Show that $\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$ Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $<n$. Once again, the task is to show
$$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$
For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$.
I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?
 A: Hint (without L'Hôpital):  assume WLOG $Q$ is monic (otherwise cancel out the leading term between the two sides). Then $Q(z)=\prod_{k=1}^n(z-\alpha_k)$ and, by the product rule of differentiation, $\,Q'(z)=\sum_{k=1}^n \prod_{j \ne k}(z-\alpha_j)\,$ so in particular $\,Q'(\alpha_k)=\prod_{j \ne k}(\alpha_k-\alpha_j)\,$.
Multiplying by $Q(z)$ and using the above, the equality to prove becomes:
$$
\begin{align}
P(z) = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} & = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} \\[5px]
 & = \sum_{k=1}^{n} \frac{P(\alpha_k)\,\prod_{j \ne k}(z-\alpha_j)}{\prod_{j \ne k}(\alpha_k-\alpha_j)} \\[5px]
\end{align}
$$
It follows that the equality holds for all $\,z=\alpha_k\,$ and, since the two sides of the equality are polynomials of degree $\le n-1\,$ that are equal at $n$ points, it further follows that the equality holds in general, for all $z$.
A: Use Partial Fractions and L'Hôpital:
$$
\frac{P(x)}{Q(x)}=\sum_{k=1}^n\frac{A_k}{x-a_k}\\
$$
where $A_k=\lim\limits_{x\to a_k}(x-a_k)\frac{P(x)}{Q(x)}=\frac{P(a_k)}{Q'(a_k)}$.
