Search the corpus

    Laurent expansion on an annulus

    T-4XPWL

    A function holomorphic on an annulus has a unique Laurent expansion there; on a punctured disc, the number of nonzero negative-index terms determines the type of the isolated singularity.Suggest a slogan

    Let \(z_0\in\CC\), \(0\le r<R\le\infty\), and let \(f\) be holomorphic on the annulus \[ A\coloneqq\ts{z\in\CC\st r<\abs{z-z_0}<R}. \] Then \(f\) has a two-sided expansion \[ f(z)=\sum_{n\in\ZZ}c_n(z-z_0)^n \] converging locally uniformly on \(A\), with coefficients \[ c_n=\frac{1}{2\pi i}\int_{\abs{z-z_0}=\rho}\frac{f(z)}{(z-z_0)^{n+1}}\dz \] for any \(r<\rho<R\), independent of \(\rho\). The expansion is unique.

    If \(r=0\), so that \(z_0\) is an isolated singularity of \(f\), then the principal part \(\sum_{n<0}c_n(z-z_0)^n\) has no nonzero terms if and only if \(z_0\) is removable, finitely many if and only if \(z_0\) is a pole, and infinitely many if and only if \(z_0\) is essential.

    See [@Ahl79, Chapter 5, Section 1.3, p. 184].

    Guide Appearances

    What links to this