Let $p$ be an $nth$ degree polynomial with zeros $a_1,a_2, \dots , a_n$. If $R$ is real number such that $|a_i| < R$ for $i = 1,2,3, \dots , n$ evaluate
$$\frac{1}{2\pi i} \int_{|z|=R} z \frac{p'(z)}{p(z)}dz$$
This, to me, screams argument principle. But that requires there be no extra $z$ there and does not require the point on the zeroes. How do we go about this?
If $r$ is a root of $p(z)$, then $p(z)$ can be written as $q(z)(z-r)^k$ for some polynomial $q(z)$ such that $q(r)\neq0$ and then\begin{align}\frac{zp'(z)}{p(z)}&=\frac{kzq(z)(z-r)^{k-1}+zq'(z)(z-r)^k}{q(x)(z-r)^k}\\&=\frac{kzq(z)+zq'(z)(z-r)}{q(z)(z-r)}.\end{align}So,$$\operatorname{res}_r\left(\frac{zp'(z)}{p(z)}\right)=\frac{krq(r)}{q(r)}=kr.$$Therefore, by the residue theorem, your integral is equal to the sum of the roots of $p(z)$, counted with their multiplicities.
