= Solution
\b[The assertion is false for $k\ge2$, but true for $k=1$.] For $k\ge2$, take the jointly <holomorphic> right-hand side $A_\alpha(z,w)=kz^{k-1}+\alpha$. Its solution with zero initial value is
$$
f_\alpha(z)=z^k+\alpha z=z(z^{k-1}+\alpha).
$$
At $\alpha=0$, the value zero has <multiplicity> $k$ at the origin. For $\alpha\ne0$, however, every zero is a <simple zero>: $f_\alpha'(0)=\alpha$, and at a nonzero zero $z^{k-1}=-\alpha$,
$$
f_\alpha'(z)=kz^{k-1}+\alpha=-(k-1)\alpha\ne0.
$$
Thus there is no zero of <multiplicity> $k$ anywhere, even though the $k$ simple zeros stay close to zero for small $\alpha$. When $k=1$, part (a) gives exactly one zero counted with <multiplicity> in a small disc, so that zero is simple. Alternatively, the <holomorphic implicit function theorem> gives its <holomorphic> location as a function of $\alpha$.
Back to article page
