Contour integration of a function $(z-a)^{-m}(z-b)^{-n}$

Question

Let $m, n \in \mathbb{N}$. Moreover let's find $a, b$ such that both are inside a contour $C$.

I am asked to prove that $$\int_C\frac1{(z-a)^m}\frac1{(z-b)^n}\,\mathrm{d}z=0.$$

Cauchy's theorem doesn't work here unfortunately because partial fraction decomposition seems pretty impossible.

How can it be done?

Answer 1

Since all singularities of the integrand $$f(z)=\frac1{(z-a)^m(z-b)^n}$$ are inside the integration contour the simplest way to evaluate the integral is to use:

$$\oint_C f(z)dz=-2\pi i\operatorname{Res}(f(z),\infty) =2\pi i \operatorname{Res}\left[\frac1{w^2}f\left(\frac1w\right),0\right]\\ =2\pi i \operatorname{Res}\left[\frac1{w^2}\frac{w^{m+n}}{(1-aw)^m(1-bw)^n},0\right] =2\pi i\delta_{m+n,1}. $$

Thus the integral is zero provided that $m+n\ne1$. If $m,n$ are both positive this is the case.

Answer 2

We have $\int_C {1 \over (z-a)^m (z-b)^n } dz = 2 \pi i (\operatorname{res} (f,a)+\operatorname{res} (f,b))$.

$\operatorname{res} (f,a) = {1 \over (m-1)!} \lim_{z \to a} {d^{m-1} \over dz^{m-1}}{1 \over (z-b)^n } = (-1)^n { (n+m-2)! \over (m-1)! (n-1)!} (b-a)^{-n-m+1}$

$\operatorname{res} (f,b) = {1 \over (n-1)!} \lim_{z \to b} {d^{n-1} \over dz^{n-1}}{1 \over (z-a)^m } = (-1)^m { (n+m-2)! \over (m-1)! (n-1)!} (a-b)^{-n-m+1}$

Adding the two together yields zero.