= Hyperbolic-cosine map of a half-strip
{title2=$\zeta=\cosh(\pi z/a)$}
For $a>0$, this <conformal map> takes $\operatorname{Re}z>0$, $0<\operatorname{Im}z<a$ 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 $|s|>1$, $\operatorname{Im}s>0$. The equation $s+s^{-1}=2\zeta$ has exactly one root outside the <unit circle> when $\operatorname{Im}\zeta>0$: the roots have product one and neither has unit modulus. The identity $\operatorname{Im}\zeta=(1-|s|^{-2})\operatorname{Im}s/2$ 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>.
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