Let \(f\) be holomorphic on a punctured disc \(D_r(z_0)\setminus\{z_0\}\), so that \(z_0\) is an isolated singularity of \(f\), and let \[ f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k \qquad (0<\abs{z-z_0}<r) \] be its Laurent expansion there. The principal part of \(f\) at \(z_0\) is \(\sum_{k\le -1}a_k(z-z_0)^k\), and the residue of \(f\) at \(z_0\) is \(\Res_{z=z_0}f\coloneqq a_{-1}\) [@Ahl79].
For every \(0<\rho<r\) one has \(\Res_{z=z_0}f=\frac{1}{2\pi i}\int_{\abs{z-z_0}=\rho}f(z)\dz\). If \(z_0\) is a pole of order $n\ge1$, then \(a_k = 0\) for \(k<-n\) and the principal part is \[ P(z)=\frac{a_{-n}}{(z-z_0)^n}+\cdots+\frac{a_{-1}}{z-z_0}; \] the function \(f-P\) equals \(\sum_{k\ge0}a_k(z-z_0)^k\) on \(D_r(z_0)\setminus\{z_0\}\), so it extends to a holomorphic function on \(D_r(z_0)\), and \(P\) is the unique polynomial in \((z-z_0)^{-1}\) without constant term with this property.