Solution of Homework 1
Problem (6.6):
Solution:
The domain
Ω
is conformally equivalent to
D
(0
,
1)
\{
0
}
.
f
(
z
) = 1
/z
is a
conformal map from
D
(0
,
1)
\{
0
}
to
Ω
.
Suppose
g
is a conformal mapping from
Ω
to
Ω
. Then
f

1
◦
g
◦
f
is a
biholomorphcself map of
D
(0
,
1)
\{
0
}
.
And
Aut
(
D
(0
,
1)
\{
0
}
) =
{
e
iψ
z

ψ
∈
R
}
So we know that
g
=
e

iψ
z
.
Above all:
Aut
(Ω) =
{
e
iϕ
z

ϕ
∈
R
}
¥
Problem (6.11):
Solution:
Observe that
φ
1
◦
φ

1
2
:
D
→
D
is a conformal map. According to
Aut
(
D
(0
,
1)) =
{
e
iθ
a

z
1

¯
az

a
∈
D
(0
,
1)
, θ
∈
[0
,
2
π
)
}
So there exist a
θ
∈
[0
,
2
π
)
,
ϕ
∈
Aut
(
D
(0
,
1))
such that
φ
1
=
ϕ
◦
φ
2
. Then
we know
φ
1
(
z
) =
e
iθ
a

φ
2
(
z
)
1

¯
aφ
2
(
z
)
¥
Problem (6.17)
Solution:
Since
φ
(
P
) =
P
,
φ
(
P
) = 1
, from the hint in the book, we know that
φ
(
z
) =
P
+ (
z

P
) +
h
(
z
)
, where
h
(
z
)
contains the higher order terms.
1
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Suppose
φ
(
z
)
6
=
z
. then
h
(
z
)
has anonzero term, say
a
n
(
z

P
)
n
. Since
φ
maps
Ω
to itself, so

φ
(
z
)
 ≤
M
, where
M
=
sup
{
z

, z
∈
Ω
}
. By cauchy
estimate,

a
n
 ≤
M
r
n
.
And by induction
φ
j
(
z
) =
φ
◦
φ
◦
φ
···◦
φ
=
P
+(
z

P
)+
ja
n
(
z

P
)
n
+
····
But
φ
j
maps
Ω
to itself. So the same cauchy estimate gives
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 Spring '09
 Compact space, Möbius transformation, Conformal geometry, Conformal map, linear fractional transformation, cauchy estimate

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