Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 12 / 12.2A

For distinct $z,w$ in the convex disc, integrate along the segment:The second term has modulus strictly below $|f^{\prime }(z_0)||z-w|$, so the sum cannot vanish. Thus $f$ is one-to-one.

Harmonic means twice continuously differentiable with $u_{xx}+u_{yy}=0$. On the simply connected plane, choose an entire harmonic conjugate so $F=u+iv$ is entire. If $u\ge 0$, then $e^{-F}$ is bounded; Liouville's theorem makes it, and hence $u$, constant.

Choose an entire $F$ with real part $u$. The bound implies that $e^F$ has at most polynomial growth, so Cauchy's estimates make it a polynomial. It has no zeros, hence that polynomial is constant. Therefore $F$, and in particular $u$, is constant.