Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$ 
Evaluate 
  $$\int_{|z|=3}  {1 \over P(z)}dz$$ 
  where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.

I proved that  $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are inside the ball.
$$P(z)=(z-9)(2z^4+2z^3+z^2-2z+2)=(z-9)(z-a_1)(z-\bar a_1)(z-a_2)(z-\bar a_2).$$ 

The roots of $P(z)= 2z^4 + 2z^3+ z^2-2z+2 $ are $a_1 = -1-i$, $a_2 = \frac{-a_1}{2}$ , $\overline{a_1}$ and $\overline{a_2}$. Now by the residue Theorem, you get,
\begin{equation*}
\int_{|z|=3} \frac{dz}{P(z)} = 2i\pi \left( \frac{1}{(a_1 -9)P'(a_1)} + \frac{1}{(\overline{a_1}-9)P'(\overline{a_1})} + \frac{1}{(a_2-9)P'(a_2)} + \frac{1}{(\overline{a_2}-9)P'(\overline{a_2})} \right)
\end{equation*}
Since $P'$ has real coefficients, this sum is easier to compute....
