Competencia Iberoamericana Interuniversitaria Let $f$ a rational function with complex coeficients and without mutiple roots in the denominator. Let $u_0,u_1,...,u_n$ ($n \ge 1$) complex roots of $f$ and $w_1,w_2,...,w_n$ roots of $f'$ (each root are considered many times as the multiplicity). Suppose that $u_0$ is a root of $f$ with multiplicity one. Prove that
$$\sum_{k=1}^{m}\dfrac{1}{w_k-u_0}=2\sum_{k=1}^{n}\dfrac{1}{u_k-u_0}$$
I wanna some ideas to solve this problem. Thank's.
 A: Put $f=P/Q$, with $P,Q$ with no commun zeros. We have $\displaystyle f^{\prime}=\frac{P^{\prime}Q-Q^{\prime}P}{Q^2}$.
A) We have $$ \frac{f^{\prime}}{f}=\frac{P^{\prime}}{P}-\frac{Q^{\prime}}{Q}=\sum_{k=0}^n \frac{1}{z-u_k} -\frac{Q^{\prime}}{Q}$$Hence:
$$(z-u_0)f^{\prime}-f=(z-u_0)f(\sum_{k=1}^n \frac{1}{z-u_k})-(z-u_0)f(\frac{Q^{\prime}}{Q})$$
We take the derivative:
$$(z-u_0)f^{\prime\prime}=((z-u_0)f)^{\prime}(\sum_{k=1}^n \frac{1}{z-u_k})-(z-u_0)f(\sum_{k=1}^n \frac{1}{(z-u_k)^2})-((z-u_0)f)^{\prime}(\frac{Q^{\prime}}{Q})-((z-u_0)f)(\frac{Q^{\prime}}{Q})^{\prime}$$
We divide by $(z-u_0)f^{\prime}$:
$$\frac{f^{\prime\prime}}{f^{\prime}}=\frac{((z-u_0)f)^{\prime}}{(z-u_0)f^{\prime}}(\sum_{k=1}^n \frac{1}{z-u_k})-\frac{f}{f^{\prime}}(\sum_{k=1}^n \frac{1}{(z-u_k)^2})-\frac{((z-u_0)f)^{\prime}}{{(z-u_0)f^{\prime}}}(\frac{Q^{\prime}}{Q})-\frac{f}{f^{\prime}}(\frac{Q^{\prime}}{Q})^{\prime}$$
Now if $z\to u_0$, $\displaystyle \frac{((z-u_0)f)^{\prime}}{(z-u_0)f^{\prime}}\to 2$, and  $\displaystyle \frac{f}{f^{\prime}}\to 0$. Hence:
$$\frac{f^{\prime\prime}(u_0)}{f^{\prime}(u_0)}=2\sum_{k=1}^n\frac{1}{u_0-u_k}-2\frac{Q^{\prime}(u_0)}{Q(u_0)}$$
B) We have $\displaystyle f^{\prime}(x)=\frac{P^{\prime}Q-Q^{\prime}P}{Q^2}$, and by the hypothesis, $P^{\prime}Q-Q^{\prime}P$ have no commun zero with $Q^2$. As $P^{\prime}Q-Q^{\prime}P=U=c(x-w_1)\cdots (x-w_m)$, we get 
$$\frac{f^{\prime\prime}}{f^{\prime}}=\frac{U^{\prime}}{U}-2\frac{Q^{\prime}}{Q}=\sum_{k=1}^m\frac{1}{z-w_k}-2\frac{Q^{\prime}}{Q}$$
Now we put $z=u_0$, and we use A), and we are done.
