The problem:

Let $\displaystyle P(z)$ be a polynomial of degree greater than 1, with all of it's zeroes inside the circle $\displaystyle C_{R}=\{z:|z|=R\}$. show that $\displaystyle \int_{C_{R}} \frac{dz}{P(z)} = 0$.

My idea for a proof:

Since every polynomial over $\displaystyle C$ can be writen as $\displaystyle P(z)=(z-a_1)(z-a_2)...(z-a_k)$, where $\displaystyle a_1,...a_k$ are all the zeroes of $\displaystyle P(z)$ (not necessarily unique). Than we can write $\displaystyle \int_{C_{R}} \frac{dz}{P(z)} = \int_{C_{R}} \frac{b_1}{z-a_1} dz + ... + \int_{C_R}} \frac{b_k}{z-a_k} dz$.

Now we can say that for every $\displaystyle a_i$ we can find a circle $\displaystyle C_{a_{i}}$ such that $\displaystyle C_{a_{i}}\subset C_{R}, a_i\notin C_{a_{i}}$

and therfore $\displaystyle \frac {1}{z-a_i}$ is analytic in $\displaystyle C_{a_i}}$ so we get $\displaystyle \int_{C_{a_{i}}} \frac{dz}{z-a_i} =0$, but since $\displaystyle C_{a_{i}}\subset C_{R}$ we have $\displaystyle \int_{C_{R}} f = \int_{C_{a_i}} f = 0$ and that shows what we needed to show.

I would appreciate some thoughts on this proof as I am not 100% sure it is valid.