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  "body": "\n## Proof\n\nLet $c$, $r$, $R$, $A$ and $f$ be as in the statement. For $r \u003c \\rho \u003c R$, let $C_\\rho$ be the circle $\\abs{w - c} = \\rho$. All integrals over $C_\\rho$ are [[circle integrals|0T7ABX4CWE]].\nFor $n \\in \\Z$ and $r \u003c \\rho \u003c R$, let $I_n(\\rho) = \\frac{1}{2\\pi i} \\oint_{C_\\rho} \\frac{f(w)}{(w - c)^{n+1}} \\, dw$.\n\n*Lemma: integrals over circles do not depend on the radius.*\n\n1. Let $\\varphi$ be [[holomorphic|0DG0Z72CQD]] on the [[annulus|0AJ0YVK743]] $A$, and let $J(\\rho) = \\oint_{C_\\rho} \\varphi(w) \\, dw = \\int_0^{2\\pi} \\varphi(c + \\rho e^{i\\theta}) \\, i \\rho e^{i\\theta} \\, d\\theta$.\n2. By the definition of [[complex differentiability|0D1C9Y1EXA]], as in the real chain rule, $\\partial_\\rho \\varphi(c + \\rho e^{i\\theta}) = \\varphi'(c + \\rho e^{i\\theta}) e^{i\\theta}$ and $\\partial_\\theta \\varphi(c + \\rho e^{i\\theta}) = \\varphi'(c + \\rho e^{i\\theta}) i \\rho e^{i\\theta}$.\n3. Let $\\Phi(\\rho, \\theta) = \\varphi(c + \\rho e^{i\\theta}) i \\rho e^{i\\theta}$ and $G_\\rho(\\theta) = \\varphi(c + \\rho e^{i\\theta}) \\rho e^{i\\theta}$. By step 2, $\\rho \\, \\partial_\\rho \\Phi = \\partial_\\theta G_\\rho$. Both sides equal $i \\varphi' \\rho^2 e^{2i\\theta} + i \\varphi \\rho e^{i\\theta}$.\n4. By [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]], $\\varphi'$ is holomorphic, hence continuous. So $\\partial_\\rho \\Phi$ is continuous on $[\\rho_a, \\rho_b] \\times [0, 2\\pi]$ for each $r \u003c \\rho_a \u003c \\rho_b \u003c R$.\n5. By [[differentiation under the integral sign|0AR06W1E1B]] for the real and imaginary parts, and by the [[fundamental theorem of calculus|0NHGY0N0JB]], $J'(\\rho) = \\frac{1}{\\rho} \\bigl( G_\\rho(2\\pi) - G_\\rho(0) \\bigr) = 0$.\n6. By the [[mean value theorem|0BARQY52N5]] for the real and imaginary parts of $J$, $J$ is constant on $(r, R)$.\n7. Apply this to $\\varphi(w) = f(w)/(w - c)^{n+1}$, which is holomorphic on $A$ since $c \\notin A$. So $I_n(\\rho)$ does not depend on $\\rho$. Write $a_n = I_n(\\rho)$.\n\n*Existence.*\n\n8. Fix $z \\in A$ and choose $\\rho_1, \\rho_2$ with $r \u003c \\rho_1 \u003c \\abs{z - c} \u003c \\rho_2 \u003c R$.\n9. Let $q(w) = \\frac{f(w) - f(z)}{w - z}$ for $w \\in A$, $w \\ne z$, and $q(z) = f'(z)$. It is holomorphic on $A \\setminus \\set{z}$ as a quotient.\n10. Near $z$: on a disc around $z$ inside $A$, $f(w) = \\sum_{k \\ge 0} b_k (w - z)^k$ by [[holomorphic functions are analytic|0826SGV3R0]], with $b_0 = f(z)$ and $b_1 = f'(z)$ by [[term-by-term differentiation|0YMP521ZR5]].