Mapping the upper half plane conformally onto a semi-infinite strip, Map the upper half y>0 of the z-plane conformally onto the semi-infinite strip u>0, $-\pi<v<\pi$ in the w-plane.
I would like some hints for now, please.  
I'm not sure how to even get started on this problem, to be honest.  
Thanks,
 A: Let $H$ be the half plane $\{z \in \mathbb{C} \ | \ \Im z > 0 \}$ and :
\begin{align}
\cosh : \mathbb{C} & \rightarrow \mathbb{C} \\
z & \mapsto \frac{\exp(z) + \exp(-z)}{2}.
\end{align}
We have :
\begin{equation}
\forall z = x + iy \in \mathbb{C}, \cosh z = \cosh x \cos y + i \sinh x \sin y.\qquad(1)
\end{equation}
Thus, the preimage of $H$ under $\cosh$ is an union of semi-infinite strips :
\begin{equation}
\cosh^{-1} (H) = (\bigcup_{n \in 2 \mathbb{Z}} \{z \in \mathbb{C} \ | \ \Re z > 0 ; n \pi < \Im z < (n + 1) \pi\}) \cup (\bigcup_{n \in 2 \mathbb{Z}} \{z \in \mathbb{C} \ | \ \Re z < 0 ; (n - 1) \pi < \Im z < n \pi\})
\end{equation}
Let $S$ be the semi-infinite strip $\{z \in \mathbb{C} \ | \ \Re z > 0 ; 0 < \Im z < \pi\}$. I'll show that the restriction $\cosh_{|S}$ of $\cosh$ to $S$ is a biholomorphism onto $H$, from what your problem is easy to solve.
First, let's show that $\cosh_{|S}$ is injective. Let $z, z' \in S$ such that $\cosh z = \cosh z'$. We have $\cosh^2 z = \cosh^2 z'$, so $\sinh^2 z = \sinh^2 z'$, and finally $\sinh z = \pm \sinh z'$. If $\sinh z = \sinh z'$, since $\cosh z = \cosh z'$, we have $\exp z = \exp z'$, and since $z \in S$ and $ z' \in S$, this implies $z = z'$. If $\sinh z = - \sinh z'$, we have $\exp z = \exp (-z')$, which is impossible since $\Re z > 0$ and $\Re z' > 0$. Thus $\cosh_{|S}$ is injective.
Next, let's show that $\cosh_{|S}$ is surjective. First, being a holomorphic function, its image is open. Then, thanks to the formula (1), $\cosh$ sends the boundary of $S$ onto the boundary of $H$, and tends towards infinity as its argument grows inside $S$. Thus $\cosh_{|S}$ is a proper function, which implies that its image is closed. Since $H$ is connected, any subset of $H$ which is non-empty, open and closed must equal $H$. Finally, $\cosh_{|S}$ is surjective, and the proof is finished.
