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  "body": "\n## Proof\n\n1. Let $U$, $f$ and $z_0$ be as in the statement. Choose $R \u003e 0$ with $D = \\set{z : \\abs{z - z_0} \u003c R} \\subseteq U$.\n2. By [[holomorphic functions are analytic|0826SGV3R0]], there are $a_0, a_1, \\dots \\in \\C$ with $f(z) = \\sum_{n \\ge 0} a_n (z - z_0)^n$ for each $z \\in D$.\n3. First, the two cases cannot both hold. If they held with radii $r_1$ and $r_2$, the point $z = z_0 + \\min(r_1, r_2)/2$ would give $f(z) = 0$ and $f(z) \\ne 0$.\n4. If $a_n = 0$ for each $n$, then $f = 0$ on $D$, and case 1 holds with $r = R$.\n5. Otherwise let $m$ be the smallest index with $a_m \\ne 0$. Then $f(z) = \\sum_{n \\ge m} a_n (z - z_0)^n$ on $D$.\n6. Consider the [[power series|0VARVQ9ZW1]] $\\sum_{k \\ge 0} a_{m+k} (z - z_0)^k$. For $z \\in D$ with $z \\ne z_0$, its partial sums are $(z - z_0)^{-m}$ times partial sums of the series in step 5. So it converges at each $z \\in D$.\n7. If a power series converges at a point $w$, its terms tend to $0$ and so are bounded. By the definition of the [[radius of convergence|045EWAXY9P]], the radius is at least $\\abs{w - z_0}$.\n8. By steps 6 and 7, the series of step 6 has radius of convergence at least $R$. Let $g(z)$ be its sum on $D$.\n9. By [[power series are holomorphic|0YMP521ZR5]], $g$ is [[holomorphic|0DG0Z72CQD]] on $D$, hence continuous. Also $g(z_0) = a_m \\ne 0$.\n10. By step 6, $f(z) = (z - z_0)^m g(z)$ for each $z \\in D$, including $z = z_0$ where both sides are $0$ if $m \\ge 1$, and $a_0$ if $m = 0$.\n11. By continuity of $g$ at $z_0$, there is $r \\in (0, R]$ with $g(z) \\ne 0$ for $\\abs{z - z_0} \u003c r$.\n12. For $0 \u003c \\abs{z - z_0} \u003c r$, both factors $(z - z_0)^m$ and $g(z)$ are nonzero. So $f(z) \\ne 0$, and case 2 holds.\n13. By steps 3, 4 and 12, exactly one of the two cases holds.\n\n## Idea\n\nNear $z_0$ the function is a power series. Either every coefficient is zero, so $f$ vanishes near $z_0$, or there is a first nonzero coefficient. In the second case $f(z) = (z - z_0)^m g(z)$ with $g(z_0) \\ne 0$, and continuity keeps $g$ away from zero near $z_0$.\n",
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