Suppose that I'm trying to construct Laurent series for the following complex function:
$$f(z)=(z^2+4)^{\frac{1}{3}}$$
Since I have a 1/3 exponent, this means that my branch points must extend to infinity when I define my branch cut (to make f single-valued). So, even though I may define a Laurent series such that |z|>2, it is not valid because the branch cut introduces discontinuities, correct? Is this reasoning the right way to think about it? I ask because I constructed the Laurent series in the region |z| > 2, programmed it in Matlab, and I'm getting the correct results.
Here's what I calculated for |z| > 2:
$$f(z)=z^{\frac{2}{3}}\sum\limits_{n=0}^{\infty} \frac{\Gamma(4/3)}{n!\Gamma(4/3-n)}(\frac{2}{z})^{2n}$$
The function $$ F(z) = z^{-2/3} f(z) = (1+4z^{-2})^{1/3} $$ admits a single-valued holomorphic branch in $|z|>2$, which you can indeed expand into a Laurent series. This is the series in your post.
However, multiplying by $z^{2/3}$ you do not get a Laurent series for the original $f$. By its definition, a Laurent series does not have fractional powers of $z$.
So: you have a representation for (multi-valued) $f$ in terms of (multi-valued) $z^{2/3}$. It's just not a Laurent series of $f$.