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Solutions Homework 14.
(1) (Schwarz’s lemma) Let
f
be a holomorphic function on
B
(0
,
1) with

f
(
z
)
 ≤
1 for
all

z

<
1 and
f
(0) = 0.
a.
Deﬁne
f
1
(
z
) =
f
(
z
)
z
for
z
6
= 0 in
B
(0
,
1). Prove that
z
= 0 is a removable
singularity of
f
1
.
Solution:
Since
f
is holomorphic at
z
= 0, lim
z
→
0
f
(
z
)
z
=
f
0
(0) exists, so
z
= 0
is a removable singularity of
f
1
(
z
).
b.
Prove that

f
1
(
z
)
 ≤
1
r
on
B
(0
,r
) for all 0
< r <
1. (Hint: use the maximum
modulus principle.)
Solution:
If 0
< r <
1, then

f
1
(
z
)

=
1
r

f
(
z
)
 ≤
1
r
on

z

=
r
. Hence by the
maximum modulus principle

f
1
(
z
)
 ≤
1
r
for all

z
 ≤
r
.
c.
Conclude that

f
(
z
)
 ≤ 
z

for all
z
∈
B
(0
,
1). Moreover if equality holds for some
z
0
6
= 0, then there exists
c
with

c

= 1 such that
f
(
z
) =
cz
for all
z
∈
B
(0
,
1).
Solution:
Let
z
∈
B
(0
,
1). Then for all

z
 ≤
r <
1 we have by part b)

f
(
z
)
 ≤
1
r

z

. letting
r
→
1 we get

f
(
z
)
 ≤ 
z
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 Fall '11
 Schep

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