Rational Function Residue Formula Proof I came across a theorem used to calculate residues of rational functions that states that if f and g are analytic functions at $z_k$ and $g'(z_k)$ isn't 0, then the residue of $f(z)/g(z)$ at $z_k$ is $f(z_k)/g'(z_k)$. I want to prove this using Laurent series (taking out the coefficient of $(z-z_k)^-1$) but I can't get around the algebra of actually dividing the two Laurent series. Can someone guide me through this?
 A: Well, the theorem you are describing works only for simple poles.  But the proof is straightforward:
$$\operatorname*{Res}_{z=z_k} \frac{f(z)}{g(z)} = \lim_{z\to z_k} (z-z_k) \frac{f(z)}{g(z)} = \frac{\displaystyle \lim_{z\to z_k} f(z)}{\displaystyle \lim_{z\to z_k} \frac{g(z)}{z-z_k}} = \frac{f(z_k)}{g'(z_k)} $$
That last step was possible because, by definition, $g(z_k)=0$.
Analogous formulae for higher-order poles become increasingly unwieldy.  Frequently, it is better to use the Laurent expansion.
EDIT
The OP wanted to see a proof using the Laurent expansion.  Well, in that case, the respective expansions in a neighborhood about $z=z_k$ are
$$f(z) = f(z_k) + f'(z_k) (z-z_k) + \cdots $$
$$g(z) = g'(z_k) (z-z_k) + \frac12 g''(z_k) (z-z_k)^2 + \cdots $$
where the ellipses indicate higher orders that can be neglected.  Then
$$\begin{align}\frac{f(z)}{g(z)} &= \frac1{z-z_k} \frac{f(z_k)+f'(z_k) (z-z_k)+\cdots}{g'(z_k) + \frac12 g''(z_k) (z-z_k)+\cdots} \\  &=\frac1{z-z_k} \frac{f(z_k)}{g'(z_k)} \left [1+\left (\frac{f'(z_k)}{f(z_k)} - \frac12 \frac{g''(z_k)}{g'(z_k)}\right )(z-z_k) \right ]+\cdots \\ &=  \frac{f(z_k)}{g'(z_k)} \frac1{z-z_k} + \left [\frac{f'(z_k)}{g'(z_k)} - \frac12 \frac{f(z_k) g''(z_k)}{g'(z_k)^2} \right ] + \cdots \end{align}$$ 
As the residue is the coefficient of $(z-z_k)^{-1}$, QED.
