Question

What is the integral of 1/(z-1) (z-2) (2-4) dz where C is |z| = 3?

Answer 1

$\displaystyle \int_{|z|=3}\frac{1}{(z-1) (z-2) (z-4)}\ dz$

$\displaystyle =\frac{1}{3}\int_{|z|=3}\frac{1}{z-1}\ dz+\frac{1}{6}\int_{|z|=3}\frac{1}{z-4}\ dz-\frac{1}{2}\int_{|z|=3}\frac{1}{z-2}\ dz$

But,

$\displaystyle \int_{|z|=3}\frac{1}{z-1}\ dz=2\pi i$, by Cauchy integral theorem.

$\displaystyle \int_{|z|=3}\frac{1}{z-4}\ dz=0$, by Cauchy-Goursat theorem.

$\displaystyle \int_{|z|=3}\frac{1}{z-2}\ dz=2\pi i$, by Cauchy integral theorem.

Thus,

$\displaystyle \int_{|z|=3}\frac{1}{(z-1) (z-2) (z-4)}\ dz$

$\displaystyle =\frac{1}{3}\int_{|z|=3}\frac{1}{z-1}\ dz+\frac{1}{6}\int_{|z|=3}\frac{1}{z-4}\ dz-\frac{1}{2}\int_{|z|=3}\frac{1}{z-2}\ dz$

$\displaystyle =\frac{1}{3}\times 2\pi i+\frac{1}{6}\times 0-\frac{1}{2}\times2\pi i=-\frac{\pi i}{3}$