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  "body": "\n## Definition\n\nLet $\\Omega \\subseteq \\C$ be an [[open set|01ZQQDCERP]], and let $\\Gamma$ be a family of locally [[rectifiable curves|0C28RT5BG7]] in $\\Omega$.\nA Borel measurable function $\\rho \\colon \\Omega \\to [0, \\infty]$ is *admissible* for $\\Gamma$ if $\\int_\\gamma \\rho \\, \\abs{dz} \\ge 1$ for every $\\gamma \\in \\Gamma$.\nThe *conformal modulus* of $\\Gamma$ is\n$$M(\\Gamma) = \\inf_{\\rho} \\int_\\Omega \\rho(x + iy)^2 \\, dx\\, dy \\in [0, \\infty],$$\nthe infimum taken over all admissible $\\rho$.\n\n## Remarks\n\nThe modulus is the reciprocal of the [[extremal length|HX72KT9D0G]]: $M(\\Gamma) = 1/\\lambda(\\Gamma)$, with $1/0 = \\infty$ and $1/\\infty = 0$.\nIt is invariant under conformal maps of $\\Omega$.\nExample: for an [[annulus|0AJ0YVK743]] $\\set{r \u003c \\abs{z} \u003c R}$ and the family $\\Gamma$ of curves in it that join the two boundary circles, $\\rho(z) = \\frac{1}{\\abs{z}\\log(R/r)}$ is admissible, and $M(\\Gamma) = \\frac{2\\pi}{\\log(R/r)}$.\nThe check of the lower bound integrates the inequality $\\int_r^R \\rho(te^{i\\theta})\\, dt \\ge 1$ over $\\theta$ and applies the Cauchy–Schwarz inequality.\nNon-example: $\\rho = \\frac{1}{2\\abs{z}\\log(R/r)}$ is not admissible, since along a radial segment its integral is $\\frac{1}{2}$.\nA quadrilateral is a Jordan domain $Q$ with four marked boundary points, which cut the boundary into four sides.\nIts modulus is the modulus of the family of curves in $Q$ that join one pair of opposite sides.\nMap $Q$ conformally onto a rectangle $(0, a) \\times (0, b)$, with the marked points at the corners and the chosen sides vertical; then the modulus is $b/a$.\nConventions for ring domains differ: the modulus of $\\set{r \u003c \\abs{z} \u003c R}$ is often defined as $\\frac{1}{2\\pi}\\log(R/r)$, which is the extremal length of the family above, or as $\\log(R/r)$.\n",
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