= Solution
\b[The assertion is true.] The assumed finite <multiplicity> means that $f_0-w_1$ is not identically zero and has an isolated zero of order $k$ at $z_1$. Choose $\rho>0$ so that $\overline{D(z_1,\rho)}\subset D(0,r)$ and $f_0-w_1$ has no other zeros in this closed disc. On its boundary,
$$
\eta=\min_{|z-z_1|=\rho}|f_0(z)-w_1|>0.
$$
The joint <holomorphy> established above gives $f_\alpha\to f_0$ uniformly on that circle as $\alpha\to0$. For sufficiently small $|\alpha|$, therefore, $|f_\alpha-f_0|<\eta$. By <Rouché's theorem>, $f_\alpha-w_1$ has exactly $k$ zeros in $D(z_1,\rho)$, counted with <multiplicity>. In particular it has at least one zero in $D(0,r)$. This preserves the total local zero count, without asserting that the zeros remain together.
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