How to you evaluate ${\int \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$ using the Residue Theorem

Question

How to you evaluate $${\int\limits_{|z|=2} \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$$ using the Residue Theorem.

This problem has been driving me crazy, I am supposed to apply the Residue formula for double poles, but I can't figure out how. I would appreciate help.

Answer 1

You don't have a double pole. You have simple poles at $0,1$ and $3$. The residues are $1/3,-1/2$ and $1/6$, respectively.

If the contour encloses all three poles, you get $2\pi i (1/3-1/2+1/6)=0$.

Answer 2

By Cauchy intgeral theorem: $$\mathcal{I}:=\oint_{\lvert z\rvert =2}\frac{1}{z\left(z-1\right)\left(z-3\right)}dz$$ $$=2\pi i\left(\operatorname{Res}_{z=0}\left[\frac{1}{z\left(z-1\right)\left(z-3\right)}\right]+\operatorname{Res}_{z=1}\left[\frac{1}{z\left(z-1\right)\left(z-3\right)}\right]\right).$$

Evaluating these residues yields $$\mathcal{I}=2\pi i(1/3-1/2)=-\frac{\pi i}{3}$$