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  "body": "\n## Definition\n\nLet $U \\subseteq \\mathbb{C}$ be an open set, let $f : U \\to \\mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $U$, and let $z_0 \\in U$.\nAssume that $f$ is not identically zero on any open disc with center $z_0$.\nThe order of $f$ at $z_0$ is the smallest integer $m \\ge 0$ with $f^{(m)}(z_0) \\neq 0$, where $f^{(m)}$ is the $m$-th complex derivative of $f$.\nIf $f(z_0) = 0$, this order $m \\ge 1$ is the order of the [[zero|0CEGEASY49]] $z_0$ of $f$. A zero of order $1$ is a simple zero.\nIf $f$ is identically zero on an open disc with center $z_0$, the order of $f$ at $z_0$ is $\\infty$.\n\n## Remarks\n\nThe order is $m$ if and only if $f(z) = (z - z_0)^m g(z)$ near $z_0$, where $g$ is holomorphic near $z_0$ and $g(z_0) \\neq 0$.\nThe order exists because [[holomorphic functions are analytic|0826SGV3R0]]: if all the derivatives at $z_0$ are $0$, then the Taylor series is $0$ and $f = 0$ near $z_0$.\nFor example, $z^3 (z - 1)$ has a zero of order $3$ at $0$ and a simple zero at $1$. The function $\\sin z - z$ has a zero of order $3$ at $0$.\n\n## Explanations\n\n### L1 human\n\nThe order of a zero says how fast the function goes to $0$ at that point.\nNear a zero of order $m$, the function behaves like a constant times $(z - z_0)^m$.\n",
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