(1) Determine all conformal maps of
D
(0
,
1)
\{
0
} →
D
(0
,
1)
\{
0
}
.
Solution:
Note that a conformal map
f
from
U
=
D
(0
,
1)
\{
0
} →
U
has a singularity at 0. As the map is bounded, it is clearly bounded near
0 and thus a removable singularity. Thus
f
extends to a holomorphic map
ˆ
f
:
D
(0
,
1)
→
C
. The image of
D
(0
,
1) under this map is an open set,
and it is also contained in the closure
U
=
D
(0
,
1) of
U
. The image of
0 can’t be in
∂D
(0
,
1), as then
ˆ
f
(
D
(0
,
1)) would not be open (another
argument is that

ˆ
f

can’t have a maximum in an internal point), so
ˆ
f
(0) =
0. Thus conformal maps
U
→
U
are the restrictions to
U
of conformal
maps
D
(0
,
1)
→
D
(0
,
1) which are 0 at 0. By Schwarz’s lemma these are
only the rotations.
We conclude that the only conformal maps
D
(0
,
1)
\{
0
} →
D
(0
,
1)
\{
0
}
are
f
(
z
) =
ωz
for some
ω
∈
C
with

ω

= 1.
(2) Deﬁne the automorphism group
Aut
(
H
) =
{
f
:
H
→
H

f
holomorphic and bijective
}
.
Here
H
=
{
z
 =
(
z
)
>
0
}
denotes the upper half plane. Now show that
Aut
(
H
) is isomorphic to
PSL
2
(
R
) =
{
M
=
±
a b
c d
²

a,b,c,d
∈
R
,ad

bc
= 1
}
/
∼
where
M
∼
N
if and only if
N
=
M
or
N
=

M
. Finally show that
Aut
(
D
(0
,
1)) is also isomorpic to
PSL
2
(
R
).
Solution:
Recall the Cayley map
ψ
(
z
) = (
z

i
)
/
(
z
+
i
) which maps
H
conformally to
D
(0
,
1). In particular given a conformal map
f
:
H
→
H
we
ﬁnd that
ψ
◦
f
◦
ψ

1
is a conformal map of
D
(0
,
1)
→
D
(0
,
1). This creates
an isomorphism between
Aut
(
H
) and
Aut
(
D
(0
,
1)) (note the map is clearly
invertible). In particular any conformal map in
Aut
(
H
) can be written as
ψ

1
◦
g
◦
ψ
for a conformal map in
Aut
(
D
(0
,
1)). By the classiﬁcation of
Aut
(
D
(0
,
1)) we know that
g
is a fractional linear transformation, and so
is
ψ
. Thus all elements of
Aut
(
H
) are fractional linear transformations.
Moreover we see that an element of
Aut
(
H
) must preserve
∂
H
=
R
. Let
us therefore consider the FLTs which preserve
R
. They must satisfy the
following equation for all
z
∈
R
az
+
b
cz
+
d
=
az
+
b
cz
+
d
=
¯
az