---
tag: 1328NH5NKD
kind: proof
name: Proof of the argument principle by the residue theorem
msc: [30C15]
proves: 0PN2BXRS8G
uses: [0NSD1TPHR0, 0WPQYW8MTE, 0K8SGQSV7G, 07PFYV65MB, 0K83KEJ43P, 0ZMT1GP4KM, 0YBVKMYX0W, 064N002TX4, 0826SGV3R0, 0YMP521ZR5, 045EWAXY9P, 0YK27TJ4WF, 0QEBN8DGVE, 01991NCP7H, 032HKYS329, 0X7SBS8E0F, 0DG10TVYPN, 02AC7JR6G6, 0YQ458D7PF, 0T7ABX4CWE, 0DG0Z72CQD]
sources:
  - url: https://en.wikipedia.org/wiki/Argument_principle
    locator: Proof of the argument principle
---

## Proof

1. 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 $W$.
2. There is $r_2 > r$ with $K_2 = \set{z : \abs{z - c} \le r_2} \subseteq W$. Otherwise there are $w_j \notin W$ with $\abs{w_j - c} < r + 1/j$. By [[sequential compactness|0K8SGQSV7G]], a subsequence tends to a point of the closed disc, which lies in the open set $W$, a contradiction.
3. $Q \cap K_2$ 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 $W$.
4. Let $\Omega = \set{z : \abs{z - c} < r_2} \setminus Q$. It is open, since it removes finitely many points from a disc, by step 3.
5. $\Omega$ 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.
6. Choose a segment in the disc $\abs{z - c} < r_2$ 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.
7. 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]].
8. 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 $K_1 \setminus Q$.
9. 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 $p$.
10. 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.
11. 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]].
12. 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 $n < m$ and $b_m \ne 0$.
13. 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 $h(a) = b_m \ne 0$.
14. 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 $\varphi = m + (z - a) h'/h$.
15. 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 $f'/f$.
16. By the [[residue at a simple pole|01991NCP7H]], $\operatorname{Res}(f'/f, a) = \lim_{z \to a} \varphi(z) = m$.
17. 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 $f' = g'/(z - b)^m - m g/(z - b)^{m+1}$.
18. 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, $\operatorname{Res}(f'/f, b) = -m$.
19. Let $U_1 = \set{z : \abs{z - c} < r_1}$. It is simply connected: it is convex, so the straight-line homotopy deforms each closed path to a constant one.
20. 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.
21. 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.
22. If $\abs{a - c} < r$, the [[winding number|0X7SBS8E0F]] $n(\gamma, a)$ is $1$, by [[Cauchy's integral formula|0DG10TVYPN]] for the constant function $1$.
23. If $\abs{a - c} > r$, then $1/(z - a)$ is holomorphic on the disc $\abs{z - c} < \abs{a - c}$, which contains $\gamma$. By [[primitives on a disc|02AC7JR6G6]] and the [[fundamental theorem of contour integration|0YQ458D7PF]], $n(\gamma, a) = 0$.
24. By steps 16, 18, 22 and 23, only the zeros and poles in $\abs{z - c} < r$ 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 $Z - P$.

## Idea

At 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.