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?
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1 Answer
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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 $\alpha\in \mathbb D$.
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$\begingroup$ No I am sure about the minus sign. Are there any other known conformal maps from $\mathbb{H}$ to $\mathbb{D}$ rather than $G$? $\endgroup$– RuzayqatCommented Mar 9, 2014 at 9:18
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1$\begingroup$ Um..... I am a bit unsure about your $G$. Let's use another one. $\endgroup$– user99914Commented Mar 9, 2014 at 9:34
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1$\begingroup$ @Hamza: I use another $G$, please see the answer. $\endgroup$– user99914Commented Mar 9, 2014 at 9:49
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1$\begingroup$ In general if $z_1, z_2$, $w_1, w_2$ are arbitrary, there might not be $f$ with that property.@mnmn1993 $\endgroup$– user99914Commented Dec 14, 2017 at 16:43
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1$\begingroup$ No that's still not true in general, you need that they $d_H(z_1, z_2) = d_D(w_1, w_2)$ @mnmn1993 $\endgroup$– user99914Commented Dec 18, 2017 at 16:14
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