Residue

The integral formula comes from integrating the Laurent series term by term over the circle \(|z-z_0|=\rho\): \(\oint (z-z_0)^n\,dz\) equals \(0\) if \(n\neq-1\) and \(2\pi i\) if \(n=-1\). Hence \(\oint f\,dz=2\pi i\,a_{-1}\), and the residue is exactly what \(f\) contributes to the integral. At a removable singularity the residue is \(0\), but beware: it can also be \(0\) at a pole (like \(\frac1{z^2}\)) or nonzero at an essential singularity (like \(e^{1/z}\)).

For poles there are shortcuts that avoid expanding the series. At a simple pole, \(\operatorname{Res}(f,z_0)=\lim_{z\to z_0}(z-z_0)f(z)\). If \(f=\frac gh\) with \(g(z_0)\neq0\), \(h(z_0)=0\) and \(h'(z_0)\neq0\), then \(\operatorname{Res}=\frac{g(z_0)}{h'(z_0)}\). At a pole of order \(m\), \(\operatorname{Res}(f,z_0)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m f(z)\big]\).

Guiding example: for \(\frac{1}{z^2+1}\) at \(z=i\), the pole is simple and the residue is \(\frac{1}{2z}\) evaluated at \(i\), that is, \(\frac1{2i}=-\frac i2\). When the shortcut is awkward, multiply by a known series: for \(\frac{\sin z}{z^4}\) just substitute \(\sin z=z-\frac{z^3}{6}+\cdots\) and read off the coefficient of \(z^{-1}\).

The residue depends on the point and the function, not on the curve: any circle with \(0<\rho<r\) gives the same result. The natural next step is the residue theorem, which adds the residues of all singularities inside a contour to evaluate its integral in one stroke.