{
  "aliases": [
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  "body": "\n## Statement\n\nLet $c \\in \\mathbb{C}$ and $r \u003e 0$.\nLet $f$ and $g$ be [[holomorphic|0DG0Z72CQD]] on an open set that contains the closed disc $\\{ z : |z - c| \\le r \\}$.\nSuppose that $|g(z)| \u003c |f(z)|$ for each $z$ with $|z - c| = r$.\nThen $f$ and $f + g$ have the same number of zeros in the open disc $|z - c| \u003c r$, each zero counted as many times as its [[order|064N002TX4]].\n\n## Remarks\n\nThe usual proof applies the [[argument principle|0PN2BXRS8G]] to $f$ and to $f + g$.\nFor example, on $|z| = 1$ we have $|z^5 + 1| \\le 2 \u003c 3 = |3z|$, so $z^5 + 3z + 1$ has exactly one zero in the open unit disc, as $3z$ has.\nFor 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]].\n\n## Explanations\n\n### L1 human\n\nThe theorem says that a small change does not change the number of zeros.\nOn the circle, $g$ is smaller than $f$, so $f + g$ stays close to $f$ in a relative sense.\nThen $f + g$ turns around $0$ as many times as $f$, and so it has the same number of zeros inside.\n",
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  "name": "Rouché's theorem",
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