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  "body": "\n## Proof\n\n1. Let $D$, $f$, $z_0$ and $r$ be as in the statement. Let $\\alpha = f(z_0)$ and $M = \\abs{\\alpha}$.\n2. If $M = 0$, then $\\abs{f(z)} \\le 0$ for $\\abs{z - z_0} \u003c r$, so $f = \\alpha$ near $z_0$. Go to step 11 in this case.\n3. Let $M \u003e 0$. Since $D$ is open, choose $\\rho_0 \\in (0, r)$ with $\\set{z : \\abs{z - z_0} \\le \\rho_0} \\subseteq D$. Fix $\\rho \\in (0, \\rho_0]$.\n4. By [[Cauchy's integral formula|0DG10TVYPN]] with $a = c = z_0$, written out as a [[circle integral|0T7ABX4CWE]], $$\\alpha = \\frac{1}{2\\pi i} \\int_0^{2\\pi} \\frac{f(z_0 + \\rho e^{i\\theta})}{\\rho e^{i\\theta}} \\, i \\rho e^{i\\theta} \\, d\\theta = \\frac{1}{2\\pi} \\int_0^{2\\pi} f(z_0 + \\rho e^{i\\theta}) \\, d\\theta.$$\n5. Multiply by $\\overline{\\alpha}$ and take real parts. With $q(\\theta) = \\operatorname{Re} \\bigl( \\overline{\\alpha} f(z_0 + \\rho e^{i\\theta}) \\bigr)$, this gives $M^2 = \\frac{1}{2\\pi} \\int_0^{2\\pi} q(\\theta) \\, d\\theta$.\n6. Since $\\rho \u003c r$, we have $q(\\theta) \\le \\abs{\\alpha} \\abs{f(z_0 + \\rho e^{i\\theta})} \\le M^2$. So $\\varphi = M^2 - q$ is continuous, $\\varphi \\ge 0$, and $\\int_0^{2\\pi} \\varphi = 0$ by step 5.\n7. Then $\\varphi = 0$ everywhere. Indeed, if $\\varphi(\\theta_0) \u003e 0$, continuity gives an interval of length $\\ell \u003e 0$ on which $\\varphi \u003e \\varphi(\\theta_0)/2$.\n8. By [[monotonicity of the integral|0MBWQKM8QR]], the integral of $\\varphi$ would then be at least $\\ell \\varphi(\\theta_0)/2 \u003e 0$, a contradiction.\n9. So $\\operatorname{Re}(\\overline{\\alpha} f) = M^2$ on the circle $\\abs{z - z_0} = \\rho$. Also $\\abs{\\overline{\\alpha} f} \\le M^2$ there, so $\\operatorname{Im}(\\overline{\\alpha} f) = 0$, and $\\overline{\\alpha} f = M^2 = \\overline{\\alpha} \\alpha$.\n10. Dividing by $\\overline{\\alpha} \\ne 0$ gives $f = \\alpha$ on each circle $\\abs{z - z_0} = \\rho$ with $0 \u003c \\rho \\le \\rho_0$. So $f = \\alpha$ on the disc $\\abs{z - z_0} \u003c \\rho_0$.\n11. In both cases, $f$ equals the constant function $\\alpha$ on an open disc with center $z_0$ contained in $D$. Each point of this disc is an accumulation point of the set where $f = \\alpha$.\n12. The constant function $\\alpha$ is [[holomorphic|0DG0Z72CQD]]. By the [[identity theorem|0YBVKMYX0W]] on the [[domain|097KNJFJ7V]] $D$, $f = \\alpha$ on $D$. So $f$ is constant on $D$.\n\n## Idea\n\nCauchy's integral formula says that $f(z_0)$ is the average of $f$ over each small circle around $z_0$. An average of values of modulus at most $\\abs{f(z_0)}$ can equal $f(z_0)$ only if every value equals $f(z_0)$. So $f$ is constant near $z_0$, and the identity theorem spreads this over the domain.\n",
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