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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:

$$f(z)=z^{\frac{2}{3}}\sum\limits_{n=0}^{\infty} \frac{\Gamma(4/3)}{n!\Gamma(4/3-n)}(\frac{2}{z})^{2n}$$

Incognito
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  • Yes, that is the question: whether the laurent series for |z| > 2 is valid or not. – Incognito Mar 15 '14 at 22:55
  • The answer to which is: no, it is not valid. If you'd like a more concrete answer, I suggest that you add your Laurent series for $|z|>2$ to the question. – user127096 Mar 15 '14 at 22:57
  • Alright, I added it. – Incognito Mar 15 '14 at 23:04
  • Laurent series are of the form: $\sum_{n=-\infty}^{\infty} a_nz^n$. They do not have fractional powers. Note that your function is analytic in $\lvert z\rvert>2$, and hence it can be express as $\sum_{n=-\infty}^{0} a_nz^n$ – Yiorgos S. Smyrlis Mar 15 '14 at 23:20

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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 $f$.

user127096
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