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  "body": "\n## Proof\n\n1. Let $U$, $a_1, \\dots, a_n$, $f$ and $\\gamma : [a, b] \\to U \\setminus \\set{a_1, \\dots, a_n}$ be as in the statement.\n2. For each $k$, choose $r_k \u003e 0$ so that the disc $D_k = \\set{z : \\abs{z - a_k} \u003c r_k}$ lies in $U$ and contains no $a_j$ with $j \\ne k$.\n3. By the [[existence of Laurent expansions|06WHRARZKY]] on $0 \u003c \\abs{z - a_k} \u003c r_k$, $f(z) = \\sum_{m \\in \\Z} c_{k,m} (z - a_k)^m$ there. By the definition of the [[residue|0JFGKKWWFK]], $\\operatorname{Res}(f, a_k) = c_{k,-1}$.\n4. Let $E_k(\\zeta) = \\sum_{m \\ge 1} c_{k,-m} \\zeta^m$. By the definition of a [[Laurent series|0X3XD2ATB0]], it converges at $\\zeta = 1/(z - a_k)$ for $0 \u003c \\abs{z - a_k} \u003c r_k$, that is, at each $\\zeta$ with $\\abs{\\zeta} \u003e 1/r_k$.\n5. At such points the terms are bounded. So for each $t \u003e 0$ there is $B_t$ with $\\abs{c_{k,-m}} t^m \\le B_t$ for all $m$: take $t' \u003e \\max(t, 1/r_k)$ and use $t^m \\le t'^m$.\n6. So the [[radius of convergence|045EWAXY9P]] of $E_k$ is $\\infty$, and $E_k$ is [[holomorphic|0DG0Z72CQD]] on $\\C$ by [[power series are holomorphic|0YMP521ZR5]].\n7. Let $P_k(z) = E_k\\bigl(1/(z - a_k)\\bigr)$ for $z \\ne a_k$. As a composition of [[complex differentiable|0D1C9Y1EXA]] functions, $P_k$ is holomorphic on $\\C \\setminus \\set{a_k}$.\n8. On $0 \u003c \\abs{z - a_k} \u003c r_k$, the function $T_k = f - P_k$ is the sum of the power series $\\sum_{m \\ge 0} c_{k,m} (z - a_k)^m$. This series converges on the punctured disc, so its radius is at least $r_k$, and its sum is holomorphic on all of $D_k$.\n9. Let $g = f - \\sum_{k=1}^{n} P_k$ on $U \\setminus \\set{a_1, \\dots, a_n}$. It is holomorphic there.\n10. On $0 \u003c \\abs{z - a_k} \u003c r_k$, $g = T_k - \\sum_{j \\ne k} P_j$. The right side is holomorphic on all of $D_k$, since $a_j \\notin D_k$ for $j \\ne k$.\n11. Define $g(a_k)$ as the value of this right side at $a_k$. Then $g$ is holomorphic on a disc around each $a_k$, so $g$ is holomorphic on $U$.\n12. By [[Cauchy's integral theorem|01MPZXYVYF]] on the simply connected set $U$, $\\oint_\\gamma g(z) \\, dz = 0$. By linearity of the [[contour integral|00FWNP9TXY]], $\\oint_\\gamma f(z) \\, dz = \\sum_{k=1}^{n} \\oint_\\gamma P_k(z) \\, dz$.\n13. Fix $k$. The image of $\\gamma$ is compact and does not contain $a_k$. By [[attainment of bounds|0V4DET59NW]], $d = \\min_t \\abs{\\gamma(t) - a_k} \u003e 0$.\n14. On the image of $\\gamma$, $\\abs{c_{k,-m} (z - a_k)^{-m}} \\le \\abs{c_{k,-m}} d^{-m} \\le B_{2/d} \\, 2^{-m}$, with $B_{2/d}$ from step 5.\n15. By the [[Weierstrass M-test|0NX5NWAFC9]], applied to real and imaginary parts, the series for $P_k$ converges uniformly on the image of $\\gamma$.\n16. By the [[estimation lemma|0DJMGPN8K9]] applied to the tails, $\\oint_\\gamma P_k(z) \\, dz = \\sum_{m \\ge 1} c_{k,-m} \\oint_\\gamma (z - a_k)^{-m} \\, dz$.\n17. For $m \\ge 2$, the function $(z - a_k)^{-m}$ has the primitive $-(z - a_k)^{1-m}/(m - 1)$ on $\\C \\setminus \\set{a_k}$. By the [[fundamental theorem of contour integration|0YQ458D7PF]], its integral over the closed path $\\gamma$ is $0$.\n18. For $m = 1$, the definition of the [[winding number|0X7SBS8E0F]] gives $\\oint_\\gamma \\frac{dz}{z - a_k} = 2\\pi i \\, n(\\gamma, a_k)$.\n19. By steps 16 to 18, $\\oint_\\gamma P_k(z) \\, dz = 2\\pi i \\, n(\\gamma, a_k) \\, c_{k,-1} = 2\\pi i \\, n(\\gamma, a_k) \\operatorname{Res}(f, a_k)$.\n20. Summing over $k$ in step 12 gives $\\oint_\\gamma f(z) \\, dz = 2\\pi i \\sum_{k=1}^{n} n(\\gamma, a_k) \\operatorname{Res}(f, a_k)$.\n\n## Idea\n\nNear each $a_k$, the negative powers in the Laurent series form the principal part $P_k$, which is holomorphic everywhere except at $a_k$. Subtracting all principal parts from $f$ leaves a function with only removable singularities, whose integral over $\\gamma$ is zero by Cauchy's theorem. In each principal part only the term $c_{k,-1}/(z - a_k)$ contributes, and it contributes $2\\pi i$ times the winding number times the residue.\n",
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