Solution
= Solution
The inequality implies $u(u-1)\geq v^2\geq0$, so $u\leq0$ or $u\geq1$. Since $u$ is continuous and $\mathbb C$ is <connected>, its image lies wholly in one of these intervals. If $u\geq1$, then $g$ has no zero and $|1/g|\leq1$; <Liouville theorem> makes $1/g$, hence $g$, constant. If $u\leq0$, then $|g-1|\geq1$, so the same argument applied to $1/(g-1)$ proves that \b[$g$ is constant].