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Suppose $f=P/Q$ is a rational function and suppose $f$ has a simple pole at $a$. Then a formula for calculating the residue of $f$ at $a$ is $$ \text{Res}(f(z),a)=\lim_{z\to a}(z-a)f(z)=\lim_{z\to a}\frac{P(z)}{\frac{Q(z)-Q(a)}{z-a}}=\frac{P(a)}{Q'(a)}. $$

In the second equality, how does the $Q(z)-Q(a)$ appear? I only see that it would equal $\lim_{z\to a}\frac{P(z)}{\frac{Q(z)}{z-a}}$.

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Oh damn, $Q(a)=0$... – Resty Mar 21 '12 at 3:32

Since the pole at $\,a\,$ is simple we have that $$Q(z)=(z-a)H(z)\,\,,\,H(z)\,\,\text{a polynomial}\,\,,\,P(a)\cdot H(a)\neq 0\,$$ Thus, as polynomials are defined and differentiable everywhere: $$Res_{z=a}(f)=\lim_{z\to a}\frac{P(z)}{H(z)}=\frac{P(a)}{H(a)}$$ and, of course, $$Q'(z)=H(z)+(z-a)H'(z)\xrightarrow [z\to a]{}H(a)$$

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Because $a$ is a zero of $Q(z)$; i.e., $$Q(a) = 0.$$ So, $$Res_{a}f(z)=\lim_{z→a}(z−a)f(z) =\lim_{z→a}\frac{P(z)}{\frac{Q(z)}{z−a}} =\lim_{z→a}\frac{P(z)}{\frac{Q(z) - 0}{z−a}} = \lim_{z→a}\frac{P(z)}{\frac{Q(z)−Q(a)}{z−a}}= \frac{P(a)}{Q′(a).}$$ Then this is a "trick" to compute residue for a simple pole.

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