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So, I've been studying Laurent series, and I'm fine with series such as $ \frac {1}{(z-1)(z+1)} $ for example. For these, we can just use partial fraction decomposition and then geometric series. However, I'm not even sure how to get started with the following function:

$ f(z)= (z^2+4)^\frac {1}{3}$

Obviously, I can see that there are zero's at $ +/- 2i $, but I have no idea how to even start to arrange this as a Laurent series. Do I do the same thing I would do for Taylor series? Any guidance would be appreciated.

Incognito
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  • Laurent series is related to a particular ring / annulus of convergence, which one do you want? Anyway, you need the binomial series for this one. – user127096 Mar 11 '14 at 02:37
  • I'm also trying to figure out which annulus it converges in. Alright, if I use the binomial series (I'll work it out), how does that make it a Laurent series? Feels like that's what I would do for a Taylor series. – Incognito Mar 11 '14 at 02:39
  • Ahhhhh......so, |z|<2 is for a taylor series, and |z|>2 is for a Laurent series? Would I have to factor it out somehow such that I have 1/z in the function to end up with a Laurent series? – Incognito Mar 11 '14 at 02:42
  • The Taylor series is a special case of Laurent series. See example here. But I spoke in haste: there isn't a Laurent series for $|z|>2$, because $f$ is multivalued there, and does not admit a holomorphic branch. – user127096 Mar 11 '14 at 02:50
  • If I'm thinking of it correctly, the laurent series is no different from the taylor series in this case, correct? (seems very odd to me) – Incognito Mar 11 '14 at 02:54
  • Yes. Nothing odd. The Laurent series of $e^z$ is $\sum_{n=0}^\infty z^n/n!$, for example. – user127096 Mar 11 '14 at 03:03
  • Thank you very much for explaining – Incognito Mar 11 '14 at 03:08

1 Answers1

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Write $f(z) = \sqrt[3]{4} (({z \over 2})^2+1)^{1 \over 3}$, and use the binomial theorem to expand, this gives the Laurent series for $|z| < 2$.

copper.hat
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  • Isn't that just a Taylor series though?.... – Incognito Mar 11 '14 at 02:49
  • It is, but it is also the Laurent series... – copper.hat Mar 11 '14 at 02:50
  • Hmmmmm, I'm just having trouble differentiating between the two series. Could you please explain the difference? I thought that laurent series had to go from -infinity to infinity? – Incognito Mar 11 '14 at 02:53
  • The Taylor series only has $a_n$ for $n \ge 0$. Or think of it as $a_n = 0$ for $n < 0$. The Laurent & Taylor series are unique within the annulus or radius of convergence respectively. – copper.hat Mar 11 '14 at 03:00
  • Hmmmm, well, we end up only with $n>=0$, correct? So the laurent series is just a taylor series for this example. – Incognito Mar 11 '14 at 03:01
  • In this case, the Laurent series is also a Taylor series... – copper.hat Mar 11 '14 at 03:04
  • If a function has a Taylor series, then this is also the Laurent series for the function... – copper.hat Mar 11 '14 at 03:05
  • Alright, thank you very much for clarifying this for me. Sorry about that; I don't have any resources which explain this; they just provide a standard defintion without sufficient examples. Thanks for clarifying though. – Incognito Mar 11 '14 at 03:07
  • One last question: Would it be possible to define the series in a region other than |z|<2? – Incognito Mar 11 '14 at 03:09
  • A Laurent series does not exist for $|z| \ge 2$ as $f$ is multivalued there... – copper.hat Mar 11 '14 at 03:51
  • So, I did some more reading/digging around, and I noticed that the coefficients for the (z-zo)^(-n) terms correspond to singularity points. Thus, since my function has no singularities, it only has coefficients ao, a1,...,an. So, if a function has no singularities, the Laurent expansion will be equivalent to the Taylor series expansion, correct? – Incognito Mar 11 '14 at 17:54
  • If a function is analytic on some disc, then the Laurent and Taylor series will be the same. – copper.hat Mar 11 '14 at 18:48