Odd Laurent Series

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.

• 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 2:37
• 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 2:39
• 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 2:42
• 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 2:50
• 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 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 $|z| < 2$.
• 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 3:00
• 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 3:01