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

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.

• From domain to upper half plane. – DD90 Jul 17 '19 at 0:51
• 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 $f =\frac{z-1}{z+1}$. – Parthiv Basu Jul 17 '19 at 1:10
• See part (2) of this answer: math.stackexchange.com/a/3285469/669152. If you still can't figure it out I can explain more. – trisct Jul 17 '19 at 1:52

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

• $z \mapsto z^2$ does not map $\{z: |z| > 1 \land \operatorname{Re} z > 0 \}$ to $\{z: |z| > 1 \land \operatorname{Im} z > 0\}$. – Maxim Jul 20 '19 at 23:20
• @Maxim you're right. I was thinking in the first quadrant only, I forgot the 4th quadrant to be considered. I'll fix it – user486983 Jul 21 '19 at 0:31