Conformal maps from the upper half-plane to the unit disc has the form

Prove that the conformal maps from the upper half-plane $\mathbb{H}$ to the unit disc $\mathbb{D}$ has the form

$$e^{i\theta}\dfrac{z-\beta}{z-\overline{\beta}},\quad\theta \in \mathbb{R} \text { and }\beta \in \mathbb{H}.$$

Any hints?

Solution 1:

One candidate of $G$ will be

$$G(z) = \frac{z-i}{z+i}$$

(which is the case when $\theta=0$ and $\beta=i$). Then

$$F(z) = (f\circ G)(z)= e^{i\theta_1} \frac{z(1-\alpha) - i(1+ \alpha)}{z(1-\bar \alpha) + i(1+\bar\alpha)}$$ $$\ \ \ \ \ \ \ \ \ = e^{i\theta_1} \frac{1-\alpha}{1-\bar \alpha}\frac{z -\beta}{z - \bar\beta}\ ,$$

where

$$\beta = \frac{i(1+\alpha)}{1-\alpha}$$

$$\bigg| \frac{1-\alpha}{1-\bar \alpha}\bigg|=1\Rightarrow \frac{1-\alpha}{1-\bar \alpha} = e^{i\psi}$$

for some $\psi$. Then

$$F(z) = e^{i\theta} \frac{z -\beta}{z - \bar\beta}\ ,$$

where $\theta = \theta_1 + \psi$. To be complete let me also check $\beta\in \mathbb H$: let $\alpha = a+ bi$, then

$$\beta = \frac{-2b +(1-a^2-b^2)i}{|1-\alpha|^2} \in \mathbb H$$

as $|\alpha|^2 = a^2+ b^2 <1$ as $\alpha\in \mathbb D$.