# How to calculate a residue

I have to calculate the sum of the residues at the $$n$$ m-order poles of a complex function $$f(z)=\frac{z^{np-1}}{(az^n+b)^m}\ln z$$ when applying Residue Theorem for an integral. In this function, $$m,n,p$$ are positive integers, and $$m>p$$. $$a,b\in\mathbf{R}$$ are positive.

When using the limit equation for the residue of a m-order pole, I find it hard to get the high order derivative of this function.

Can anyone help me? Thanks for any guidance.

• You could rewrite the expression as a Laurent Series in $a/b$ – C. Brendel Nov 13 '19 at 5:12

## 1 Answer

You want the sum of the residues of the poles at radius $$b/a$$. These lie on an annulus (with inner radius $$b/(2a)$$ and outer radius $$2b/a$$, say). The positively oriented path integral along its inner boundary component encloses only the (potential) singularity at $$z = 0$$; call this $$A$$. The positively oriented path integral along its outer boundary component encloses all $$n$$ of the $$m^\text{th}$$-order poles you are interested in and the singularity at $$z = 0$$; call this $$B$$. The quantity you want is $$\frac{1}{2\pi\mathrm{i}}(B - A)$$. However, $$B$$ is hard to calculate. The negatively oriented path integral along the annulus's outer boundary component encloses only the (potential) singularity at $$z = \infty$$; call this integral $$C$$. Note that $$C = -B$$, so the sum of residues you want is $$\frac{1}{2\pi\mathrm{i}}(-C-A)$$.

I observe the direct attack, computing residues of all $$n$$ of the $$m^\text{th}$$-order pole is hard. Computing the residues of the singularities at $$0$$ and $$\infty$$ might be easier...

• $ln z$ seems to have no residue at z=0… – worstcoder Nov 13 '19 at 6:28