Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 4 / 4F (Complex Analysis)

The Cauchy integral formula and its differentiated form areIf $f$ is entire and $|f|\le M$, the Cauchy estimate gives $|f^{\prime }(a)|\le M/\rho$. Letting $\rho \to \infty$ yields $f^{\prime }(a)=0$ for every $a$. Thus $f$ is constant, which is Liouville theorem.

The inequality implies $u(u-1)\ge v^2\ge 0$, so $u\le 0$ or $u\ge 1$. Since $u$ is continuous and $\mathbb{C}$ is connected, its image lies wholly in one of these intervals. If $u\ge 1$, then $g$ has no zero and $|1/g|\le 1$; Liouville theorem makes $1/g$, hence $g$, constant. If $u\le 0$, then $|g-1|\ge 1$, so the same argument applied to $1/(g-1)$ proves that $g$ is constant.