---
tag: 0K83KEJ43P
kind: statement
role: theorem
name: Isolated zeros theorem
aliases: [principle-of-isolated-zeros, zeros-are-isolated]
msc: [30C15]
uses: [0DG0Z72CQD, 0826SGV3R0, 064N002TX4]
sources:
  - url: https://en.wikipedia.org/wiki/Identity_theorem
    locator: Lemma
---

## Statement

Let $U \subseteq \mathbb{C}$ be an open set, let $f : U \to \mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] 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|064N002TX4]] of $f$ at $z_0$. By continuity, $g \neq 0$ near $z_0$.
On a [[domain|097KNJFJ7V]], 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 $0$.