Finding the value of an integral $\int_{|z|=3}\frac{2z^2-z-2}{z-\omega}dz$

Question

What is the value of $$\int_{|z|=3}\frac{2z^2-z-2}{z-\omega}dz$$ when $|\omega|>3$. I know that when $|\omega|<3$ the value is $2\pi i(2\omega^2-\omega-2)$.

Answer 1

Note that when $|\omega|>3$, the function $\displaystyle \frac{2z^2-z-2}{z-\omega}$ is holomorphic in an open set containing $\{|z|\leq3\}$. Therefore, the integral is zero by Cauchy's integral theorem.

Answer 2

(This is probably an unnecessary hint for you at this point, but I have a slow browser and I wrote this before I saw there was an accepted answer... >.<)

To make the application of Cauchy's integral theorem even more transparent, you might first calculate the partial fraction decomposition of the integrand:

$$\begin{align} \frac{2z^2-z-2}{z-w} &=\frac{2z^2-2zw}{z-w}+\frac{2zw-z-2}{z-w}\\ &=2z+\frac{2zw-2w^2}{z-w}+\frac{2w^2-z-2}{z-w}\\ &=2z+2w+\frac{w-z+2w^2-w-2}{z-w}\\ &=2z+2w-1+\frac{2w^2-w-2}{z-w}.\\ \end{align}$$

Then the integral is seen to be,

$$\begin{align} \int_{|z|=3}\frac{2z^2-z-2}{z-w}\,\mathrm{d}z &=\int_{|z|=3}\left(2z+2w-1\right)\,\mathrm{d}z+\int_{|z|=3}\frac{2w^2-w-2}{z-w}\,\mathrm{d}z\\ &=0+\left(2w^2-w-2\right)\int_{|z|=3}\frac{\mathrm{d}z}{z-w},\\ \end{align}$$

thus reducing the whole problem to evaluating the basic integral $\int_{|z|=3}\frac{\mathrm{d}z}{z-w}$.