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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Zero residue does not imply a removable singularity

Statement refuted

Refuted claim: if Res(f,a)=0, then a is a removable singularity of f.

The witness is

f(z)=1z2(a=0).

Its residue at 0 is 0, but 0 is a pole of order 2, not a removable singularity.

Facts & Assumptions

Given: The function f(z)=1/z2 on 0<z<1.

[L1]

The residue is the coefficient of (za)1 in the Laurent expansion (The residue of an isolated singularity).

[L2]

A finite nonzero principal part characterizes a pole (Characterizations of poles).

Counterexample

technique · direct
1.1

The Laurent expansion of f at 0 is just z2, so the coefficient of z1 is 0; by [L1], Res(f,0)=0.

L1algebra
2.1

The same Laurent expansion has finite nonzero principal part z2, so [L2] makes 0 a pole of order 2.

L2step 1.1algebra
3.1

Thus f has residue 0 at 0 but the singularity is not removable, refuting the claim.

step 1.1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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.