How to show this equation equals zero? Let $P(z)$ denote a complex polynomial of degree $n$ with simple zeros $a_1, . . . , a_n$. Show that $\sum\limits_{k=1}^{n}\dfrac{a_k^p}{P'(a_k)}=0$, for $p=0,1,...n-2$. 
I have no idea where to start and the naive way of solving it by brute force is just not working. Any hint would be appreciated.
 A: ${\bf Hint}$: We see that $\frac{a_k^p}{P'(a_k)}$ is the residue of $f(z) = \frac{z^p}{P(z)}$ at the simple pole $z=a_k$. By the residue theorem the sum can therefore be rewritten as a contour integral
$$\sum_{k=1}^n \frac{a_k^p}{P'(a_k)} = \frac{1}{2\pi i}\oint_\gamma \frac{z^p}{P(z)}{\rm d}z$$
where $\gamma$ is any contour enclosing all of the roots $a_1,a_2,\ldots,a_n$ of $P(z)$. Take $\gamma$ to be a circle of radius $R$ and estimate the integral using the estimation lemma in the limit $R\to \infty$.

The result can also be derived using only the tools from real analysis. Here is a proof sketch. One first show that
$$f[a_1,a_2,\ldots,a_n] = \sum_{k=1}^n \frac{f(a_k)}{P'(a_k)}$$
where $f[a_1\ldots,a_n]$ is the divided difference of $f$. Next the mean value theorem for divided differences says that 
$$f[a_1,a_2,\ldots,a_n] = \frac{f^{(n-1)}(\zeta)}{(n-1)!}$$
for some $\zeta\in[\min(\{a_k\}_{k=1}^n),\max(\{a_k\}_{k=1}^n)]$. Taking $f(z) = z^p$ with $p\leq n-2$ then this implies $f[a_1,a_2,\ldots,a_n] = 0$ and the result follows.
