Rouché's theorem

Statement

Let $c \in \mathbb{C}$ and $r > 0$. Let $f$ and $g$ be holomorphic 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.

Remarks

The usual proof applies the argument principle 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.

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.