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Suppose that $P(z)$ and $Q(z)$ are polynomials with no common roots and $\deg(P) < \deg(Q)$. Suppose that $Q(z) = \displaystyle\prod_{k=1}^t (z - z_k)^{n_k}$ then the partial fraction decomposition of $\frac{P(z)}{Q(z)}$ has the form: $$ \frac{P(z)}{Q(z)} = \sum_{k=1}^t \sum_{j=1}^{n_k} \frac{a_{kj}}{(z - z_k)^j}$$ Show that $ a_{kj} = \operatorname{Res}_{z_k} ((z - z_k)^{j-1} \frac{P(z)}{Q(z)})$

I am having a hard time answering this question could someone shed some light?

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Try changing the title of the quesition to something more informative, and less subjective, if complex means complicated and is not related to the question being from complex analysis. – Pedro Tamaroff Apr 10 '12 at 0:17
Shouldn't we write $a_{k,j}$ rather than $a_{k_j}$? – anon Apr 10 '12 at 0:20

Hint: residue is linear in the function. What would you get for each of the terms?

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From the partial fraction decomposition, $(z - z_k)^{j-1} \frac{P(z)}{Q(z)}$ has a holomorphic primitive at all points $z\ne z_k$ except for the single term $\frac{a_{kj}}{z - z_k}$. Hence the residue at $z_k$ is $a_{kj}$.

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