Complex integration using change of variable

Question

Evaluate the integral $$\int_{\mid z\mid=2} \frac{dz}{(z-3)(z^5-1)}$$ I know how to do it by residue theorem, but this problem gives a hint "change of variables", I don't see why change of variables is useful.

Answer 1

By substituting $z\mapsto1/z$, one can avoid having many poles:

$$\oint_{|z|=2}\frac{\mathrm dz}{(z-3)(z^5-1)}=-\oint_{|z|=1/2}\frac{z^4~\mathrm dz}{(3z-1)(z^5-1)}$$

which can easily be computed. Such a substitution for rational functions is often desired if there are less poles outside the enclosed region e.g.

$$\oint_{|z|=2}\frac{\mathrm dz}{(z-3)(z^n-1)}=-\oint_{|z|=1/2}\frac{z^{n-1}~\mathrm dz}{(3z-1)(z^n-1)}=\frac{2\pi i}{3^n-1}$$