Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 2A (Complex Analysis or Complex Methods)

ChooseIt is holomorphic on the right half-plane with derivative $1/z\ne 0$, and its inverse $z=e^w$ has positive real part precisely for $|\mathrm{Im}w|<\pi /2$. Thus this branch of the complex logarithm is a conformal bijection onto $S$.

For $\mathrm{Re}z>0$,so $w=(z-1)/(z+1)$ lies in the unit disc. Its inverse is $z=(1+w)/(1-w)$, whose real part is $(1-|w{|}^2)/|1-w{|}^2>0$. Both derivatives are nonzero, proving this Möbius transformation is conformal.

Composing the maps givesa conformal bijection from $S$ to the unit disc.