TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Laurent coefficients are given by contour integrals and are unique

Statement

Let

$f (z) = \sum_{n \in Z} c_{n} (z - a)^{n}$

be a convergent Laurent series on the annulus $A (a; r, R)$ (Annuli in the complex plane, Convergent Laurent series on an annulus). Then for every $ρ$ with $r < ρ < R$ and every integer $n$ ,

$c_{n} = \frac{1}{2 π i} \int_{∣ ζ - a ∣ = ρ} \frac{f (ζ)}{(ζ - a)^{n + 1}} d ζ .$

Consequently, if two Laurent series on the same annulus have the same sum, then their coefficients agree term by term.

Facts & Assumptions

Given: A Laurent expansion $f (z) = \sum_{n \in Z} c_{n} (z - a)^{n}$ on $A (a; r, R)$ and a radius $ρ$ with $r < ρ < R$ .

[L1]

The Laurent series of a holomorphic function converges locally uniformly on the annulus (Laurent expansion on an annulus).

[L2]

Uniform convergence of continuous integrands on a fixed contour permits passage of the limit through the contour integral (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L3]

Complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand).

[L4]

On the positively oriented circle $∣ ζ - a ∣ = ρ$ , the integral of $(ζ - a)^{m}$ is $2 π i$ when $m = - 1$ and $0$ otherwise (On a positively oriented circle about a, the integral of (z-a)^m is zero for every integer m except -1, and is 2 pi i for m=-1).

Proof

technique · direct

1.1

On the circle $γ_{ρ} (t) = a + ρ \exp (i t)$ , the Laurent series for $f$ converges uniformly by [L1], so [L2] gives $\int_{γ_{ρ}} \frac{f (ζ)}{(ζ - a)^{n + 1}} d ζ = \lim_{N \to \infty} \int_{γ_{ρ}} \sum_{k = - N}^{N} c_{k} (ζ - a)^{k - n - 1} d ζ .$

givenL1L2

2.1

By [L3] each finite integral in step 1.1 is $\sum_{k = - N}^{N} c_{k} \int_{γ_{ρ}} (ζ - a)^{k - n - 1} d ζ$ , and [L4] kills every summand except $k = n$ , for which the integral is $2 π i$ ; therefore every finite sum equals $2 π i c_{n}$ .

step 1.1L3L4algebra

3.1

Letting $N \to \infty$ in step 2.1 proves the contour formula for $c_{n}$ .

step 1.1step 2.1

4.1

If also $f (z) = \sum_{n \in Z} d_{n} (z - a)^{n}$ on the same annulus, the same contour formula gives $d_{n} = (2 π i)^{- 1} \int_{γ_{ρ}} f (ζ) (ζ - a)^{- n - 1} d ζ = c_{n}$ for every integer $n$ , so the coefficients are unique.

step 3.1∎

Depends on

Used by

Laurent coefficients are independent of the intermediate radius Corollary
The residue is the normalized small-circle integral Corollary
The residue of an isolated singularity Definition
Characterizations of removable singularities Theorem
Laurent series split into regular and principal parts Theorem

Dependency tree · two levels

22 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

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 5 §1.3 (standard reference, not scraped)
Jeremy Orloff, MIT 18.04 Topic 7: Taylor and Laurent Series (standard reference, not scraped)