Write down a formula that gives a conformal equivalence between the vertical half-strip \[ \mathcal{S}=\{z\in\mathbb{C}:\ 0<\operatorname{Re}(z)<1,\ \operatorname{Im}(z)>0\} \] and the upper half-plane \[ \mathbb{H}=\{z\in\mathbb{C}:\operatorname{Im}(z)>0\}. \]
Solution
The map \[ \boxed{F(z)=-\cos(\pi z)} \] is a conformal equivalence from \(\mathcal S\) onto \(\mathbb H\).
Indeed, write \(z=x+iy\) with \(0<x<1\) and \(y>0\). Then \[ \operatorname{Im}F(z)=\sin(\pi x)\sinh(\pi y)>0, \] so \(F(\mathcal S)\subset\mathbb H\).
To see the geometry and bijectivity, factor \(F\) as \[ z\longmapsto \zeta=e^{\pi i z} \longmapsto -\frac12\left(\zeta+\frac1\zeta\right). \] The exponential maps \(\mathcal S\) biholomorphically onto the upper half of the unit disk: \(0<\arg\zeta<\pi\) and \(0<|\zeta|<1\). The second map is the standard Joukowski map carrying that upper half-disk biholomorphically onto \(\mathbb H\). Their composite is \(-\cos(\pi z)\).