Problem statement: Show that $f(z) = -\cos(z)$ is a biholomorphism between the semi infinite strip $\{(x,y): 0<x<\pi, y>0\}$ to the upper half plane.
To show holomorphic, we just notice that the imaginary part magnitude of $e^{iz} = e^{-y}e^{ix}$ is strictly smaller than $e^{-iz} = e^{y}e^{-ix}$. But is there an easy way to show biholomorphism without using complicated injective/surjective argument?
Technique from Biholomorphic mapping of $\tan(z)$ does not seem to work in this case.
Best Answer
We have $f(z) = - \cos(z) = g(e^{iz})$ with $$ g(w) = -\frac{w^2+1}{2w} \, . $$ $z \mapsto e^{iz}$ is a biholomorphic mapping from the half-strip $\{(x,y): 0<x<\pi, y>0\}$ to the semi-disk $\{ z : |z| < 1, \operatorname{Im}(z) > 0 \}$. The rational function $$ g(w) = \frac{(w-1)^2+(w-1)^2}{(w-1)^2-(w-1)^2} = \frac{\left( \frac{w-1}{w+1}\right)^2+1}{\left( \frac{w-1}{w+1}\right)^2-1} $$ is a composition of “elementary” functions (Möbius transformations and squaring). These are biholomorphic mappings from the semi-disk onto the second quadrant, then onto the lower half-plane, and finally onto the upper half-plane.