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  "body": "\n## Proof\n\n1. Let $f : \\C \\to \\C$ be an [[entire function|0419794R0T]], and let $M$ be a real number with $\\abs{f(z)} \\le M$ for each $z \\in \\C$.\n2. Fix $c \\in \\C$ and a real number $r \u003e 0$. The closed disc $\\set{z : \\abs{z - c} \\le r}$ lies in $\\C$, where $f$ is [[holomorphic|0DG0Z72CQD]].\n3. On the circle $\\abs{z - c} = r$ we have $\\abs{f(z)} \\le M$. So [[Cauchy's estimates|0JJRENWMP6]] with $n = 1$ give $\\abs{f'(c)} \\le M / r$.\n4. Step 3 holds for every $r \u003e 0$. Letting $r \\to \\infty$ gives $\\abs{f'(c)} \\le 0$, so $f'(c) = 0$.\n5. Since $c$ was arbitrary, $f'(c) = 0$ for each $c \\in \\C$.\n6. View $f$ as a map $\\R^2 \\to \\R^2$ by $x + iy \\leftrightarrow (x, y)$. Fix $c \\in \\C$.\n7. By the definition of [[complex differentiability|0D1C9Y1EXA]] with $f'(c) = 0$, we have $\\abs{f(c + h) - f(c)} / \\abs{h} \\to 0$ as $h \\to 0$.\n8. Step 7 says that $f$ is differentiable at $c$ as a map of two real variables, with total derivative $Df(c) = 0$.\n9. The set $\\C = \\R^2$ is open and connected. By step 8 and the theorem that a [[zero derivative on a connected open set implies constant|0BYZTFM0DJ]], $f$ is constant.\n\n## Idea\n\nCauchy's estimates bound $f'(c)$ by the bound of $f$ on a circle divided by the radius. For an entire function the radius can be as large as we like, while the bound $M$ stays the same. So the derivative is zero everywhere, and $f$ is constant.\n",
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