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Find a conformal map that maps the planar domain, $\Omega:=\{z\in \mathbb{C} :|z|>1, Re(z)>0\}$ to the unit disk $\mathbb{D}$.

I tried this first using the composition of the map $z^2$ and then the map $\frac{1}{z}$. But it did not work. Now I am thinking of using the map from upper half plane to the unit disc. But I do not know how to get the upper half plane using the given domain. If anyone has an idea please comment.

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  • $\begingroup$ From domain to upper half plane. $\endgroup$ – DD90 Jul 17 at 0:51
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    $\begingroup$ I am not sure if I understand the question correctly. But if you want a map $f : \Omega \to \mathbb{D}$ such that $f$ is conformal, then you could just take the Möbius transformation $f =\frac{z-1}{z+1}$. $\endgroup$ – Parthiv Basu Jul 17 at 1:10
  • $\begingroup$ See part (2) of this answer: math.stackexchange.com/a/3285469/669152. If you still can't figure it out I can explain more. $\endgroup$ – trisct Jul 17 at 1:52
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Hint

Consider this three transformations

$T_1(z)=iz,$ which maps from $\{z\in \mathbb{C} :|z|>1, Re(z)>0\} $ to $\{z\in \mathbb{C} :|z|>1, Im(z)>0\}.$

$T_2(z)=\frac 12(z+\frac 1z),$ which maps from $\{z\in \mathbb{C} :|z|>1, Im(z)>0\} $ to the upper half plane.

and

$T_3(z)=\frac{z-i}{z+i},$ which maps from the upper half plane to the unit circle.

and take $T=T_3\circ T_2\circ T_1$

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  • $\begingroup$ $z \mapsto z^2$ does not map $\{z: |z| > 1 \land \operatorname{Re} z > 0 \}$ to $\{z: |z| > 1 \land \operatorname{Im} z > 0\}$. $\endgroup$ – Maxim Jul 20 at 23:20
  • $\begingroup$ @Maxim you're right. I was thinking in the first quadrant only, I forgot the 4th quadrant to be considered. I'll fix it $\endgroup$ – user486983 Jul 21 at 0:31

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