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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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Negative Laurent coefficients vanish on a Hartogs figure

Statement

In the notation of Laurent coefficients on Hartogs slices depend holomorphically on the remaining variables, one has

an(w)=0for every n<0 and every ∣w∣<1.

Facts & Assumptions

Given: A holomorphic function f on H(r,s), a radius ρ with r<ρ<1, and the Laurent coefficient functions an defined in the preceding lemma.

[L1]

Each an is holomorphic on the unit disc, and for fixed w the numbers an(w) are the Laurent coefficients of the slice z↦f(z,w) on the annulus r<∣z∣<1 (Laurent coefficients on Hartogs slices depend holomorphically on the remaining variables).

[L2]

A holomorphic function on a punctured disc has a removable singularity exactly when its Laurent expansion has no negative powers (Characterizations of removable singularities).

[L3]

A holomorphic function on a connected open set that vanishes on a nonempty open subset vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

Proof

technique · direct
1.1L1L2

Fix w with ∣w∣<s. Then (z,w)∈H(r,s) for every ∣z∣<1, so the slice z↦f(z,w) is holomorphic on the whole unit disc. By [L1], the coefficients an(w) are the Laurent coefficients of that slice on r<∣z∣<1, and [L2] therefore forces an(w)=0 for every n<0.

2.1step 1.1L3

For each fixed n<0, step 1.1 shows that the holomorphic function an is zero on the nonempty open disc {∣w∣<s}. The domain {∣w∣<1} is connected, so [L3] gives an≡0 there.

3.1step 2.1∎

As n<0 was arbitrary, every negative Laurent coefficient vanishes on the whole parameter disc {∣w∣<1}.

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