Isolated zeros theorem

Statement

Let $U \subseteq \mathbb{C}$ be an open set, let $f : U \to \mathbb{C}$ be holomorphic on $U$, and let $z_0 \in U$. Then exactly one of these two cases holds.

1. There is $r > 0$ such that $f(z) = 0$ for each $z$ with $|z - z_0| < r$.
2. There is $r > 0$ such that $f(z) \neq 0$ for each $z$ with $0 < |z - z_0| < r$.

Remarks

Proof: if case 1 does not hold, then $f(z) = (z - z_0)^m g(z)$ near $z_0$ with $g(z_0) \neq 0$, where $m$ is the order of $f$ at $z_0$. By continuity, $g \neq 0$ near $z_0$. On a domain, it follows that the zeros of a holomorphic function that is not identically zero have no accumulation point in the domain. The real function $x \mapsto e^{-1/x^2}$, with value $0$ at $0$, is infinitely differentiable and has all derivatives $0$ at $0$. The theorem has no analogue for such real functions.

Explanations

L1 human

Near a point, a holomorphic function is either zero everywhere or zero only at that point. So the zeros of a nonzero holomorphic function cannot pile up inside its domain, unlike the zeros of $\sin(1/x)$ near $0$.