Proving an interesting identity Let $Q(z)=(z-\alpha_1)\cdots(z-\alpha_n)$ be a polynomial of degree $>1$ with distinct roots outside the real line.
We have
$$\sum_{j=1}^n \frac{1}{Q'(\alpha_j)}=0.$$
I know an interesting but indirect proof using the continuity of the Fourier transform, but I want to know whether there is a proof relying on more rudimentary techniques.
 A: Let $n \geq 1$ and $\alpha_1, \cdots, \alpha_n \in \Bbb{C}$ be distinct. If we put $Q(z) = (z-\alpha_1)\cdots(z-\alpha_n)$, then from the partial fraction decomposition, it follows that
$$ \frac{1}{Q(z)} = \sum_{j=1}^{n} \frac{1}{Q'(\alpha_j)(z - \alpha_j)}. $$
From this, we have
$$ \sum_{j=1}^{n} \frac{1}{Q'(\alpha_j)}
= \lim_{|z|\to\infty} \sum_{j=1}^{n} \frac{z}{Q'(\alpha_j)(z - \alpha_j)}
= \lim_{|z|\to\infty} \frac{z}{Q(z)}
= \begin{cases}
1, & \text{if } n = 1 \\
0, & \text{if } n > 1
\end{cases}. $$
A: $f(z)=\frac{1}{Q(z)}$ is a meromorphic function with simple poles at $\alpha_1,\ldots,\alpha_n$. Since $f(z)=O\left(\frac{1}{\|z\|}\right)$ as $\|z\|\to +\infty$, by integrating $f(z)$ along a big circle centered at the origin we have:
$$ 2\pi i\sum_{k=1}^{n}\text{Res}\left(f(z),z=\alpha_k\right) = \lim_{R\to +\infty}\oint_{\|z\|=R}f(z)\,dz = 0.$$
On the other hand,
$$ \text{Res}\left(f(z),z=\alpha_k\right)=\lim_{z\to \alpha_k}\frac{z-\alpha_k}{Q(z)} \stackrel{dH}{=}\frac{1}{Q'(\alpha_k)} $$
by de l'Hospital rule, so $\sum_{k=1}^{n}\frac{1}{Q'(\alpha_k)}=0$ as wanted.
