# Strategies to finding Laurent series

Im working on Laurent series. I think I have a pretty good understanding of what they are, and why there are different ones for different domains. But one thing i really struggle with is finding Laurent series for a given function, $$f(z)$$. I feel like I don't have any strategy as for how I should approach the problem.

Currently my first step is trying to rewrite the function so that every $$z$$ is on the form $$(z-z_0)$$, when expanding about $$z_0$$, and then kind of just take it from there. But usually I just hit a wall and fail to proceed (or even express the function in terms of $$(z - z_0)$$.

So what I am wondering is, what are your guys' first steps when solving a problem of the type "find the Laurent series of a function $$f$$".

For example: $$f(z) = \frac{3-3i}{(z-i)(z-2)}$$, about $$z = 2$$.

• About which point? – saulspatz Jul 27 at 23:47
• Oopgs, forgot to mention that. z = 2. Will add it to the question. – jakvah Jul 27 at 23:49
• Use partial fractions to begin. – saulspatz Jul 27 at 23:51

There are a few major types of questions. For your $$f$$, I assume it is centered at $$z_0 = i$$? Well the denominator has one factor in the correct format, but not the other. So we can seperate them using partial fractions to see that $$f (z) = \frac{A}{z-i} + \frac{B}{z-2} \qquad A,B\in \mathbb{C}$$ Notice the $$A$$ term is already a power of $$(z-i)$$. And for the $$B$$ term, we can employ this standard trick: $$\frac{B}{z-2} = -B \frac{1}{2- i -(z-i)} = \frac{-B}{2- i} \frac{1}{1- \frac{(z-i)}{2-i}}$$ which has the laurent series of the geometric sum $$\frac{-B}{2- i} \sum_{n=0}^{\infty} w^n \qquad w=\frac{(z-i)}{2-i}$$