Laurent series

The coefficients are computed by \(a_n=\frac1{2\pi i}\oint_{|\zeta-z_0|=\rho}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta\), where \(\rho\) is any radius between \(r\) and \(R\); by Cauchy it does not depend on \(\rho\). The series splits into the regular part (\(n\ge0\)), converging on \(|z-z_0|<R\), and the principal part (\(n<0\)), converging on \(|z-z_0|>r\). Together they converge exactly on the annulus.

The expansion is unique, but it depends on the annulus: \(\frac1{z-1}\) equals \(-\sum_{n\ge0}z^n\) on \(|z|<1\) and \(\sum_{n\ge1}z^{-n}\) on \(|z|>1\). That is why a Laurent exercise must always state the region. In practice one uses known series (\(e^{w}\), \(\sin w\), the geometric series) with the substitution \(w=1/z\) or \(w=z/a\), and partial fractions for rational functions.

If \(f\) is holomorphic on the whole disc, the principal part is zero and the Taylor series is recovered. The coefficient \(a_{-1}\) plays a special role: \(\oint f\,dz=2\pi i\,a_{-1}\) over a circle in the annulus, since \(\oint z^n dz=0\) for \(n\neq-1\). Residues are born from this.

The principal part is what classifies the behaviour of \(f\) near an isolated singularity: no negative powers, finitely many, or infinitely many. Typical mistake: using the expansion for the wrong region, or expanding in \(z\) when the centre is another point.