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  "body": "\n## Statement\n\nLet $D \\subseteq \\mathbb{C}$ be a [[domain|097KNJFJ7V]] and let $f : D \\to \\mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $D$.\nAssume that $|f|$ has a local maximum at a point $z_0 \\in D$: there is $r \u003e 0$ such that $|f(z)| \\le |f(z_0)|$ for each $z$ with $|z - z_0| \u003c r$.\nThen $f$ is constant on $D$.\n\n## Remarks\n\nProof idea: the mean value formula $f(z_0) = \\frac{1}{2\\pi} \\int_0^{2\\pi} f(z_0 + \\rho e^{i\\theta}) \\, d\\theta$ for small $\\rho$ forces $|f| = |f(z_0)|$ near $z_0$. Then $f$ is constant near $z_0$, and on $D$ by the [[identity theorem|0YBVKMYX0W]].\nCorollary: if $D$ is bounded and $f$ is continuous on the closure $\\overline{D}$ and holomorphic on $D$, then $\\max_{\\overline{D}} |f|$ is attained at a point of the boundary of $D$.\nThe minimum of $|f|$ can be inside $D$, but only at a zero of $f$. For example, $|z|$ has its minimum on the unit disc at $0$.\n\n## Explanations\n\n### L1 human\n\nThe size $|f|$ of a nonconstant holomorphic function has no highest point inside its domain.\nFor example, on the closed unit disc, $|z^2 + 1|$ is largest on the boundary circle, at $z = \\pm 1$, and it equals $2$ there.\n",
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        "locator": "Using Gauss's mean value theorem",
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