Find biholomorphic function with certain property I'm working on the current problem:


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*Find a biholomorphic function $\varphi: S\to H$ where $S=\{z\in\mathbb C| 0<\Re(z)<1\}$ and $H=\{z\in \mathbb C| \Im(z)>0\}$.


*Find a biholomorphic function $\tilde\varphi: S\to S$ such that $\tilde\varphi(1/2)=1/4$.

For 1. I have an idea: First, I rotate $S$ via $\varphi_1(z)=iz$ in the complex plane to a strip that selects the same elements as $S$ just with the imaginary part. Then, by $\varphi_2(z)=\pi z$ I stretch that strip to be $\{z\in\mathbb C| 0<\Im(z)<\pi\}$. And finally, $\varphi_3(z)=e^z$ should turn that strip into the upper half-plane $H$. Thus $\varphi=(\varphi_3\circ\varphi_2\circ\varphi_1) (z)=e^{i\pi z}$ should be a biholomorphic function from $S$ to $H$.
For 2. however I'm not sure. I was thinking to go from $S$ to the unit disk $\mathbb E$ by some $\phi$ and then use a certain Moebius transform that ensures $1/2\mapsto 1/4$ and then taking $\phi^{-1}$ to go back to $S$. But neither am I sure that in the end the property still holds for the composition nor how to construct such a Moebius transform. Does it have something to do with the first part of the problem?

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*Yes. Perhaps 'conformal mapping' is a more common term.

*Yes. By the uniformization theorem you can (and should!) do just as you say. The composition of conformal mappings is conformal (analytic and with nonzero derivative). Now to map the disk into itself, use $z\mapsto \frac{z-a}{1-\overline{a}z} e^{i \theta}$ where $\theta\in\mathbb{R},\, |a|<1$. 
