CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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 $f (a) = w$ .

givenL2

2.1

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

step 1.1L1L2

3.1

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 $P (f, Γ) = 0$ .

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, §5.4 (standard reference, not scraped)
R. W. Howell and J. H. Mathews, Complex Analysis, §8.7 (standard reference, not scraped)