 Solution

The assertion is false for $k\ge 2$, but true for $k=1$. For $k\ge 2$, take the jointly holomorphic right-hand side $A_{\alpha }(z,w)=k{z}^{k-1}+\alpha$. Its solution with zero initial value isAt $\alpha =0$, the value zero has multiplicity $k$ at the origin. For $\alpha \ne 0$, however, every zero is a simple zero: $f_{\alpha }^{\prime }(0)=\alpha$, and at a nonzero zero $z^{k-1}=-\alpha$,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 $\alpha$.