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  "body": "\n## Proof\n\n1. Let $f$ be [[meromorphic|0NSD1TPHR0]] on an open set $W \\supseteq \\set{z : \\abs{z - c} \\le r}$. Let $Q \\subseteq W$ be its set of [[poles|0WPQYW8MTE]], with no accumulation point in $W$.\n2. There is $r_2 \u003e r$ with $K_2 = \\set{z : \\abs{z - c} \\le r_2} \\subseteq W$. Otherwise there are $w_j \\notin W$ with $\\abs{w_j - c} \u003c r + 1/j$. By [[sequential compactness|0K8SGQSV7G]], a subsequence tends to a point of the closed disc, which lies in the open set $W$, a contradiction.\n3. $Q \\cap K_2$ is finite. Otherwise $K_2$, which is compact by the [[Heine–Borel theorem|07PFYV65MB]], contains a sequence of distinct poles with a limit in $K_2 \\subseteq W$. That limit is an accumulation point of $Q$ in $W$.\n4. Let $\\Omega = \\set{z : \\abs{z - c} \u003c r_2} \\setminus Q$. It is open, since it removes finitely many points from a disc, by step 3.\n5. $\\Omega$ is connected. Let $z, w \\in \\Omega$. Through $z$ and each point of $Q \\cap K_2$ there is a line, and likewise for $w$; there are finitely many such lines.\n6. Choose a segment in the disc $\\abs{z - c} \u003c r_2$ that is parallel to none of these lines. Each line meets it in at most one point, so it contains a point $v$ on none of the lines.\n7. The segments $[z, v]$ and $[v, w]$ lie in the convex disc and avoid $Q$, by the choice of $v$. So $\\Omega$ is path-connected, hence connected by [[path-connected spaces are connected|0ZMT1GP4KM]].\n8. Fix $r_1 \\in (r, r_2)$ and let $K_1 = \\set{z : \\abs{z - c} \\le r_1}$. Claim: $f$ has finitely many zeros in $K_1 \\setminus Q$.\n9. Otherwise a sequence of distinct zeros has a limit $p \\in K_1$. Near a pole $\\abs{f} \\to \\infty$, so $p \\notin Q$, and $p \\in \\Omega$. By continuity $f(p) = 0$, so case 2 of the [[isolated zeros theorem|0K83KEJ43P]] fails at $p$.\n10. So $f = 0$ on a disc around $p$. By the [[identity theorem|0YBVKMYX0W]] on $\\Omega$, $f = 0$ on $\\Omega$. But the circle $\\abs{z - c} = r$ lies in $\\Omega$ and carries no zeros. This proves the claim.\n11. Let $a$ be a zero of $f$ in $K_1$, of [[order|064N002TX4]] $m$. The order is finite: $f = 0$ near $a$ would give $f = 0$ on $\\Omega$, as in step 10. Near $a$, $f(z) = \\sum_{n \\ge 0} b_n (z - a)^n$ by [[holomorphic functions are analytic|0826SGV3R0]].\n12. Differentiating term by term $n$ times at $a$, by [[power series are holomorphic|0YMP521ZR5]], gives $f^{(n)}(a) = n! \\, b_n$. So $b_n = 0$ for $n \u003c m$ and $b_m \\ne 0$.\n13. The series $h(z) = \\sum_{n \\ge m} b_n (z - a)^{n-m}$ converges where the series of $f$ does, so its [[radius of convergence|045EWAXY9P]] is no smaller, and $h$ is holomorphic. We have $f = (z - a)^m h$ and $h(a) = b_m \\ne 0$.\n14. On a small disc around $a$ where $h \\ne 0$, $f' = m (z - a)^{m-1} h + (z - a)^m h'$. So for $z \\ne a$, $\\frac{f'(z)}{f(z)} = \\frac{\\varphi(z)}{z - a}$ with $\\varphi = m + (z - a) h'/h$.\n15. The function $\\varphi$ is holomorphic near $a$, as $h'$ is by [[infinite differentiability|0QEBN8DGVE]], and $\\varphi(a) = m \\ne 0$. So $\\abs{f'/f} \\to \\infty$ at $a$, and $a$ is a [[pole of order|0YK27TJ4WF]] $1$ of $f'/f$.\n16. By the [[residue at a simple pole|01991NCP7H]], $\\operatorname{Res}(f'/f, a) = \\lim_{z \\to a} \\varphi(z) = m$.\n17. Let $b \\in Q \\cap K_1$ be a pole of order $m$. Then $f = g/(z - b)^m$ near $b$, with $g$ holomorphic and $g(b) \\ne 0$. So $f' = g'/(z - b)^m - m g/(z - b)^{m+1}$.\n18. Near $b$, $\\frac{f'}{f} = \\frac{\\psi(z)}{z - b}$ with $\\psi = -m + (z - b) g'/g$ holomorphic and $\\psi(b) = -m \\ne 0$. As in steps 15 and 16, $\\operatorname{Res}(f'/f, b) = -m$.\n19. Let $U_1 = \\set{z : \\abs{z - c} \u003c r_1}$. It is simply connected: it is convex, so the straight-line homotopy deforms each closed path to a constant one.\n20. On $U_1$, $f'/f$ is [[holomorphic|0DG0Z72CQD]] except at the finitely many zeros and poles of $f$ in $U_1$ (steps 3 and 8). The circle $\\gamma$ of radius $r$ avoids them by hypothesis.\n21. By the [[residue theorem|032HKYS329]], $\\oint_\\gamma \\frac{f'}{f} \\, dz = 2\\pi i \\sum_a n(\\gamma, a) \\operatorname{Res}(f'/f, a)$, summed over these zeros and poles, where $\\gamma$ is the [[circle integral|0T7ABX4CWE]] path.\n22. If $\\abs{a - c} \u003c r$, the [[winding number|0X7SBS8E0F]] $n(\\gamma, a)$ is $1$, by [[Cauchy's integral formula|0DG10TVYPN]] for the constant function $1$.\n23. If $\\abs{a - c} \u003e r$, then $1/(z - a)$ is holomorphic on the disc $\\abs{z - c} \u003c \\abs{a - c}$, which contains $\\gamma$. By [[primitives on a disc|02AC7JR6G6]] and the [[fundamental theorem of contour integration|0YQ458D7PF]], $n(\\gamma, a) = 0$.\n24. By steps 16, 18, 22 and 23, only the zeros and poles in $\\abs{z - c} \u003c r$ contribute to the sum in step 21. Each zero contributes its order, and each pole contributes minus its order. Dividing by $2\\pi i$ gives $Z - P$.\n\n## Idea\n\nAt a zero of order $m$, the logarithmic derivative $f'/f$ behaves like $m/(z - a)$, and at a pole of order $m$ like $-m/(z - b)$. So $f'/f$ has simple poles with residues $m$ and $-m$ there and no other singularities. The residue theorem adds these residues up for the points inside the circle.\n",
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