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I have a function $f(z)=\dfrac{12}{z(2-z)(1+z)}$, I'm trying to find the Laurent series for each of the three annuli. The singularities are at $z = 0$, $z = 2$, and $z = -1$, so I'm looking for three Laurent series valid on $(0,1)$, $(1,2)$, and $(2,\infty)$ respectively. I can by partial fractions rewrite $f(z)$ as $\dfrac{4}{z}\bigg(\dfrac{1}{1+z} + \dfrac{1}{2-z} \bigg)$.

In the book that gives this example problem, I read that for $(0,1)$ I need to expand both of those fractions In powers of $z$; for $(1,2)$ I am to expand the second fraction in powers of $\dfrac{1}{z}$, and for $(2,\infty)$ I am to expand both fractions in powers of $\dfrac{1}{z}$.

What I do not understand and would like to know, is what it means to expand any term in powers of $z$ or $\dfrac{1}{z}$, and how to tell in general which expansion to choose when calculating Laurent series. Thanks if anyone can help with my question.

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up vote 0 down vote accepted

Take the function $f(z)=\frac{1}{1-z}$. Then for $|z|<1$ you have $f(z)=\sum^\infty_{n=0}z^n$. For $|z|>1$ you can do something similar: notice that $\left|\frac{1}{z}\right| = \frac{1}{|z|}<1$, so: $f(z) = \frac{1}{1-z} = -\frac{1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}\sum^\infty_{n=0}\frac{1}{z^n}$. You can generalize easily this method to functions of the type $\frac{a}{b-cz}$, as in your case.

For other types of functions you often use known series expansions (for example $e^z=\sum^\infty_{n=0}\frac{z^n}{n!}$) or other methods.

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@DanieRobert-Nicoud Got it, thanks – user1535776 Feb 5 '13 at 0:04

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