Question

<e> Evaluate the integral ∮c= 1 /(z-1)(z-2)(z-4) dz where C is |z| = 3 ?

Answer 1

Let us evaluate the integral $∮_c\frac{dz}{(z-1)(z-2)(z-4)}$ where $C$ is $|z| = 3.$

It follows that $z_1=1$ and $z_2=2$ are the simple poles containing in the inner part of the circle $|z|=3.$

According to residue theorem

$∮_c\frac{dz}{(z-1)(z-2)(z-4)}=2\pi i(Res_{z=1}\frac{1}{(z-1)(z-2)(z-4)}+Res_{z=2}\frac{1}{(z-1)(z-2)(z-4)}) \\ =2\pi i(\frac{1}{(z-2)(z-4)}|_{z=1}+\frac{1}{(z-1)(z-4)}|_{z=2}) =2\pi i(\frac{1}{(1-2)(1-4)}+\frac{1}{(2-1)(2-4)}) \\=2\pi i(\frac{1}{3}-\frac{1}{2}) =2\pi i(-\frac{1}{6})=-\frac{\pi i}3.$