Key to Complex Analysis Homework 4
October 3, 2011
44.
Suppose
f
:
D
(0
,
1)
→
C
is a function, such that
f
2
and
f
3
are holomorphic.
Prove that
f
is holomorphic.
Proof.
Suppose
f
6≡
0; otherwise, it is trivial.
Observation
: We only need to check those points
z
0
∈
D
(0
,
1) with
f
(
z
0
) = 0
whether
f
0
(
z
0
) exists (since if
f
(
z
0
)
6
= 0, then
f
2
(
z
0
)
6
= 0, so there is
r
0
>
0
such that
f
2
has no zeros on
D
(
z
0
,r
0
)
⊂
D
(0
,
1). From
f
·
f
2
=
f
3
on
D
(
z
0
,r
0
),
∂
∂
¯
z
f
·
f
2
+
f
∂
∂
¯
z
f
2
=
∂
∂
¯
z
f
3
,
use the fact that
f
2
,f
3
are holomorphic, we get, on
D
(
z
0
,r
0
),
∂
∂
¯
z
f
·
f
2
= 0
.
Since
f
2
is never zero on
D
(
z
0
,r
0
),
∂
∂
¯
z
f
≡
0 on
D
(
z
0
,r
0
). This implies that
f
is holomorphic on
D
(
z
0
,r
0
).
Now let
z
0
∈
D
(0
,
1) with
f
(
z
0
) = 0. Then
f
2
(
z
0
) = 0 and
f
3
(
z
0
) = 0. Since
f
2
and
f
3
are holomorphic, we can write
f
2
= (
z

z
0
)
n
g
1
(
z
)
,f
3
= (
z

z
0
)
m
g
2
(
z
)
where
g
1
,g
2
are holomorphic on
D
(0
,
1), and
g
1
(
z
0
)
6
= 0
,g
2
(
z
0
)
6
= 0, thus
g
1
,g
2
have no zeros on
D
(
z
0
,r
)
⊂
D
(0
,
1) with
r >
0.
Next,
we claim that
m > n
. Indeed, assume the claim is false, and from (
z

z
0
)
m
g
2
(
z
) =
f
3
=
f
2
·
f
= (
z

z
0
)
n
g
1
(
z
)
·
f
, so
g
2
(
z
0
) = (
z

z
0
)
n

m
g
1
(
z
0
)
f
(
z
0
).
From
f
2
(
z
0
) = 0 we get
f
(
z
0
) = 0. Hence
g
2
(
z
0
) = (
z

z
0
)
n

m
g
z
(
z
0
)
f
(
z
0
) = 0
contradicts with the fact that
g
2
(
z
0
)
6
= 0. Hence
m > n
. Now for
z
∈
D
(
z
0
,r
)
⊂
D
(0
,
1) with
z
6
=
z
0
, write
f
=
f
3
f
2
= (
z

z
0
)
m

n
g
2
g
1
,
1