TITLE: Complex Conformal Mapping from $\{z\in \mathbb{C} :|z|>1, Re(z)>0\}$ to the unit disc.
QUESTION [0 upvotes]: 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.

REPLY [0 votes]: 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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