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  "body": "\n## Proof\n\n*A lemma.* Let $\\varphi$ be [[holomorphic|0DG0Z72CQD]] on a disc $D_\\delta = \\set{z : \\abs{z - c} \u003c \\delta}$ with $\\delta \\le r$, and let $f(z) = \\varphi(z)/(z - c)$ for $0 \u003c \\abs{z - c} \u003c \\delta$. We show $\\operatorname{Res}(f, c) = \\varphi(c) = \\lim_{z \\to c} (z - c) f(z)$.\n\n1. By [[holomorphic functions are analytic|0826SGV3R0]], $\\varphi(z) = \\sum_{n \\ge 0} b_n (z - c)^n$ on $D_\\delta$, with $b_0 = \\varphi(c)$.\n2. Dividing by $z - c$ gives $f(z) = \\sum_{n \\ge -1} b_{n+1} (z - c)^n$ for $0 \u003c \\abs{z - c} \u003c \\delta$. This is a [[Laurent series|0X3XD2ATB0]] with coefficient $b_0$ at $(z - c)^{-1}$.\n3. The Laurent expansion of $f$ on $0 \u003c \\abs{z - c} \u003c r$ also converges to $f$ on the smaller annulus $0 \u003c \\abs{z - c} \u003c \\delta$.\n4. By the uniqueness part of the [[existence of Laurent expansions|06WHRARZKY]] on $0 \u003c \\abs{z - c} \u003c \\delta$, the two expansions have the same coefficients.\n5. By the definition of the [[residue|0JFGKKWWFK]], $\\operatorname{Res}(f, c) = b_0 = \\varphi(c)$.\n6. Also $(z - c) f(z) = \\varphi(z) \\to \\varphi(c)$ as $z \\to c$, since $\\varphi$ is continuous. This proves the lemma.\n\n*The first formula.*\n\n7. Since $c$ is a [[pole of order|0YK27TJ4WF]] $1$, there are $\\delta \u003e 0$ and $\\varphi$ holomorphic on $D_\\delta$ with $f(z) = \\varphi(z)/(z - c)$ for $0 \u003c \\abs{z - c} \u003c \\delta$. We may take $\\delta \\le r$.\n8. By the lemma, $\\operatorname{Res}(f, c) = \\lim_{z \\to c} (z - c) f(z)$.\n\n*The formula for $g/h$.*\n\n9. Let $g$ and $h$ be holomorphic on a disc around $c$, with $g(c) \\ne 0$, $h(c) = 0$ and $h'(c) \\ne 0$, and let $f = g/h$.\n10. By analyticity, $h(z) = \\sum_{n \\ge 1} d_n (z - c)^n$ on a disc around $c$, with $d_1 = h'(c)$ by [[term-by-term differentiation|0YMP521ZR5]].\n11. The series $k(z) = \\sum_{n \\ge 1} d_n (z - c)^{n-1}$ converges on the same disc, since for $z \\ne c$ its partial sums are those of $h(z)$ divided by $z - c$. Its [[radius of convergence|045EWAXY9P]] is at least the radius of the disc, because a convergent series has bounded terms.\n12. By [[power series are holomorphic|0YMP521ZR5]], $k$ is holomorphic there. We have $h(z) = (z - c) k(z)$ and $k(c) = h'(c) \\ne 0$.\n13. By continuity, $k \\ne 0$ and $g$ is defined on some disc $D_\\delta$ with $\\delta \\le r$. On it, $\\varphi = g/k$ is holomorphic, and $f(z) = \\varphi(z)/(z - c)$ for $0 \u003c \\abs{z - c} \u003c \\delta$.\n14. By the lemma, $\\operatorname{Res}(f, c) = \\varphi(c) = g(c)/h'(c)$.\n\n## Idea\n\nNear a simple pole, $f$ is a holomorphic function $\\varphi$ divided by $z - c$. Dividing the Taylor series of $\\varphi$ by $z - c$ gives the Laurent series of $f$, whose coefficient at $(z - c)^{-1}$ is $\\varphi(c)$. For $f = g/h$, factor $h(z) = (z - c) k(z)$ with $k(c) = h'(c)$, and take $\\varphi = g/k$.\n",
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