= Solution
{c}
Suppose $w\in\mathbb C$ were omitted by $g$. Then
$$
F(z)=\frac1{g(z)-w}
$$
would be an <entire function>. The <reverse triangle inequality> gives $|g(z)-w|\geq|g(z)|-|w|\to\infty$ as $|z|\to\infty$, so $F(z)\to0$ 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 $F\equiv0$.
That is impossible, since $1/(g(z)-w)$ is nonzero at every finite $z$. No value $w$ can be omitted, and hence
$$
\boxed{g\text{ is surjective onto }\mathbb C.}
$$
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