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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 $z = 2$.

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  • $\begingroup$ About which point? $\endgroup$
    – saulspatz
    Jul 27, 2019 at 23:47
  • $\begingroup$ Oopgs, forgot to mention that. z = 2. Will add it to the question. $\endgroup$
    – jakvah
    Jul 27, 2019 at 23:49
  • $\begingroup$ Use partial fractions to begin. $\endgroup$
    – saulspatz
    Jul 27, 2019 at 23:51

1 Answer 1

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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 $$ \frac{-B}{2- i} \sum_{n=0}^{\infty} w^n \qquad w=\frac{(z-i)}{2-i} $$

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