 Solution

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.