 Hyperbolic-cosine map of a half-strip

For $a>0$, this conformal map takes $\mathrm{Re}z>0$, $0<\mathrm{Im}z<a$ bijectively onto the upper half-plane. Its imaginary part is $\sinh (\pi x/a)\sin (\pi y/a)>0$, and its derivative is nonzero in the interior. The end wall maps to $[-1,1]$ and the two long walls to the complementary real rays. To prove the map is onto and injective, first put $s=e^{\pi z/a}$, which maps onto $|s|>1$, $\mathrm{Im}s>0$. The equation $s+s^{-1}=2\zeta$ has exactly one root outside the unit circle when $\mathrm{Im}\zeta >0$: the roots have product one and neither has unit modulus. The identity $\mathrm{Im}\zeta =(1-|s{|}^{-2})\mathrm{Im}s/2$ puts that root in the correct half-plane. This permits planar potential flow problems in a closed-end channel to be solved by the method of images.