{
  "aliases": [
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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{C}$ be an open set and let $f : U \\to \\mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $U$.\nLet $c \\in \\mathbb{C}$ and $r \u003e 0$ be such that the closed disc $\\{z : |z - c| \\le r\\}$ is contained in $U$.\nThen for each integer $n \\ge 0$ and each $a \\in \\mathbb{C}$ with $|a - c| \u003c r$,\n$$f^{(n)}(a) = \\frac{n!}{2 \\pi i} \\oint_{|z - c| = r} \\frac{f(z)}{(z - a)^{n+1}} \\, dz,$$\nwhere $f^{(n)}$ is the $n$-th complex derivative of $f$. This derivative exists by [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]].\n\n## Remarks\n\nFor $n = 0$, this is [[Cauchy's integral formula|0DG10TVYPN]].\nFormally, the formula comes from differentiating $\\frac{1}{z - a}$ $n$ times under the integral sign with respect to $a$.\nFor example, with $f(z) = e^z$, $c = a = 0$ and $r = 1$: $\\oint_{|z| = 1} \\frac{e^z}{z^{n+1}} \\, dz = \\frac{2 \\pi i}{n!}$.\nIt gives [[Cauchy's estimates|0JJRENWMP6]] for the derivatives.\n\n## Explanations\n\n### L1 human\n\nThe values of a holomorphic function on a circle determine not only its values inside, but also all its derivatives inside.\nTo get the $n$-th derivative, you put a higher power of $z - a$ in the denominator and multiply by $n!$.\n",
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    "name": "Cauchy's integral formula for derivatives",
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  "msc": [
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  "name": "Cauchy's integral formula for derivatives",
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