Conformal Mappings¶

A bijective holomorphic function \(f: U\rightarrow V\) is called a conformal map or biholomorphism. Given such a mapping \(f\), we call \(U\) and \(V\) are conformally equivalent, or simply biholomorphic.

If A holomorphic function is injective, whose inverse is automatically holomorphic.

Holomorphism of the inverse of a conformal mapping

Assume \(f:U\rightarrow V\) is a holomorphic function which is injective, then \(f'(z)\neq 0\) for all \(z\in U\).

Proof

Show by contrsdiction.

\(\square\)

The disc & Upper half-plane¶

Denote the upper half-plane \(\mathbb{H}\) to be

\[ \mathbb{H}=\{z\in \mathbb{C}:\text{Im}(z)>0\} \]

Conformal function between disc & upper half-plane

Let

\[ F(w)=\frac{i-z}{i+z},\quad G(w)=i\frac{1-w}{1+w} \]

Then \(F: \mathbb{H}\rightarrow \mathbb{D}\) is a conformal map with inverse \(G: \mathbb{D}\rightarrow \mathbb{H}\).

Mappings of the form

\[ z\mapsto \frac{az+b}{cz+d} \]

where \(a,b,c,d\) are complex-valued number, and \(bc\neq ad\), are called fractional linear transformations.

Schwarz Lemma¶

Schwarz Lemma

Assume \(f:\mathbb{D} \rightarrow \mathbb{D}\) and \(f(0)=0\), then

(i) \(|f(z)|\leq |z|\) holds for all \(z\in \mathbb{D}\),

(ii) If there exists \(z_0\in \mathbb{D}\) such that \(f(z_0)=z_0\), then \(f\) is a rotation.

(iii) \(|f'(0)|\leq 1\), and if the equation holds, then \(f\) is a rotation.

Proof

\[ \frac{|f(z)|}{|z|}\leq 1 \]

A conformal map from an open set \(\Omega\) to itself is called Automorphism. The set of all automorphisim of \(\Omega\) is called \(Aut(\Omega)\), and carries the structure of group.

Automorphisms of the disc¶

In homework, we have known that

\[ F_\alpha (z)=\frac{\alpha -z}{1-\overline{\alpha}z},\quad \alpha \in \mathbb{D} \]

is a bijection from \(\mathbb{D}\) to itself, and interchanges \(0\) and \(\alpha\) by

\[ F_\alpha(0)=\alpha,\quad F_\alpha(\alpha)=0. \]

Expression of all automorphism of disc

If \(f\) is a automorphism of the disc, then there exist \(\theta\in \mathbb{R}\) and \(\alpha \in \mathbb{D}\), such that

\[ f(z)=e^{i\theta}\frac{\alpha -z}{1-\overline{\alpha }z}. \]

Proof

Using Schwarz Lemma.

\(\square\)

Automorphism of the upper half-plane¶

By using Conformal function between disc & upper half-plane, we could determine the automorphism of the upper half-plane \(\text{Aut}(\mathbb{H})\).