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So this is an exercise from my complex analysis course.

Let $P(z)$ be a polynomial of degree $n\geq 2$ with simple zeros $a_1, \dots, a_n$. I'm trying to prove that $$\sum_{k=1}^n\frac{{a_k}^m}{P'(a_k)}=0$$ for $m=0, \dots, n-2$

My first thought is to use induction. When $n=2$ we obtain the desired identity by straightforward computation.

Assume that the identity holds for $n=N$. Consider a polynomial $P$ of degreee $N+1$ with simple zeros $a_1, \dots, a_{N+1}$. Since $P$ has at most $N+1$ zeros, the $a_k$'s must be all of them, each with multiplicity $1$. Therefore we can write $$P(z)=c\prod_{k=1}^{N+1} (z-a_k)$$ where $c\in \mathbb{C}$ is a constant.

Letting $Q(z)=c\prod_{k=1}^{N} (z-a_k)$, we have $P(z)=(z-a_{N+1})Q(z)$. Therefore $$P'(z)=Q(z)+(z-a_{N+1})Q'(z)$$ Hence $$P'(a_k)=Q(a_k)+(a_k-a_{N+1})Q'(a_k)=(a_k-a_{N+1})Q'(a_k)$$ for $k=1, \dots, N$.

Moreover, $P'(a_{N+1})=Q(a_{N+1})$.

By induction hypothesis, we have $$\sum_{k=1}^N\frac{{a_k}^m}{Q'(a_k)}=0$$ for $m=1, \dots, N-2$.

We want to show that $$\sum_{k=1}^{N+1}\frac{{a_k}^m}{P'(a_k)}=0$$ for $m=1, \dots, N-1$.

But I just can't simplify the first one to get the second. Can someone help me? Am I on the right track (do we really have to use induction)? Any effort is appreciated!

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    $\begingroup$ Consider $\oint_{|z|=R} \frac{z^m{\rm d}z}{P(z)}$. Evaluate this using the residue theorem and estimate the integral using the estimation lemma. Take $R\to\infty$ in the end. $\endgroup$ – Winther Dec 2 '16 at 3:12
  • $\begingroup$ @Winther Thanks! I didn't realize it was that easy! :-) So there is nothing special about $n-2$, and we need it only because the integral vanishes for $0\leq m\leq n-2$ (by estimation lemma). Is that correct? $\endgroup$ – Liebster Jugendtraum Dec 2 '16 at 3:18
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    $\begingroup$ Yes this argument only works when $m \leq n-2$, but it's easy to show by example that it can fail when $m = n-1$. Take for example if $n=2$ and $P(z) = (z-1)(z+1)$ then $\sum_{x_i=\pm 1} \frac{x_i^m}{P'(x_i)} = \frac{1}{2}\sum_{x_i=\pm 1} x_i^{m-1} \not = 0$ if $m = 1 = n-1$. $\endgroup$ – Winther Dec 2 '16 at 3:22

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