Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 5 / a / iii

Suppose $w\in \mathbb{C}$ were omitted by $g$. Thenwould be an entire function. The reverse triangle inequality gives $|g(z)-w|\ge |g(z)|-|w|\to \infty$ as $|z|\to \infty$, so $F(z)\to 0$ there. On a sufficiently large closed disk, $F$ is bounded by continuity and the extreme value theorem; outside it, the limit gives a bound. The Liouville theorem therefore makes $F$ constant, and the limit forces $F\equiv 0$.

That is impossible, since $1/(g(z)-w)$ is nonzero at every finite $z$. No value $w$ can be omitted, and hence