{
  "aliases": [
    "riemann-removable-singularity-theorem"
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  "body": "\n## Statement\n\nLet $c \\in \\mathbb{C}$, let $r \u003e 0$, and let $f$ be [[holomorphic|0DG0Z72CQD]] on the punctured disc $0 \u003c |z - c| \u003c r$.\nAssume that $f$ is bounded near $c$: there are $M \\ge 0$ and $\\delta \\in (0, r)$ with $|f(z)| \\le M$ for $0 \u003c |z - c| \u003c \\delta$.\nThen $c$ is a [[removable singularity|0P18JC72RE]] of $f$. Moreover, the limit $\\lim_{z \\to c} f(z)$ exists, and the extension of $f$ by this value at $c$ is holomorphic on $|z - c| \u003c r$.\n\n## Remarks\n\nThe following are equivalent: $c$ is removable; $f$ has a continuous extension to $c$; $f$ is bounded near $c$; $\\lim_{z \\to c} (z - c) f(z) = 0$.\nProof idea: $h(z) = (z - c)^2 f(z)$, with $h(c) = 0$, is complex differentiable at $c$ with $h'(c) = 0$. So $h(z) = \\sum_{n \\ge 2} a_n (z - c)^n$, and $f(z) = \\sum_{n \\ge 2} a_n (z - c)^{n-2}$.\nFor example, $\\frac{\\sin z}{z}$ is bounded near $0$, so $0$ is removable.\n\n## Explanations\n\n### L1 human\n\nIf a holomorphic function stays bounded near a single bad point, then the point is not really bad.\nYou can give the function a value there, and the function becomes holomorphic at that point.\n",
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    "name": "Riemann's theorem on removable singularities",
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        "url": "https://en.wikipedia.org/wiki/Removable_singularity"
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  "msc": [
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  "name": "Riemann's theorem on removable singularities",
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  "url": "https://math2.tamnd.com/wiki/riemanns-theorem-on-removable-singularities/",
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