CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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A Laurent series on a punctured disc can have infinitely many negative powers

Statement refuted

Refuted claim: every Laurent series on a punctured disc has only finitely many negative powers.

The witness is the Laurent expansion of $e^{1 / z}$ on $0 < ∣ z ∣ < 1$ .

Facts & Assumptions

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

[L1]

Every holomorphic function on a punctured disc has a Laurent expansion there (Laurent expansion on an annulus).

[L2]

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

[L3]

The function $e^{1 / z}$ has an essential singularity at $0$ (e^{1/z} has an essential singularity at 0 and omits the value 0).

Counterexample

technique · direct

1.1

By [L1], the function $e^{1 / z}$ has a Laurent expansion on $0 < ∣ z ∣ < 1$ .

2.1

If that Laurent expansion had only finitely many negative powers, its principal part would be finite. It cannot be zero, because then the singularity would be removable and hence not essential. So the principal part would be finite and nonzero.

step 1.1L3

3.1

By [L2], a finite nonzero principal part would make the singularity a pole, contradicting [L3]. Therefore the Laurent expansion of $e^{1 / z}$ has infinitely many negative powers.

step 2.1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

David Greenfield, Rutgers Math 403 diary (standard reference, not scraped)