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Let $P$ and $Q$ be complex polynomials with $deg(P) < deg(Q)$.

Let $Q(z) = (z-z_1)^{k_1} (z-z_2)^{k_2}...(z-z_m)^{k_m}, z_i \in \mathbb{C} \text{ and } k_i \in \mathbb{N}$ be a complete decomposition. Then, I have to prove that there exists a unique exposition of the following form:

$\frac{P(z)}{Q(z)} = \sum\limits_{i=1}^m \sum\limits_{j=1}^{k_i} \frac{a_{ij}}{(z-z_i)^j}, a_{ij} \in \mathbb{C}$

I started with the proof of the uniqueness of this exposition and succeeded. However, I don't feel like I am doing well now, while trying to prove its existence. Basically, I want to do two inductions:

First I want to do an induction over $deg(Q)$ with $deg(P) = 0$, then I want to go on with an induction over $deg(P)$ with an arbitrary $deg(Q)$. The start of the first induction is easy, but I can't get to an end at the induction step. I have something like:

$\frac{P}{Q} = \frac{P}{(z-z*)Q'} = \frac{1}{z-z*} \frac{P}{Q'}$, where $Q'$ is a polynomial with $deg(Q') = n$.

What do I do next? I know that a unique exposition exists for the latter fraction and if $z* = z_i$ for some $i$, it's proven I guess. But what if this $z*$ is a completely new complex number?

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up vote 5 down vote accepted

By the inductive hypothesis, $\frac{P}{Q'}$ has a partial fraction decomposition. Show that by multiplying this decomposition by $\frac{1}{z - z^{\ast}}$ you can write down a partial fraction decomposition for $\frac{P}{Q}$.

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HINT $\rm\quad\displaystyle 1 = gcd(F,G)\ \ \Rightarrow\ \ 1 = AF+BG\ \ \Rightarrow\ \ \frac{1}{FG}\ = \ \frac{A}G + \frac{B}F$

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