# What does “vanish identically” mean?

This is from Stein's Complex Analysis:

Suppose that $f$ is holomorphic in a connected open set $\Omega$, has a zero at a point $z_0 \in \Omega$, and does not vanish identicallly in $\Omega$. Then, there exists a neighborhood $U\in\Omega$ of $z_0$, a non-vanishing holomorphic function $g$ on $U$, and a unique positive integer $n$ such that $$f(z) = (z - z_0)^n g(z) \text{ for all } z\in U$$

I thought that $f$ vanishes identically in $\Omega$ means that $f(z) = 0$ for all $z\in \Omega$? But then in the proof, he wrote

Since $\Omega$ is connected and $f$ is not identically zero, we conclude that $f$ is not identically zero in a neighborhood of $z_0$.

Why? Why cannot $f$ be identically zero in a neighborhood of $z_0$? It does not contradict the fact that $f$ is not identically zero in $\Omega$.

• It is a theorem (probably proven somewhere in the book, but I don't own it) that if the set of zeros of $f$ has an accumulation point in its domain of definition, then it is identically zero in this domain. Also, you are right - "vanish identically", "identically zero" and "zero for all $z\in\Omega$" mean the same thing. – Wojowu Nov 25 '16 at 22:19
• Adding to Wojowu's comment: the mentioned theorem often goes by Identity theorem. – Hirshy Nov 25 '16 at 22:20

This is due to the Principle of isolated zeros of analytic functions: all zeroes of a (non identically zero) analytic function on a connected open set $U$ of $\mathbf C$ are isolated, i.e. for each zero $a\in U$ there exists an open neighbourhood $V\subset U$ of $a$ in which $a$ is the only zero of $f$.
If $f\equiv 0$ in a full neighborhood of $z_0,$ then $f\equiv 0$ in $\Omega$ from the identity principle, a contradiction.