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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27
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An indented arc around a simple singularity contributes the expected residue fraction

Statement

Let f have a simple pole at a, let 0<ε<R, and for arbitrary real angles α,β let γε(t)=a+εei((1−t)α+tβ)(0≤t≤1) be the circular arc oriented from angle α to angle β inside 0<∣z−a∣<R. Then

lim⁡ε↓0∫γεf(z) dz=i(β−α)Res⁡(f,a).

In particular, an upper indentation from left to right contributes −iπ Res⁡(f,a) and a lower indentation contributes +iπ Res⁡(f,a).

Facts & Assumptions

Given: A simple pole of f at a and the oriented arc γε(t)=a+εei((1−t)α+tβ).

[L1]

At a simple pole, the Laurent principal part is c−1/(z−a), where c−1=Res⁡(f,a); hence f(z)=c−1/(z−a)+h(z) with h holomorphic near a (Simple poles, The residue of an isolated singularity).

Proof

technique · direct
1.1L1algebra

Write f(z)=c−1(z−a)−1+h(z) as in [L1]. Along the arc, put θ(t)=(1−t)α+tβ. Then z−a=εeiθ(t) and dz=i(β−α)εeiθ(t)dt, so ∫γεc−1z−a dz=i(β−α)c−1=i(β−α)Res⁡(f,a).

2.1step 1.1

The holomorphic function h is bounded on a small closed disc around a, say by M. Hence ∣∫γεh(z) dz∣≤Mε∣β−α∣, which tends to 0 with ε.

3.1step 1.1step 2.1∎

Adding the two parts from steps 1.1 and 2.1 proves the limit formula. The two indentation special cases are the choices (α,β)=(π,0) and (α,β)=(π,2π).

Depends on

Used by

Dependency tree · two levels

11 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