Why is the sum of the inverse derivatives of all roots of a polynomial equal to zero? [duplicate]

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.

Solution 1:

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

$$ \sum_{j=1}^{n} \frac{1}{Q'(\alpha_j)} = \lim_{|z|\to\infty} \sum_{j=1}^{n} \frac{z}{Q'(\alpha_j)(z - \alpha_j)} = \lim_{|z|\to\infty} \frac{z}{Q(z)} = \begin{cases} 1, & \text{if } n = 1 \\ 0, & \text{if } n > 1 \end{cases}. $$

Solution 2:

$f(z)=\frac{1}{Q(z)}$ 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.