Let $f(z)=a_nz^n +a_{n-1} z^{n-1} +...+a_1z+a_0$ be a polynomial of degree $n \ge 2$. Prove that the sum of the residues of $\frac{1}{f(z)}$ is zero.
Ok, so here is my thinking process so far: At first I had no clue what to do, but then I thought about the fact that this is a polynomial, which is analytic throughout and if that is so, then the residue is 0. But this seemed too simple. Taking $\frac{1}{f(z)}$ into consideration, I knew that there had to be some singularities, if nothing else, 0. Then i got stuck again. Im thinking that it has something to do with the deformation of paths??
Further: Textbook definition gives me something like this $\,\,\,\,\int_{C_2} \frac{dz}{f(z)} + \int_{-C_2} \frac{dz}{f(z)} =0$
And since C is a positively oriented simple closed contour with center at the origin, with $z_0$ entirely inside, then $\int_C \frac{dz}{f(z)}=2\pi i$ and since $\frac{dz}{f(z)}$ is analytic everywhere except at $z=0$, then the sum of the integrals equals 0.
If the degree of our polynomial $f(x)$ is greater than $1$, we have: $$ f(x) \gg |x|^2 $$ as long as $|x|\to +\infty$. This gives: $$\frac{1}{f(x)}\ll\frac{1}{|x|^2},$$ so the integral of $f(x)$ over the set $\{x:|x|=R\}$ is $O\left(\frac{1}{R}\right)$.
Now we just need to apply the Residue theorem and consider the limit as $R\to +\infty$.