
)) =
C
(here
S
denotes the closure of the set
S
). Recall that if
g
is any continuous
function, then
g
(
S
)
⊇
g
(
S
). Since exp :
C
→
C
is a continuous function, let
S
=
f
(
D
*
(
a,ε
))
and we have
exp(
S
)
⊇
exp(
S
) = exp(
C
) =
C
×
.
Hence, (exp
◦
f
)(
D
*
(
a,ε
)) is dense in
C
. By part (a),
a
is an essential singularity of exp
◦
f
.
(b) Let Ω
⊆
C
be a region. Let
a
∈
Ω and
f
: Ω
\{
a
} →
C
be a function with an isolated
singularity at
a
. Suppose for some
m
∈
N
and
ε >
0,
Re
f
(
z
)
≤ 
m
log

z

a

for all
z
∈
D
*
(
a,ε
). Show that
a
is a removable singularity of
f
.
Solution.
Note that the condition implies that

e
f
(
z
)

=
e
Re
f
(
z
)
≤
e

m
log

z

a

=
1

z

a

m
.
Date
: December 5, 2009.
1
Now that you have learnt about Laurent series and its relation with poles, you could of course just write this
down without going through the preceding arguments.
1