TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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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

1.1givenL1L2

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

2.1step 1.1L3L4algebra

Fix the integer $n$ and take $N \geq ∣ n ∣$ . By [L3], the finite integral in step 1.1 is $\sum_{k = - N}^{N} c_{k} \int_{γ_{ρ}} (ζ - a)^{k - n - 1} d ζ$ . Now $k = n$ lies in the truncated range, and [L4] kills every summand except that one, for which the integral is $2 π i$ . Therefore the finite integral equals $2 π i c_{n}$ for every $N \geq ∣ n ∣$ .

3.1step 1.1step 2.1

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

4.1step 3.1∎

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.

Laurent coefficients are given by contour integrals and are unique

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources