{
  "aliases": [
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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{C}$ be a simply connected open set. This means that $U$ is connected, and each closed path in $U$ can be continuously deformed in $U$ to a constant path.\nLet $f : U \\to \\mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $U$.\nLet $\\gamma : [a, b] \\to U$ be a closed path that is continuous and piecewise continuously differentiable.\nThen the [[contour integral|00FWNP9TXY]] of $f$ along $\\gamma$ is\n$$\\oint_\\gamma f(z) \\, dz = 0.$$\n\n## Remarks\n\nOpen discs, open half-planes, $\\mathbb{C}$ and each convex open set are simply connected. So the theorem holds on each of them.\nA special case: if $f$ is holomorphic on an open set that contains the closed disc $|z - c| \\le R$, then $\\oint_{|z - c| = R} f(z) \\, dz = 0$.\nThe hypothesis on $U$ is necessary. On $\\mathbb{C} \\setminus \\{0\\}$, the [[integral of 1/z|0SXZ68HV6F]] over the unit circle is $2 \\pi i$.\nIt follows that the integral of $f$ along a path in $U$ depends only on the end points of the path.\n\n## Explanations\n\n### L1 human\n\nThe integral of a holomorphic function around a closed loop is zero, as long as the loop does not go around a hole of the domain.\nIt is like walking around a hill and coming back to the start: the total change of height is zero.\n",
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  "name": "Cauchy's integral theorem",
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