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I've just been searching Laurent series expansion online I found an old exam papers where the question is:

Let $\ f(z) = \frac 1 {z(z-2)}$ Give the Laurent series expansion of f in each of the following annuli:

ann(0,0,2) ann(2,0,2) ann(0,2,$\ \infty$)

Any examples I've seen online have used equalities to display annuli so I'm unsure how to solve this type of Laurent series expansion. Any help would be appreciated. Thanks

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  • $\begingroup$ Do you also have (z-1) in the denominator? $\endgroup$
    – user371838
    Nov 30, 2016 at 14:54
  • $\begingroup$ @Rohan No that's the question given word for word $\endgroup$
    – user6122081
    Nov 30, 2016 at 14:56
  • $\begingroup$ I just thought if not it would otherwise be very easy. $\endgroup$
    – user371838
    Nov 30, 2016 at 14:56
  • $\begingroup$ @Rohan as you can see I'm teaching myself mind explaining? $\endgroup$
    – user6122081
    Nov 30, 2016 at 15:01

1 Answer 1

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So I assume the annuli are the following ones $$ \operatorname{ann}(c,a,b) = \{ z \in \mathbb C : a < |z| < b \} $$ You would not use equalities or less than or equals here, expansion is going to happen on open sets so the open annuli given above.

To do this first do partial fractions. That is write $$ \frac{1}{z(z-2)} = \frac{A}{z} + \frac{B}{z-2} = \frac{-1/2}{z} + \frac{1/2}{z-2} $$ And expand each separately then add. Let's do one of the annuli, the $\operatorname{ann}(0,0,2)$

$\frac{-1/2}{z}$ is already a Laurent series around zero, so do nothing with it. And use geometric series for the second. $$ \frac{1/2}{z-2} = \frac{-1}{4} \frac{1}{1-z/2} = \frac{-1}{4} \sum_{n=0}^\infty (z/2)^n = \sum_{n=0}^\infty \frac{-1}{2^{n+2}} z^n $$ So the series is $$ \frac{1}{z(z-2)} = \frac{-1/2}{z} + \sum_{n=0}^\infty \frac{-1}{2^{n+2}} z^n = \sum_{n=-1}^\infty \frac{-1}{2^{n+2}} z^n $$ Yay, we even got lucky that the general term in the second series happened to work out to be the one for the $n=-1$ term. That won't happen in general.

The convergence has to be in the annulus given since the first term conveges in $\operatorname{ann}(0,0,\infty)$ (well it's a finite series, it'd better), and te second converges in the disc of radius 2. So the intersection of those two is the right convergence region.

The others are similar. In the annulus between 2 and $\infty$ you have to be slightly trickier with how you use the geometric series, but not too bad. You want to use that $\left|\frac{2}{z}\right| < 1$ whenever $|z| > 2$.

The point in all of these is to use put the terms in the right form so that you can just apply the geometric series that converges in the right domain.

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  • $\begingroup$ Hey, thanks a mill for your response, I can see from your answer that it's really well lay-out. I just can't really follow what's happening? Thanks again $\endgroup$
    – user6122081
    Nov 30, 2016 at 15:49
  • $\begingroup$ @JiriLebl Would it be necessary to expand $\frac{-1}{2z}$ if the annulus was not centered at zero, say $0<|z-2|<1$? If I understand your answer correctly, is not expanded because it already has the form $\frac{a_{-1}}{z-z_0}$, where $z_0=0$ in this case? I think yes (since the Laurent series are all about expansion around a singular point, which is the center of the open annulus ), but lets see. $\endgroup$
    – Alexander Cska
    Feb 10, 2020 at 14:29
  • $\begingroup$ @AlexanderCska : Yes that's the idea. Really $\frac{-1}{2z}$ is not expanded because it already is a Laurent series at the origin, so there is nothing left to do. If things were not centered at zero, then yes, you'd need to expand it around that point. $\endgroup$
    – Jiri Lebl
    Feb 11, 2020 at 20:48

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