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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 $c \\in U$ and $r \u003e 0$ be such that the open disc $D = \\{z : |z - c| \u003c r\\}$ is contained in $U$.\nThen there are complex numbers $a_0, a_1, a_2, \\dots$ such that the [[power series|0VARVQ9ZW1]] $\\sum_{n \\ge 0} a_n (z - c)^n$ converges to $f(z)$ for each $z \\in D$.\nFor each $\\rho$ with $0 \u003c \\rho \u003c r$, the coefficients are\n$$a_n = \\frac{1}{2 \\pi i} \\oint_{|z - c| = \\rho} \\frac{f(z)}{(z - c)^{n+1}} \\, dz.$$\nIn particular, $f$ is analytic on $U$: near each point of $U$, $f$ is the sum of a convergent power series.\n\n## Remarks\n\nProof idea: in [[Cauchy's integral formula|0DG10TVYPN]], expand $\\frac{1}{z - a}$ as a geometric series in $\\frac{a - c}{z - c}$, and integrate term by term.\nThe radius of convergence of the Taylor series at $c$ is at least the distance from $c$ to the boundary of $U$.\nFor example, the Taylor series of $1/(1 - z)$ at $0$ has radius $1$, the distance from $0$ to the point $1$ where the function is not defined.\nThe converse is also true: [[power series are holomorphic in their disc of convergence|0YMP521ZR5]].\n\n## Explanations\n\n### L1 human\n\nA function with one complex derivative everywhere on an open set is automatically equal to its Taylor series near each point.\nThis is very different from real functions, where a differentiable function can fail to have a second derivative.\n",
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    "name": "Holomorphic functions are analytic",
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