Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 11 (Complex Analysis OR Complex Methods)

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.

Off the slit choose the stated analytic square-root branch. Differentiation gives $f\prime (z)=c^{-1}(1+z/\sqrt{z^2-c^2})$, which does not vanish there, so the map is conformal.

Multiplying $(z+\sqrt{z^2-c^2})(z-\sqrt{z^2-c^2})=c^2$ gives the reciprocal identity and hence $z=c(\zeta +{\zeta }^{-1})/2$.

For $\zeta =R{e}^{i\theta }$,
$x=\frac{c}{2}(R+R^{-1})\cos \theta$ and $y=\frac{c}{2}(R-R^{-1})\sin \theta$. Thus $a=c(R+R^{-1})/2$, $b=c(R-R^{-1})/2$. Nested level curves and the behaviour at infinity show that exteriors correspond.

The Möbius map sends $|\zeta |=R$ to the imaginary axis and $\infty$ to $1$; testing one exterior point shows that $|\zeta |>R$ maps to $\Re w>0$.

The required harmonic function isOn the upper and lower boundary its argument is respectively $\pi /2$ and $-\pi /2$, while at infinity the logarithm tends to $0$.