CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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The argument principle counts preimages of a target value

Statement

Let $Ω \subseteq C$ be open, let $f$ be meromorphic on $Ω$ , let $Γ$ be admissible for the residue theorem in $Ω$ , and let $w \in C$ satisfy $f (z) \neq w$ for every $z \in Γ^{*}$ . Then

$\frac{1}{2 π i} \int_{Γ} \frac{f^{'} (z)}{f (z) - w} d z = N_{w} (f, Γ) - P (f, Γ),$

where

$N_{w} (f, Γ) : = \sum_{f (a) = w} n (Γ, a) {ord}_{a} (f - w)$

is the weighted multiplicity count of the preimages of $w$ , and $P (f, Γ)$ is the weighted pole count of $f$ .

In particular, if $f$ is holomorphic on $Ω$ , then the pole term vanishes and the integral counts the preimages of $w$ with multiplicity.

Facts & Assumptions

Given: A meromorphic function $f$ on an open set $Ω$ , an admissible cycle $Γ$ , and a complex number $w$ with $f (z) \neq w$ on $Γ^{*}$ .

[L1]

The argument principle applied to a meromorphic function $g$ gives $\frac{1}{2 π i} \int_{Γ} \frac{g^{'} (z)}{g (z)} d z = Z (g, Γ) - P (g, Γ)$ (The argument principle for an admissible null-homologous cycle).

[L2]

Derivatives ignore constants, so $(f - w)^{'} = f^{'}$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · direct

1.1givenL2

Put $g : = f - w$ . Then $g$ is meromorphic on $Ω$ , has the same poles as $f$ , and has no zero on $Γ^{*}$ by the hypothesis on $w$ . Its zeros are exactly the points $a$ with $f (a) = w$ .

2.1step 1.1L1L2

Applying [L1] to $g$ and then using [L2] gives $\frac{1}{2 π i} \int_{Γ} \frac{f^{'} (z)}{f (z) - w} d z = \frac{1}{2 π i} \int_{Γ} \frac{g^{'} (z)}{g (z)} d z = Z (g, Γ) - P (g, Γ) .$

3.1step 1.1step 2.1∎

By step 1.1, the zero count $Z (g, Γ)$ is exactly $N_{w} (f, Γ)$ and the pole count $P (g, Γ)$ is exactly $P (f, Γ)$ . Substituting that into step 2.1 proves the formula. If $f$ is holomorphic, then it has no poles, so $P (f, Γ) = 0$ .

The argument principle counts preimages of a target value

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources