Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 3 / c

Write $\alpha =a+ib$ with $b>0$. ThenIt preserves the upper half-plane, has inverse $(z-a)/b$, and sends $i$ to $\alpha$. Its nonzero derivative $b$ verifies that this bijection is conformal.

The map $h(z)=e^{\pi z/2}$ is a conformal bijection from the width-two strip to the upper half-plane, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.

Write $e^{\pi \beta /2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:Every argument of $L$ lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, andThe final equality uses $0<\mathrm{Im}(\pi \beta /2)<\pi$, so no multiple of $2\pi i$ is lost. This is a strip automorphism with a prescribed image of its midpoint.