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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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  • $\begingroup$ Do you know anything about conformal map from $\mathbb H \to \mathbb H$ or from $\mathbb D$ to $\mathbb D$? $\endgroup$ – user99914 Mar 9 '14 at 7:56
  • $\begingroup$ Yes I know about them. $\endgroup$ – Ruzayqat Mar 9 '14 at 7:58
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    $\begingroup$ Then you can first find ONE map $G$ from $\mathbb H$ to $\mathbb D$, then all others $F :\mathbb H \to D$ are of the form $F = f\circ G$, where $f:\mathbb D \to \mathbb D$. $\endgroup$ – user99914 Mar 9 '14 at 8:00
  • $\begingroup$ By considering $G=\dfrac{z-1}{z+1}$ and $f=e^{i\theta}\dfrac{\alpha-z}{1-\overline{\alpha}z}$ I get: $F=e^{i\theta}\dfrac{z(1-\alpha)-(1+\alpha)}{z(1-\overline{\alpha})+(1+\overline{\alpha})}$ $\endgroup$ – Ruzayqat Mar 9 '14 at 8:14
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    $\begingroup$ A little tip: instead of using \,\,\,\,\,\,\, to adjust horizontal spacing, you can use \quad and \qquad. $\endgroup$ – TZakrevskiy Mar 9 '14 at 9:52
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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 $G$? $\endgroup$ – Ruzayqat Mar 9 '14 at 9:18
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    $\begingroup$ Um..... I am a bit unsure about your $G$. Let's use another one. $\endgroup$ – user99914 Mar 9 '14 at 9:34
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    $\begingroup$ @Hamza: I use another $G$, please see the answer. $\endgroup$ – user99914 Mar 9 '14 at 9:49
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    $\begingroup$ In general if $z_1, z_2$, $w_1, w_2$ are arbitrary, there might not be $f$ with that property.@mnmn1993 $\endgroup$ – user99914 Dec 14 '17 at 16:43
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    $\begingroup$ 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 $\endgroup$ – user99914 Dec 18 '17 at 16:14

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