Find the conformal map that takes the upper half-plane comformally onto the half-strip \(\{ w=x+iy:\ -\pi/2<x<\pi/2\ y>0\}\).
Solution
Let \[ S=\left\{w:\ -\frac\pi2<\operatorname{Re}w<\frac\pi2, \ \operatorname{Im}w>0\right\}. \] Consider \[ \Phi(w)=\sin w. \] If \(w=x+iy\in S\), then \[ \operatorname{Im}\sin w=\cos x\,\sinh y>0, \] so \(\Phi(S)\subset\mathbb H\).
The three boundary pieces map to the real axis as follows: \[ \sin x\in(-1,1)\quad(y=0), \] while \[ \sin\left(\frac\pi2+iy\right)=\cosh y\in(1,\infty), \qquad \sin\left(-\frac\pi2+iy\right)=-\cosh y\in(-\infty,-1). \] Moreover \(\Phi'(w)=\cos w\) has no zero in \(S\). The standard identity \[ \sin w_1=\sin w_2 \iff w_1-w_2\in2\pi\mathbb Z \quad\text{or}\quad w_1+w_2\in\pi+2\pi\mathbb Z \] shows that neither alternative can occur for two distinct points of \(S\). Thus \(\Phi\) is injective. The boundary correspondence, or equivalently the argument principle applied on exhausting rectangles in \(S\), shows that its image is all of \(\mathbb H\).
Hence \(\sin:S\to\mathbb H\) is biholomorphic, and the required map is its inverse branch \[ \boxed{z\longmapsto\arcsin z,} \] chosen so that \(-\pi/2<\operatorname{Re}(\arcsin z)<\pi/2\) and \(\operatorname{Im}(\arcsin z)>0\) for \(z\in\mathbb H\).