[proofplan] We combine [Rouche's theorem](/theorems/357) (which counts zeros with multiplicity) with a perturbation argument showing that, for $w \neq f(a)$ sufficiently close to $f(a)$, no zero of $f - w$ can be a multiple zero. The key observation is that a multiple zero of $f - w$ must simultaneously satisfy $f(z) = w$ and $f'(z) = 0$. Since the zeros of $f'$ near $a$ are isolated, shrinking $r$ avoids them for the perturbed equation. [/proofplan] [step:Set up the local factorisation and count zeros with multiplicity via Rouche's theorem] Write $f(z) - f(a) = (z - a)^k h(z)$ with $h$ holomorphic and $h(a) \neq 0$. Choose $r > 0$ small enough that: 1. $\overline{B(a, r)} \subseteq U$, 2. $h(z) \neq 0$ on $\overline{B(a, r)}$, 3. $a$ is the only zero of $f - f(a)$ in $\overline{B(a, r)}$. Define $\delta = \min_{|z-a|=r} |f(z) - f(a)| > 0$. By [Rouche's theorem](/theorems/357) (as in the proof of the [Open Mapping Theorem](/theorems/358)), for every $w$ with $|w - f(a)| < \delta$, the function $f - w$ has exactly $k$ zeros in $B(a, r)$, counted with multiplicity. [/step] [step:Show the $k$ zeros are distinct for $w \neq f(a)$ by avoiding critical points of $f$] A zero $z^*$ of $f - w$ has multiplicity $\geq 2$ if and only if $f(z^*) = w$ and $f'(z^*) = 0$ simultaneously (since $f(z) - w$ has a zero of order $\geq 2$ at $z^*$ precisely when both $f(z^*) - w = 0$ and $f'(z^*) = 0$). Since $f$ is non-constant, $f' \not\equiv 0$, so the zeros of $f'$ in $\overline{B(a, r)}$ are isolated. By shrinking $r$ if necessary (and redefining $\delta$ accordingly), we may assume that $f'$ has no zeros in $\overline{B(a, r)} \setminus \{a\}$. For $w \neq f(a)$ with $|w - f(a)| < \delta$, the $k$ zeros of $f - w$ in $B(a, r)$ are all distinct from $a$ (since $f(a) \neq w$). These zeros lie in $B(a, r) \setminus \{a\}$, where $f' \neq 0$. Therefore no zero of $f - w$ can be a simultaneous zero of $f'$, so every zero of $f - w$ is simple. Since all $k$ zeros are simple, they are distinct. [guided] Why must a multiple zero of $f - w$ also be a zero of $f'$? If $f(z^*) = w$ and $f$ has a zero of order $m \geq 2$ at $z^*$ for the function $f - w$, then $f(z) - w = (z - z^*)^m \tilde{h}(z)$ with $\tilde{h}(z^*) \neq 0$. Differentiating: $f'(z) = m(z - z^*)^{m-1} \tilde{h}(z) + (z - z^*)^m \tilde{h}'(z)$, so $f'(z^*) = 0$. Conversely, if $f(z^*) = w$ and $f'(z^*) \neq 0$, then $f - w$ has a simple zero at $z^*$ (the first derivative does not vanish). So the zeros of $f'$ are precisely the locations where multiple zeros can occur. By ensuring that the only possible critical point ($f' = 0$) in $B(a, r)$ is $a$ itself, and noting that $f(a) \neq w$ for $w \neq f(a)$, we guarantee that none of the $k$ roots of $f(z) = w$ in $B(a, r)$ can be multiple. [/guided] [/step]