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  "body": "\n## Proof\n\nWe expand the Cauchy kernel in a geometric series and integrate term by term.\n\n1. Fix $\\rho$ with $0 \u003c \\rho \u003c r$. The closed disc $\\set{w : \\abs{w - c} \\le \\rho}$ lies in $D \\subseteq U$. For $n \\ge 0$ put $a_n(\\rho) = \\frac{1}{2 \\pi i} \\oint_{\\abs{w - c} = \\rho} \\frac{f(w)}{(w - c)^{n+1}} \\, dw$, a [[circle integral|0T7ABX4CWE]] of a continuous function.\n2. Fix $z$ with $\\abs{z - c} \u003c \\rho$ and put $q = \\abs{z - c}/\\rho \u003c 1$. By [[Cauchy's integral formula|0DG10TVYPN]], $f(z) = \\frac{1}{2 \\pi i} \\oint_{\\abs{w - c} = \\rho} \\frac{f(w)}{w - z} \\, dw$.\n3. For $w$ on the circle put $u = (z - c)/(w - c)$, so $\\abs{u} = q$ and $w - z = (w - c)(1 - u)$. By the [[finite geometric sum|05APWSX53S]], $\\frac{1}{1 - u} = \\sum_{n=0}^{N} u^n + \\frac{u^{N+1}}{1 - u}$.\n4. Hence $\\frac{f(w)}{w - z} = \\sum_{n=0}^{N} \\frac{f(w)}{(w - c)^{n+1}} (z - c)^n + T_N(w)$, with $T_N(w) = \\frac{f(w) \\, u^{N+1}}{(w - c)(1 - u)}$.\n5. The circle is closed and bounded, hence [[compact|07PFYV65MB]], so $\\abs{f} \\le M$ on it for some $M$ by the [[extreme value theorem|0V4DET59NW]]. Since $\\abs{1 - u} \\ge 1 - q$ by the [[triangle inequality|0FJZMBX0ZZ]], $\\abs{T_N(w)} \\le \\frac{M q^{N+1}}{\\rho (1 - q)}$.\n6. Integrate step 4 over the circle and divide by $2 \\pi i$. We get $f(z) = \\sum_{n=0}^{N} a_n(\\rho) (z - c)^n + \\frac{1}{2\\pi i} \\oint T_N(w) \\, dw$.\n7. The circle has length $2 \\pi \\rho$. By the [[estimation lemma|0DJMGPN8K9]], the last term has modulus at most $\\frac{M q^{N+1}}{1 - q}$. This tends to $0$ as $N \\to \\infty$, since [[$q^N \\to 0$|0Q7K4AXE53]].\n8. So the [[power series|0VARVQ9ZW1]] $\\sum_{n \\ge 0} a_n(\\rho) (z - c)^n$ converges to $f(z)$ whenever $\\abs{z - c} \u003c \\rho$.\n9. Now let $0 \u003c \\rho_1 \u003c \\rho_2 \u003c r$ and $b_n = a_n(\\rho_1) - a_n(\\rho_2)$. By step 8 for both radii, $\\sum_{n \\ge 0} b_n w^n = 0$ for every $w$ with $\\abs{w} \u003c \\rho_1$.\n10. Suppose some $b_n \\ne 0$, and let $m$ be the least such $n$. Put $s = \\rho_1 / 2$. The series $\\sum b_n s^n$ converges, so its terms tend to $0$. Hence $\\abs{b_n} s^n \\le K$ for all $n$, for some $K$.\n11. Let $0 \u003c \\abs{w} \u003c s$. Then $\\abs{b_n w^{n-m}} \\le K s^{-m} (\\abs{w}/s)^{n-m}$ for $n \\ge m$. So $T(w) = \\sum_{n \u003e m} b_n w^{n-m}$ converges and $\\abs{T(w)} \\le K s^{-m} \\frac{\\abs{w}/s}{1 - \\abs{w}/s}$.\n12. By steps 9 and 10, $0 = \\sum_{n \\ge m} b_n w^n = w^m (b_m + T(w))$. Since $w \\ne 0$, $b_m = -T(w)$.\n13. Letting $w \\to 0$ in step 11 gives $\\abs{b_m} \\le \\lim \\abs{T(w)} = 0$. This contradicts $b_m \\ne 0$. So $a_n(\\rho_1) = a_n(\\rho_2)$ for all $n$.\n14. Thus $a_n = a_n(\\rho)$ does not depend on $\\rho \\in (0, r)$, and it is given by the stated formula.\n15. Let $z \\in D$. Choose $\\rho$ with $\\abs{z - c} \u003c \\rho \u003c r$. By steps 8 and 14, $\\sum_{n \\ge 0} a_n (z - c)^n$ converges to $f(z)$.\n16. Finally, let $c \\in U$ be any point. Since $U$ is open, some disc $\\set{z : \\abs{z - c} \u003c r}$ with $r \u003e 0$ lies in $U$. By step 15, $f$ is the sum of a convergent power series near $c$. So $f$ is analytic on $U$.\n\n## Idea\n\nCauchy's formula writes $f(z)$ as an average of $f(w)/(w - z)$ over a circle. The kernel $1/(w - z)$ is a geometric series in $(z - c)/(w - c)$, which converges uniformly on the circle. Integrating term by term gives the power series, and the uniqueness of coefficients shows they do not depend on the radius.\n",
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