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The Problem

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My Work

We know that:

$\frac{Q'(z)}{Q(z)}=\sum\limits_{k=1}^{n}\frac{1}{(z-\alpha_k)}$

Thus

$\frac{P(z)}{Q(z)}=\sum\limits_{k=1}^{n}\frac{P(z)}{Q'(z)(z-\alpha_k)}$

This is about as far as I have traveled any ideas?

Possible Ideas

I get the feeling that I should use :

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But I am not sure of it's relevancy.

José Carlos Santos
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user160110
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2 Answers2

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Let $$R(z):=Q(z)\cdot\sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}=\sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)}\cdot\frac{Q(z)}{z-\alpha_k}.$$ Then $R$ is a polynomial of degree $\leq n-1$ such that for $j=1,\dots,n$, $$\lim_{z\to \alpha_j}R(z)=\sum_{k\not=j} \frac{P(\alpha_k)Q(\alpha_j)}{Q'(\alpha_k)(\alpha_j-\alpha_k)}+ \frac{P(\alpha_j)}{Q'(\alpha_j)}\lim_{z\to \alpha_j}\frac{Q(z)}{z-\alpha_j} = \frac{P(\alpha_j)Q'(\alpha_j)}{Q'(\alpha_j)}=P(\alpha_j)$$ (recall that $Q(\alpha_k)=0$ and $Q'(\alpha_k)\not=0$ for $k=1,\dots,n$).

Hence the polynomials $P$ and $R$ coincide in $n$ points. What may we conclude?

Robert Z
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We know that, for some $K\in\mathbb{C}\setminus\{0\}$, $Q(z)=K(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n)$. So, by partial fraction decomposition, $\frac{P(z)}{Q(z)}$ can be written as$$\sum_{k=1}^n\frac{\beta_k}{z-\alpha_k}.$$Besides,$$\beta_k=\lim_{z\to\alpha_k}(z-\alpha_k)\sum_{k=1}^n\frac{\beta_k}{z-\alpha_k}=\lim_{z\to\alpha_k}\frac{P(z)}{\frac{Q(z)}{z-\alpha_k}}=\frac{P(\alpha_k)}{Q'(\alpha_k)}.$$

José Carlos Santos
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  • Why does $\beta_k=\lim\limits_{z\to \alpha_k}(z-\alpha_k)\sum\limits_{m=1}^{n}\frac{\beta_m}{z-\alpha_m}$? – user160110 Sep 29 '17 at 08:29
  • I apologise I am on the mathstack app anyways I understand your answer. – user160110 Sep 29 '17 at 08:32
  • @user160110 Because $\displaystyle\lim_{z\to\alpha_k}(z-\alpha_k)\frac{\beta_k}{z-\alpha_k}=\beta_k$ and, if $j\neq k$, $\displaystyle\lim_{z\to\alpha_k}(z-\alpha_k)\frac{\beta_j}{z-\alpha_j}=0\times\frac{\beta_j}{\alpha_k-\alpha_j}=0$. – José Carlos Santos Sep 29 '17 at 08:33