Solution
= Solution
In the <Laurent series>
$$
f(z)=\sum_{j=-m}^{\infty}c_j(z-z_0)^j,\qquad 0<|z-z_0|<R,
$$
about a <pole> of order $m$, the <residue> is the coefficient
$$
\boxed{\operatorname{Res}_{z=z_0}f=c_{-1}}
=\frac1{2\pi i}\oint_{|z-z_0|=\rho}f(z)\,dz,\qquad 0<\rho<R,
$$
where the circle is traversed anticlockwise. Equivalently, the <residue> is $\frac1{(m-1)!}\left.\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]\right|_{z=z_0}$.