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  "body": "\n## Definition\n\nLet $c \\in \\mathbb{C}$, let $r \u003e 0$, and let $f$ be [[holomorphic|0DG0Z72CQD]] on the punctured disc $0 \u003c |z - c| \u003c r$, so that $c$ is an [[isolated singularity|0TYB3HM9KE]] of $f$.\nLet $f(z) = \\sum_{n \\in \\mathbb{Z}} a_n (z - c)^n$ be the [[Laurent expansion|06WHRARZKY]] of $f$ on this punctured disc.\nThe residue of $f$ at $c$ is the coefficient\n$$\\operatorname{Res}(f, c) = a_{-1}.$$\n\n## Remarks\n\nFor each $\\rho$ with $0 \u003c \\rho \u003c r$, $\\operatorname{Res}(f, c) = \\frac{1}{2 \\pi i} \\oint_{|z - c| = \\rho} f(z) \\, dz$. This follows from the coefficient formula for $n = -1$.\nThe residue is $0$ at a [[removable singularity|0P18JC72RE]]. It can be $0$ at a pole too: $\\operatorname{Res}(1/z^2, 0) = 0$.\nFor example, $\\operatorname{Res}(1/z, 0) = 1$ and $\\operatorname{Res}(e^{1/z}, 0) = 1$.\nResidues at simple poles are easy to compute, see [[residue at a simple pole|01991NCP7H]].\n\n## Explanations\n\n### L1 human\n\nWhen you integrate a Laurent series term by term around a small circle, every term gives $0$ except the term $a_{-1}/(z - c)$, which gives $2 \\pi i a_{-1}$.\nSo the residue is the one coefficient that the integral remembers.\n",
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  "name": "Residue",
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