CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

$f (z) = \frac{1}{z^{2}} (a = 0) .$

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

Facts & Assumptions

Given: The function $f (z) = 1 / z^{2}$ on $0 < ∣ z ∣ < 1$ .

[L1]

The residue is the coefficient of $(z - a)^{- 1}$ 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

1.1L1algebra

The Laurent expansion of $f$ at $0$ is just $z^{- 2}$ , so the coefficient of $z^{- 1}$ is $0$ ; by [L1], $Res (f, 0) = 0$ .

2.1L2step 1.1algebra

The same Laurent expansion has finite nonzero principal part $z^{- 2}$ , so [L2] makes $0$ a pole of order $2$ .

3.1step 1.1step 2.1∎

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

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.