Proposition 27.2.3.label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be connected, and $f \in H(U; E) \setminus \bracs{0}$, then for each $z_{0} \in \bracs{f = 0}$, there exists $n \ge 1$ and $g \in H(U; \complex)$ such that:
- (1)
$g(a) \ne 0$.
- (2)
For each $z \in U$, $f(z) = (z - z_{0})^{n}g(z)$.
Proof. Since $f \ne 0$, $n = \min\bracs{n \in \natp| D^nf(z_0) \ne 0}< \infty$ by Proposition 27.2.2. Let
then $g$ is continuous on $U$ and holomorphic on $U \setminus \bracs{z_0}$. As $f \in H(U; E)$, there exists $r > 0$ and $\bracs{a_k}_{0}^{\infty} \subset E$ such that $f(z) = \sum_{k = 0}^{\infty} a_{k} (z - z_{0})^{k}$ for all $z \in B(z_{0}, r)$, where the series has a radius of convergence of at least $r$. By assumption on $n$, for each $z \in B(z_{0}, r)$,
where the series $\sum_{k = 0}^{\infty} a_{k+n}(z - z_{0})^{k}$ has the same radius of convergence. Therefore $g$ is holomorphic at $z_{0}$.$\square$
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