TITLE: What does "vanish identically" mean?
QUESTION [5 upvotes]: 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 $\Omega$.

REPLY [2 votes]: If $f\equiv 0$ in a full neighborhood of $z_0,$ then $f\equiv 0$ in $\Omega$ from the identity principle, a contradiction.