Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 11 / 11.1G

Apply the maximum modulus principle to $f(z)/z$, with its removable value $f\prime (0)$, to obtain $|f(z)|\le |z|$. Equality at a nonzero interior point makes $f(z)/z$ constant, so $f(z)=az$ with $|a|=1$.

Let $a=e^{i\pi /3}$ and $b=e^{-i\pi /3}$. The Möbius map $T(z)=(z-a)/(z-b)$ sends the two circular arcs through $a,b$ to rays bounding a wedge of angle $2\pi /3$. After choosing the branch and a unimodular constant so the interior has positive argument, $f(z)=e^{i\varphi }T(z{)}^{3/2}$ maps that wedge bijectively onto the upper half-plane.