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.

  • $\begingroup$ 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. $\endgroup$ – user127096 Mar 11 '14 at 2:37
  • $\begingroup$ 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. $\endgroup$ – Incognito Mar 11 '14 at 2:39
  • $\begingroup$ 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? $\endgroup$ – Incognito Mar 11 '14 at 2:42
  • $\begingroup$ 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. $\endgroup$ – user127096 Mar 11 '14 at 2:50
  • $\begingroup$ 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) $\endgroup$ – Incognito Mar 11 '14 at 2:54

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 $|z| < 2$.

  • $\begingroup$ Isn't that just a Taylor series though?.... $\endgroup$ – Incognito Mar 11 '14 at 2:49
  • $\begingroup$ It is, but it is also the Laurent series... $\endgroup$ – copper.hat Mar 11 '14 at 2:50
  • $\begingroup$ 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? $\endgroup$ – Incognito Mar 11 '14 at 2:53
  • $\begingroup$ 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. $\endgroup$ – copper.hat Mar 11 '14 at 3:00
  • $\begingroup$ Hmmmm, well, we end up only with $n>=0$, correct? So the laurent series is just a taylor series for this example. $\endgroup$ – Incognito Mar 11 '14 at 3:01

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