{
  "aliases": [
    "laurents-theorem",
    "laurent-expansion-theorem"
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  "body": "\n## Statement\n\nLet $c \\in \\mathbb{C}$, let $0 \\le r \u003c R \\le \\infty$, and let $A = \\{z : r \u003c |z - c| \u003c R\\}$ be an [[annulus|0AJ0YVK743]].\nLet $f : A \\to \\mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $A$.\nThen there are unique complex numbers $(a_n)_{n \\in \\mathbb{Z}}$ such that the [[Laurent series|0X3XD2ATB0]] $\\sum_{n \\in \\mathbb{Z}} a_n (z - c)^n$ converges to $f(z)$ for each $z \\in A$.\nFor each $\\rho$ with $r \u003c \\rho \u003c R$ and each $n \\in \\mathbb{Z}$,\n$$a_n = \\frac{1}{2 \\pi i} \\oint_{|z - c| = \\rho} \\frac{f(z)}{(z - c)^{n+1}} \\, dz.$$\nThe series converges absolutely on $A$ and uniformly on each closed annulus $\\rho_1 \\le |z - c| \\le \\rho_2$ with $r \u003c \\rho_1 \\le \\rho_2 \u003c R$.\n\n## Remarks\n\nUniqueness means: each Laurent series that converges to $f$ on $A$ has these coefficients. So we can find them by any method, for example with known power series.\nFor example, on $0 \u003c |z| \u003c 1$, $\\frac{1}{z(1 - z)} = \\frac{1}{z} + 1 + z + z^2 + \\cdots$. On $|z| \u003e 1$, the same function is $-\\frac{1}{z^2} - \\frac{1}{z^3} - \\cdots$.\nIf $f$ is holomorphic on the whole disc $|z - c| \u003c R$, then $a_n = 0$ for $n \u003c 0$, and the Laurent series is the Taylor series.\n\n## Explanations\n\n### L1 human\n\nOn a ring around $c$, each holomorphic function is a sum of powers of $z - c$, with negative and positive exponents.\nThe coefficients come from circle integrals, as for Taylor series. The same function can have different Laurent series on different rings.\n",
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  "msc": [
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  "name": "Existence of Laurent expansions on an annulus",
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