Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 4G / Solution

Use the one-sided product bound for an entire function by consideringThis is an entire function, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\le M$, then $|F|\le e^{2M}$ everywhere. The Liouville theorem therefore makes $F$ constant.

Because an exponential never vanishes, differentiating yields $0=F^{\prime }=-2if{f}^{\prime }F$, and hence $(f^2{)}^{\prime }=2f{f}^{\prime }=0$. Thus $f^2$ is constant. If that constant is zero, $f=0$; otherwise $f$ takes values in the two square roots of one nonzero constant. Continuity on the connected space $\mathbb{C}$ prevents switching between those two values. Therefore $f$ is constant.