CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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A positively oriented circle integral is the sum of the enclosed residues

Statement

Let $C (a, r)$ be a positively oriented circle, and let $f$ be meromorphic on a neighbourhood of the closed disc $\overline{D (a, r)}$ with no pole on the circle. Then

$\int_{C (a, r)} f (z) d z = 2 π i \sum_{∣ b - a ∣ < r} Res (f, b),$

the sum being over the poles of $f$ inside the circle.

Facts & Assumptions

Given: A positively oriented circle $C (a, r)$ and a meromorphic $f$ on a neighbourhood of $\overline{D (a, r)}$ with no pole on the circle.

[L1]

The residue theorem holds for an admissible cycle (The residue theorem for a null-homologous cycle).

[L2]

A positively oriented circle has index $1$ at interior points and $0$ at exterior points (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

Proof

technique · direct

1.1L1

The circle $C (a, r)$ is null-homologous in any open set containing the closed [given, L1] disc it bounds, so applies to it.

2.1L1

By [L2], every pole $b$ with $∣ b - a ∣ < r$ contributes the factor [step 1.1, L2] ∎ $n (C (a, r), b) = 1$ , while every pole outside the circle contributes the factor $0$ . Substituting those indices into gives the formula.

Depends on

Used by

Weak solutions of a degree-zero divisor and the logarithmic-derivative identity Lemma

Dependency tree · two levels

33 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

E. Stein and R. Shakarchi, Complex Analysis, Ch. 3, Corollary 2.3 (standard reference, not scraped)