---
tag: 0D430QR02W
kind: statement
role: theorem
name: Rouché's theorem
aliases: [rouche-theorem]
msc: [30C15]
uses: [0DG0Z72CQD, 064N002TX4, 0PN2BXRS8G]
sources:
  - url: https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem
    locator: lead section
  - url: https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem
    locator: Usage
same_as:
  wikidata: Q1191862
  wikipedia: Rouché's theorem
---

## Statement

Let $c \in \mathbb{C}$ and $r > 0$.
Let $f$ and $g$ be [[holomorphic|0DG0Z72CQD]] on an open set that contains the closed disc $\{ z : |z - c| \le r \}$.
Suppose that $|g(z)| < |f(z)|$ for each $z$ with $|z - c| = r$.
Then $f$ and $f + g$ have the same number of zeros in the open disc $|z - c| < r$, each zero counted as many times as its [[order|064N002TX4]].

## Remarks

The usual proof applies the [[argument principle|0PN2BXRS8G]] to $f$ and to $f + g$.
For example, on $|z| = 1$ we have $|z^5 + 1| \le 2 < 3 = |3z|$, so $z^5 + 3z + 1$ has exactly one zero in the open unit disc, as $3z$ has.
For a polynomial $p(z) = a_n z^n + \dots + a_0$ with $a_n \neq 0$, compare $p$ with $a_n z^n$ on a large circle; this gives $n$ zeros of $p$ and proves the [[fundamental theorem of algebra|0JP0W07Y9A]].

## Explanations

### L1 human

The theorem says that a small change does not change the number of zeros.
On the circle, $g$ is smaller than $f$, so $f + g$ stays close to $f$ in a relative sense.
Then $f + g$ turns around $0$ as many times as $f$, and so it has the same number of zeros inside.