= Solution
For a closed piecewise <smooth curve> $\gamma$ avoiding $w$, its <winding number> about $w$ is
$$
\boxed{
\operatorname{wind}(\gamma,w)
=\frac1{2\pi i}\int_\gamma\frac{dz}{z-w}}.
$$
The <argument principle> says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic <function> $F$ has no zeros or poles on $\gamma$, then
$$
\boxed{
\frac1{2\pi i}\int_\gamma\frac{F'(z)}{F(z)}\,dz
=N_U(F)-P_U(F)},
$$
where zeros and poles are counted with multiplicity.
Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline U$ and
$$
|G(z)|<|F(z)|
\qquad(z\in\gamma).
$$
For $0\leq t\leq1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus <Rouché's theorem> states that
$$
\boxed{F\text{ and }F+G\text{ have the same number of zeros in }U}.
$$
Now let $w_0=f(z_0)$. Since $|w_0|<1\leq|f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. Therefore
$$
\boxed{\mathbb D\subseteq f(\mathbb D)}.
$$
This is the <unit-disc image from a boundary modulus lower bound>.
Finally, take a sufficiently small positively oriented circle $C$ around zero. The residue
$$
k=\operatorname{res}_{0}\frac{g'}g
=\frac1{2\pi i}\int_C\frac{g'}g\,dz
$$
is the winding number of $g(C)$ around zero, so $k\in\mathbb Z$. The pole is simple, hence its residue is nonzero and $k\ne0$. For
$$
h(z)=z^{-k}g(z)
$$
we have
$$
\frac{h'}h=\frac{g'}g-\frac{k}{z}.
$$
The second term cancels the complete principal part at zero, so the <integer residue of a logarithmic derivative> proves that $h'/h$ has a removable singularity there.
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