 Solution

The hypothesis says that $f$ has a zero of order $k$ at $a$, so writewith $g$ holomorphic near $a$. Differentiating givesChoose $\delta >0$ so small that the closed disc $\overline{D}(a,\delta )$ lies in the domain, $g$ never vanishes there, and the parenthesized factor never vanishes there. Thus $a$ is the only critical point of $f$ in that disc.

On the boundary circle, $f$ is nonzero. SetIf $0<|b|<\epsilon$, then $|b|<|f(z)|$ on the boundary. Rouché's theorem applied to $f$ and $f-b$ says that $f-b$ has the same number of zeros in $D(a,\delta )$ as $f$, namely $k$, counted with multiplicity.

None of these zeros is $a$ because $b\ne 0$, and none is a critical point because $f^{\prime }$ has no other zero in the disc. Every zero of $f-b$ is therefore simple. Hence there are exactly