•
f
:
U
→
V
is said to be
conformal
or
biholomoprhic
if
f
is holomorphic, one
toone and onto.
•
Examples of conformal mappings
:
1.
f
:
C
→
C
is conformal if and only if
f
is linear.
2.
f
:
D
(0
,
1)
→
D
(0
,
1) is conformal if and only if
f
is a Mobius transformation
(a consequence of SchwarzPick Lemma).
3.
f
:
{
z

Imz >
0
} →
D
(0
,
1) is conformal if and only if
f
(
z
) =
r
iθ
z

a
z

¯
a
,
Ima >
0
.
4.
f
:
{
z

Imz >
0
} → {
z

Imz >
0
}
is conformal if and only if
f
(
z
) =
az
+
b
cz
+
d
,
where
a, b, c, d
are real numbers, and
ad

bc >
0.
17
•
Normal Families
:
–
Let
F
=
{
f
α
}
α
∈
Λ
be a family of holomorphic functions on
U
⊂
C
. We say
that
F
is
normal family
if every sequence
{
f
n
} ⊂ F
has a sequence that
converges uniformly on every compact subsets of
U
.
–
Montel’s Theorem
Let
F
=
{
f
α
}
α
∈
Λ
be a family of holomorphic functions
on
U
⊂
C
that is bounded on compact subsets, i.e. for every compact set
K
⊂
U
, there is a
M
K
>
0
such that

f
(
z
)
 ≤
M
k
,
z
∈
K, f
∈ F
.
Then
F
is normal.
•
Riemann Mapping Theorem (Analytic Version)
If
U
is a holomorphically simply
connected open set in
C
and
U
6
=
C
, then
U
is conformally equivalent to the
unit disc
.
Idea of pf
: Set, for a fixed
P
∈
U
, 1.
F
=
{
f

f
:
U
→
D
(0
,
1) is holomorphic, and onetoone
, f
(
P
) = 0
}
.
2. Use the normality to show that there is
f
0
∈ F
such that

f
0
0
(
P
)

= sup
f
∈F

f
0
(
P
)

.
3. Show that
f
0
maps
U
onto
the unit disc
D
(0
,
1).
•
If
U
is simply connected, then
U
is a holomorphically simply connected.
•
So the
Riemann Mapping Theorem
:
If
U
is simply connected and
U
6
=
C
. Then
U
is conformally equivalent to the unit disc
. So the beauty of this theorem is
that it basically says that
the topological property of
U
implies the analytic
property of
U
.
•
Little Picard Theorem (Generalization of Liouville’s theorem)
If
f
is entire and
omits 2 values, then it is constant
.
Idea of pf
:
Step 1
.
From omitting 2 values to omitting a lot of values (from
spotting 2 cockroaches to a lot of cockroaches), this can be achieved by proving
the following lemma:
If
U
is simply connected and
f
is holomorphic on
U
and
does not assume the values of
0
and
1
. Then there is a holomorphic curve on
U
such that
f
(
z
) =

exp(
iπ
cosh[2
g
(
z
)])
.
18
Furthermore, it can be verified that
g
does not assume, on
U
, the values of
π
log(
√
n
+
√
n

1)+
1
2
mπi
for positive integer
n
and all integer
m
.
Since these
points from a vertices of a grid of rectangles with diagonal
<
2, we have that
g
(
U
) contains no disc of radius 1
.
Step 2
. We prove the socalled
Bloch’s theorem
Let
f
be analytic on a region containing
D
(0
,
1)
and
f
(0) =
0
, f
0
(0) = 1
. Then there
f
(
D
(0
,
1))
constains a disk of radius
1
/
72
(which is
independent of
f
!)
.
Step 3
. The proof of little picard theorem can thus be proved by considering
g
(
z
+
z
0
)
/g
0
(
z
0
) (by choosing some
z
0
with
g
0
(
z
0
)
6
= 0 and by Bloch’s theorem,
g
(
D
(
z
0
, R
)) contains a disc of redius (1
/
72)
R
, which contains a disc of radius 1
when
R
is bigger enough, which contradicts with that fact that
g
(
U
) contains
no disc of radius 1
derived in step 1.
The proof of Bloch’s theorem
: We first prove the following lemma:
Lemma
Let
f
be holomorphic on
D
(0
,
1)
with
f
(0) = 0
, f
0
(0) = 1
.
Assume that

f
(
z
)
 ≤
M
for all
z
∈
D
(0
,
1)
. Then
M
≥
1
and
f
(
D
(0
,
1))
⊃
D
(0
,
1
/
(6
M
))
.
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 Fall '08
 Staff
 Equations, Complex Numbers, dζ, holomorphic functions