{
  "aliases": [
    "residue-formula-for-simple-poles"
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  "body": "\n## Statement\n\nLet $c \\in \\mathbb{C}$, let $r \u003e 0$, and let $f$ be [[holomorphic|0DG0Z72CQD]] on $0 \u003c |z - c| \u003c r$.\nAssume that $f$ has a simple pole at $c$, that is, a [[pole of order|0YK27TJ4WF]] $1$.\nThen\n$$\\operatorname{Res}(f, c) = \\lim_{z \\to c} (z - c) f(z),$$\nwhere $\\operatorname{Res}(f, c)$ is the [[residue|0JFGKKWWFK]] of $f$ at $c$.\nIn particular, let $f = g / h$, where $g$ and $h$ are holomorphic near $c$, $g(c) \\neq 0$, $h(c) = 0$ and $h'(c) \\neq 0$. Then $\\operatorname{Res}(f, c) = \\frac{g(c)}{h'(c)}$.\n\n## Remarks\n\nProof: near $c$, $f(z) = \\frac{a_{-1}}{z - c} + a_0 + a_1 (z - c) + \\cdots$, so $(z - c) f(z) \\to a_{-1}$.\nFor example, $\\operatorname{Res}\\left(\\frac{1}{z^2 + 1}, i\\right) = \\frac{1}{2i}$, with $g = 1$ and $h = z^2 + 1$.\nAt a pole of order $m$, the formula is $\\operatorname{Res}(f, c) = \\frac{1}{(m-1)!} \\lim_{z \\to c} \\frac{d^{m-1}}{dz^{m-1}} \\left( (z - c)^m f(z) \\right)$.\n\n## Explanations\n\n### L1 human\n\nAt a simple pole, $f$ looks like $\\frac{a}{z - c}$ plus something harmless. Multiplying by $z - c$ and letting $z \\to c$ gives $a$, the residue.\nFor example, the residue of $\\frac{1}{z}$ at $0$ is $1$, and the residue of $\\frac{1}{z - 2}$ at $2$ is also $1$.\n",
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    "aliases": [
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    "name": "Residue at a simple pole",
    "role": "proposition",
    "sources": [
      {
        "locator": "Simple poles",
        "url": "https://en.wikipedia.org/wiki/Residue_(complex_analysis)"
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  "markdown": "https://math2.tamnd.com/node/01991NCP7H.md",
  "msc": [
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  "name": "Residue at a simple pole",
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  "role": "proposition",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/019/01991NCP7H.md",
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  "tag": "01991NCP7H",
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    "tag": "01991NCP7H",
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