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  "body": "\n## Statement\n\nLet $\\Omega \\subsetneq \\C$ be a [[simply connected domain|0QW64ARSHB]] and let $a \\in \\Omega$.\nLet $\\varphi \\colon \\Omega \\to \\mathbb{D}$ be a [[conformal map|0YYT36R4YD]] of $\\Omega$ onto the unit disc $\\mathbb{D}$ with $\\varphi(a) = 0$; it exists by the [[Riemann mapping theorem|02WHMEG5ZR]].\nThen the [[Green's function|M36SD56MWQ]] of $\\Omega$ with pole at $a$ is\n$$g(z, a) = -\\log\\abs{\\varphi(z)}, \\qquad z \\in \\Omega \\setminus \\set{a}.$$\n\n## Remarks\n\nProof idea: since $\\varphi$ is injective, $\\varphi'(a) \\ne 0$, so $\\varphi(z) = (z - a) k(z)$ with $k$ holomorphic and zero-free on $\\Omega$.\nHence $-\\log\\abs{\\varphi(z)} = -\\log\\abs{z - a} - \\log\\abs{k(z)}$, and $\\log\\abs{k}$ is harmonic on $\\Omega$.\nAway from $a$, $\\log\\abs{\\varphi}$ is harmonic because $\\varphi$ is holomorphic and has no zero.\nA conformal map onto $\\mathbb{D}$ is proper, so $\\abs{\\varphi(z)} \\to 1$ as $z$ tends to the boundary, and $g(z, a) \\to 0$ there.\nThese are the defining properties, and the Green's function is unique by the maximum principle.\nExample: for $\\Omega = \\mathbb{D}$ and $\\varphi(z) = (z - a)/(1 - \\bar a z)$, one gets $g(z, a) = \\log\\abs{(1 - \\bar a z)/(z - a)}$.\nNear the pole, $g(z, a) = -\\log\\abs{z - a} + \\log\\big(1/\\abs{\\varphi'(a)}\\big) + o(1)$, and $1/\\abs{\\varphi'(a)}$ is the [[conformal radius|GYEX4BH5WM]] of $\\Omega$ at $a$.\nThe hypothesis $\\Omega \\ne \\C$ is needed: the plane has no Green's function, since a positive harmonic function on $\\C \\setminus \\set{a}$ with a logarithmic pole at $a$ does not exist.\n\n## Explanations\n\n### L1 human\n\nFor a simply connected region, the Green's function and the Riemann map carry the same information.\nThe level curves of the Green's function are the images of the circles $\\abs{w} = r$ under the inverse of the Riemann map.\n\n### L2 human\n\nThe formula can also be read backwards.\nIf $g^*$ is a harmonic conjugate of $g(\\cdot, a)$ on $\\Omega \\setminus \\set{a}$, then $\\exp(-(g + i g^*))$ is single-valued and gives the Riemann map.\nThis is one classical route to the Riemann mapping theorem via potential theory, through the solution of a Dirichlet problem.\nOn multiply connected domains the conjugate $g^*$ has periods, so $\\exp(-(g + i g^*))$ is no longer single-valued.\n",
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