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  "body": "\n## Proof\n\n1. Let $c$, $r$, $f$, $M$ and $\\delta$ be as in the statement, so $c$ is an [[isolated singularity|0TYB3HM9KE]] of $f$. Let $D = \\set{z : \\abs{z - c} \u003c r}$.\n2. Define $h(z) = (z - c)^2 f(z)$ for $0 \u003c \\abs{z - c} \u003c r$, and $h(c) = 0$.\n3. On the punctured disc, $h$ is a product of [[holomorphic|0DG0Z72CQD]] functions, so it is complex differentiable there.\n4. At $c$: for $0 \u003c \\abs{z - c} \u003c \\delta$, $\\abs{\\frac{h(z) - h(c)}{z - c}} = \\abs{z - c} \\abs{f(z)} \\le M \\abs{z - c}$. This tends to $0$ as $z \\to c$.\n5. So $h$ is [[complex differentiable|0D1C9Y1EXA]] at $c$ with $h'(c) = 0$. By steps 3 and 4, $h$ is holomorphic on $D$.\n6. By [[holomorphic functions are analytic|0826SGV3R0]], $h(z) = \\sum_{n \\ge 0} a_n (z - c)^n$ for each $z \\in D$.\n7. We have $a_0 = h(c) = 0$. By [[term-by-term differentiation|0YMP521ZR5]] at the center, $a_1 = h'(c) = 0$.\n8. Consider the power series $\\sum_{k \\ge 0} a_{k+2} (z - c)^k$. For $z \\in D$ with $z \\ne c$, its partial sums are $(z - c)^{-2}$ times partial sums of the series in step 6. So it converges at each point of $D$.\n9. A power series that converges at a point $w$ has bounded terms there. By the definition of the [[radius of convergence|045EWAXY9P]], the series of step 8 has radius at least $r$.\n10. Let $g(z) = \\sum_{k \\ge 0} a_{k+2} (z - c)^k$ for $z \\in D$. By [[power series are holomorphic|0YMP521ZR5]], $g$ is holomorphic on $D$.\n11. By steps 7 and 8, $g(z) = h(z)/(z - c)^2 = f(z)$ for $0 \u003c \\abs{z - c} \u003c r$.\n12. By step 11 and the definition, $c$ is a [[removable singularity|0P18JC72RE]] of $f$.\n13. The function $g$ is continuous at $c$, since it is complex differentiable there. By step 11, $f(z) = g(z) \\to g(c)$ as $z \\to c$. So $\\lim_{z \\to c} f(z)$ exists and equals $g(c)$.\n14. The extension of $f$ by the value $g(c)$ at $c$ is the function $g$, which is holomorphic on $D$ by step 10.\n\n## Idea\n\nMultiplying by $(z - c)^2$ turns the bounded function $f$ into a function $h$ that is complex differentiable even at $c$, with $h(c) = h'(c) = 0$. So the power series of $h$ at $c$ starts with $(z - c)^2$. Dividing by $(z - c)^2$ gives a power series that agrees with $f$ off $c$ and fills in the missing value.\n",
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