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CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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Treating a double pole as simple gives the wrong answer

Statement refuted

Refuted claim: the simple-pole residue rule can be used unchanged at a pole of order two.

Facts & Assumptions

Given: The function f(z)=ez/(z−1)2.

[L1]

A simple pole has residue lim⁡z→a(z−a)f(z) (At a simple pole the residue is the limit of (z-a)f(z)).

[L2]

A pole of order 2 has residue ddz((z−a)2f(z))∣z=a (Residue formula for a pole of order m).

Counterexample

technique · computation
1.1L2algebra

The point a=1 is a pole of order 2 of f, and [L2] gives the correct residue Res⁡(f,1)=ddz(ez)∣z=1=e.

2.1L1step 1.1∎

If one incorrectly applies the simple-pole rule from [L1], one obtains lim⁡z→1(z−1)ez(z−1)2=lim⁡z→1ezz−1, which does not exist as a finite complex number. So the simple-pole rule does not recover the residue at a double pole.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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