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Laurent coefficients are independent of the intermediate radius

Statement

Let $f$ be holomorphic on the annulus $A (a; r, R)$ (Annuli in the complex plane) and let $(c_{n})_{n \in Z}$ be its Laurent coefficients. If $r < ρ_{1}, ρ_{2} < R$ , then for every integer $n$ ,

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

Facts & Assumptions

Given: A holomorphic function on $A (a; r, R)$ , its Laurent coefficients $(c_{n})$ , and radii $ρ_{1}, ρ_{2}$ with $r < ρ_{1}, ρ_{2} < R$ .

[L1]

Every Laurent coefficient is given by the contour integral on every intermediate circle inside the annulus (Laurent coefficients are given by contour integrals and are unique).

Proof

technique · direct

1.1L1

By [L1], the integral over $∣ ζ - a ∣ = ρ_{1}$ equals the coefficient $c_{n}$ , and so does the integral over $∣ ζ - a ∣ = ρ_{2}$ .

2.1step 1.1∎

Therefore the two integrals are equal to each other.

Depends on

Dependency tree · 2 levels

Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 5 §1.3