\n11. So $q(w) = \\sum_{k \\ge 1} b_k (w - z)^{k-1}$ on that disc. This series converges at each point of the disc, so its [[radius of convergence|045EWAXY9P]] is at least the radius of the disc.\n12. Indeed, a power series that converges at a point has bounded terms there. By step 11 and [[power series are holomorphic|0YMP521ZR5]], $q$ is holomorphic near $z$, hence on $A$.\n13. By the lemma, $\\oint_{C_{\\rho_2}} q = \\oint_{C_{\\rho_1}} q$.\n14. By [[Cauchy's integral formula|0DG10TVYPN]] for the constant function $1$, $\\oint_{C_{\\rho_2}} \\frac{dw}{w - z} = 2\\pi i$, since $\\abs{z - c} \u003c \\rho_2$.\n15. The function $w \\mapsto 1/(w - z)$ is holomorphic on the disc $\\abs{w - c} \u003c \\abs{z - c}$, which contains $C_{\\rho_1}$. By [[primitives on a disc|02AC7JR6G6]] and the [[fundamental theorem of contour integration|0YQ458D7PF]], $\\oint_{C_{\\rho_1}} \\frac{dw}{w - z} = 0$.\n16. Writing $q$ out in step 13 and using steps 14 and 15 gives $$f(z) = \\frac{1}{2\\pi i} \\oint_{C_{\\rho_2}} \\frac{f(w)}{w - z} \\, dw - \\frac{1}{2\\pi i} \\oint_{C_{\\rho_1}} \\frac{f(w)}{w - z} \\, dw.$$\n17. Let $w \\in C_{\\rho_2}$ and $t = (z - c)/(w - c)$, so $\\abs{t} = q_2 = \\abs{z - c}/\\rho_2 \u003c 1$. By the [[finite geometric sum|05APWSX53S]], $\\frac{1}{w - z} - \\sum_{n=0}^{N} \\frac{(z - c)^n}{(w - c)^{n+1}} = \\frac{t^{N+1}}{w - z}$.\n18. Let $M_2$ be the maximum of $\\abs{f}$ on $C_{\\rho_2}$, which exists by [[attainment of bounds|0V4DET59NW]]. Also $\\abs{w - z} \\ge \\rho_2 - \\abs{z - c}$ on $C_{\\rho_2}$.\n19. By the [[estimation lemma|0DJMGPN8K9]] and step 17, $\\abs{\\frac{1}{2\\pi i} \\oint_{C_{\\rho_2}} \\frac{f(w)}{w - z} \\, dw - \\sum_{n=0}^{N} a_n (z - c)^n} \\le \\frac{\\rho_2 M_2 \\, q_2^{N+1}}{\\rho_2 - \\abs{z - c}}$. Here we used step 7 to write $I_n(\\rho_2) = a_n$.\n20. As $N \\to \\infty$ the bound tends to $0$. So the first integral in step 16 equals $\\sum_{n \\ge 0} a_n (z - c)^n$.\n21. Let $w \\in C_{\\rho_1}$ and $s = (w - c)/(z - c)$, so $\\abs{s} = q_1 = \\rho_1/\\abs{z - c} \u003c 1$. Then $-\\frac{1}{w - z} - \\sum_{m=0}^{N} \\frac{(w - c)^m}{(z - c)^{m+1}} = \\frac{s^{N+1}}{z - w}$, and $\\abs{z - w} \\ge \\abs{z - c} - \\rho_1$.\n22. As in steps 18 to 20, $-\\frac{1}{2\\pi i} \\oint_{C_{\\rho_1}} \\frac{f(w)}{w - z} \\, dw = \\sum_{m \\ge 0} I_{-m-1}(\\rho_1) (z - c)^{-m-1} = \\sum_{n \\ge 1} a_{-n} (z - c)^{-n}$.\n23. By steps 16, 20 and 22, both parts of the [[Laurent series|0X3XD2ATB0]] $\\sum_{n \\in \\Z} a_n (z - c)^n$ converge, and its sum is $f(z)$. The coefficients are $a_n = I_n(\\rho)$ for any $\\rho \\in (r, R)$, by step 7.\n\n*Absolute and uniform convergence.*\n\n24. For $r \u003c \\rho \u003c R$, let $M(\\rho)$ be the maximum of $\\abs{f}$ on $C_\\rho$. By the estimation lemma, $\\abs{a_n} \\le \\frac{1}{2\\pi} \\cdot 2\\pi\\rho \\cdot M(\\rho) \\rho^{-n-1} = M(\\rho) \\rho^{-n}$ for each $n \\in \\Z$.