Finding the value of an integral $\int_{|z|=3}\frac{2z^2-z-2}{z-\omega}dz$ 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 $2\pi i(2\omega^2-\omega-2)$.
 A: 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.
A: (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 $\int_{|z|=3}\frac{\mathrm{d}z}{z-w}$.
