Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 15C / Solution

Choose the principal branch of the complex logarithm on $H$, with $-\pi <\arg z<\pi$. ThenThus $-\pi <\mathrm{Re}w<\pi$ and the imaginary part ranges over all real numbers. Conversely, for $w=x+iy$ in the strip, $z=e^{iw}=e^{-y}{e}^{ix}$ is in $H$ and its principal logarithm is $iw$. Therefore $\boxed{w=-i\log z:H⟶S}$ is an analytic bijection, with analytic inverse $z=e^{iw}$. Every other continuous branch on the connected domain is $\log z+2\pi ik$ for some fixed integer $k$, so its image is the translated strip $\boxed{S+2\pi k}$.

For $G$, introduce the Möbius transformation $\zeta =(z-1)/(z+1)$, whose inverse is $z=(1+\zeta )/(1-\zeta )$. Writing $\zeta =a+ib$, its defining circle inequalities transform as follows:andHence $G$ maps bijectively onto the wedge $\pi /2<\arg \zeta <3\pi /4$. The principal branch of the complex logarithm is analytic throughout this wedge. Multiplying its logarithm by $-8i$ expands the interval of real parts to $(4\pi ,6\pi )$, and subtracting $5\pi$ centers it at zero. The required conformal map is thereforeIts real part is $8\arg \zeta -5\pi \in (-\pi ,\pi )$, and its imaginary part $-8\log |\zeta |$ takes every real value. More explicitly, its inverse on $S$ is obtained from $\zeta =\exp (i(W+5\pi )/8)$ and $z=(1+\zeta )/(1-\zeta )$, which verifies both surjectivity and injectivity.