\n25. Let $r \u003c \\sigma_1 \\le \\sigma_2 \u003c R$, and choose $s_1 \\in (r, \\sigma_1)$ and $s_2 \\in (\\sigma_2, R)$. Let $\\sigma_1 \\le \\abs{z - c} \\le \\sigma_2$.\n26. By step 24, $\\abs{a_n (z - c)^n} \\le M(s_2) (\\sigma_2/s_2)^n$ for $n \\ge 0$, and $\\abs{a_{-n} (z - c)^{-n}} \\le M(s_1) (s_1/\\sigma_1)^n$ for $n \\ge 1$.\n27. Both bounds are terms of convergent geometric series. By the [[Weierstrass M-test|0NX5NWAFC9]], applied to real and imaginary parts, both parts converge absolutely and uniformly on the closed annulus.\n28. Each $z \\in A$ lies in such a closed annulus, with $\\sigma_1 = \\sigma_2 = \\abs{z - c}$. So the series converges absolutely on $A$.\n\n*Uniqueness.*\n\n29. Let $\\sum_{n \\in \\Z} b_n (z - c)^n$ be a Laurent series that converges to $f(z)$ at each $z \\in A$. Fix $\\rho \\in (r, R)$, $s_1 \\in (r, \\rho)$ and $s_2 \\in (\\rho, R)$.\n30. Convergence at points with $\\abs{z - c} = s_2$ and $\\abs{z - c} = s_1$ makes the terms bounded there: $\\abs{b_n} s_2^n \\le B_2$ for $n \\ge 0$ and $\\abs{b_{-n}} s_1^{-n} \\le B_1$ for $n \\ge 1$.\n31. On $C_\\rho$, this gives $\\abs{b_n (z - c)^n} \\le B_2 (\\rho/s_2)^n$ and $\\abs{b_{-n} (z - c)^{-n}} \\le B_1 (s_1/\\rho)^n$. As in step 27, both parts converge uniformly on $C_\\rho$.\n32. Fix $k \\in \\Z$. Multiplying by $(z - c)^{-k-1}$, which has modulus $\\rho^{-k-1}$ on $C_\\rho$, keeps the convergence uniform. By the estimation lemma applied to the tails, we may integrate term by term: $$a_k = \\frac{1}{2\\pi i} \\oint_{C_\\rho} \\frac{f(z)}{(z - c)^{k+1}} \\, dz = \\sum_{n \\in \\Z} b_n \\frac{1}{2\\pi i} \\oint_{C_\\rho} (z - c)^{n - k - 1} \\, dz.$$\n33. For $m \\in \\Z$, $\\oint_{C_\\rho} (z - c)^m \\, dz = i \\rho^{m+1} \\int_0^{2\\pi} e^{i(m+1)\\theta} \\, d\\theta$. This is $2\\pi i$ if $m = -1$.\n34. If $m \\ne -1$, the integrand has the primitive $e^{i(m+1)\\theta}/(i(m+1))$, which takes equal values at $0$ and $2\\pi$ by [[periodicity|02P2PYH3X3]]. So the integral is $0$.\n35. By steps 32 to 34, only the term $n = k$ survives, and $a_k = b_k$. So the coefficients are unique.\n\n## Idea\n\nCauchy's integral formula on the region between two circles writes $f(z)$ as an integral over the outer circle minus an integral over the inner circle. On the outer circle $1/(w - z)$ expands in powers of $z - c$, and on the inner circle in powers of $1/(z - c)$. Integrating term by term gives the two halves of the Laurent series. Integrating a Laurent series against $(z - c)^{-k-1}$ over a circle picks out the single coefficient $a_k$, which gives uniqueness.\n",
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