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 $\Omega$.
Best Answer
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 $f$.