MATH 185: COMPLEX ANALYSIS
FALL 2008/09
PROBLEM SET 9
For
f
: Ω
→
C
and
n
∈
N
, recall that
g
=
f
n
is the function defined by
g
(
z
) = [
f
(
z
)]
2
for all
z
∈
Ω.
A function on Ω
⊆
C
is said to be
meromorphic
if it has only removable singularities or poles in Ω
(i.e. no essential singularities or nonisolated singularities).
1.
Let Ω
⊆
C
be a region and let
f
: Ω
→
C
. Suppose
g
=
f
2
and
h
=
f
3
are both analytic on Ω.
Show that
f
is analytic on Ω.
Solution.
Note that
g
(
z
) = 0 if and only if
h
(
z
) = 0. So the zeros of
g
and
h
are in common.
Let
Z
⊆
Ω be these common zeros. If
Z
contains a limit point, then
g
and
h
are identically zero
by the identity theorem and
f
≡
0 is of course analytic. We will assume that all zeros in
Z
are
isolated. Let
z
0
∈
Z
. So there exist analytic functions
g
1
and
h
1
and some
ε >
0 such that for
all
z
∈
D
(
z
0
, ε
),
g
(
z
) = (
z

z
0
)
k
g
1
(
z
)
,
g
1
(
z
0
)
6
= 0
,
h
(
z
) = (
z

z
0
)
l
h
1
(
z
)
,
h
1
(
z
0
)
6
= 0
,
where
k
and
l
are the orders of zero of
g
and
h
respectively at
z
0
. But
g
3
=
f
6
=
h
2
and so
(
z

z
0
)
3
k
g
1
(
z
)
3
= (
z

z
0
)
2
l
h
1
(
z
)
2
.
So we get
g
1
(
z
)
3
h
1
(
z
)
2
= (
z

z
0
)
3
k

2
l
.
Since the
lhs
is nonzero for
z
=
z
0
, the
rhs
is also nonzero for
z
=
z
0
and this is only possible
if 3
k

2
l
= 0. In other words,
l > k
, i.e. the order of zero of
h
is larger than the order of zero
of
g
. So
h/g
has a removable singularity at
z
0
. Since this is true for all
z
0
∈
Z
,
h/g
may be
extended to an analytic function
e
f
on Ω. But since for all
z
∈
Ω
\
Z
,
e
f
(
z
) =
h
(
z
)
g
(
z
)
=
f
(
z
)
,
and Ω
\
Z
clearly has limit points,
e
f
≡
f
by the identity theorem.
2.
Is there a polynomial
p
(
z
) such that
p
(
z
)
e
1
/z
is an entire function?
Solution.
By the Taylor expansion of exp,
e
1
/z
=
∞
X
n
=0
1
n
!
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 Fall '07
 Lim
 Math, Taylor Series, n=N +1

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