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This question already has an answer here:

Problem: Find a conformal map from $\left\{ z \in \mathbb{C} \mid -1 < \text{Re} (z) < 1 \right\}$ to the unit disk.

Attempt: What I did, was to first shift everything to the right with one unit. So I used the map $F_1(z) = z + 1$. Then I rotated this vertical strip into a horizontal strip by 90 degrees, using the map $F_2(z) = i z$.

Now I have the infinite horizontal strip which goes from $0i$ to $2i$ in vertical direction. I wanted to map this into the unit disk.

I would use $F_3(z) = \frac{z - i}{z + i}$. But I'm not sure if this will give me a bijection, because I know already that this function maps the whole upper plane into the unit disk. And now I only have a part of the upper half plane.

Is there any way I can 'extend' my infinite horizontal strip to the whole upper half plane?

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marked as duplicate by Nosrati, Brahadeesh, hardmath, José Carlos Santos, Andres Mejia Aug 18 '18 at 18:45

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You made a good start. First rotate by the function $f_1(z)=iz$ to get to $\{-1<Im(z)<1\}$. Then you can use $f_2(z)=z+i$ to get to $\{0<Im(z)<2\}$. Then using $f_3(z)=\pi z$ will take you to $\{0<Im(z)<2\pi\}$. And now it's time to use the exponential function. $f_4(z)=e^z$ will take you to the domain $K=\{z \in \mathbb{C}:0<arg(z)<2\pi\}$, I hope you understand why. Now, in $K$ there is a branch of logarithm and hence a branch of square root. So use $f_5(z)=\sqrt z$ and you will get to $\{z \in \mathbb{C}:0<arg(z)<\pi\}$ which is the upper half plane.

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