Laurent expansion of principal root

**bernardbb** · May 2nd 2009, 04:53 AM

How do I find the Laurent expansion of a function containing the principal branch cut of the nth root?

Example:
$f(z)=-iz\cdot\mathrm{pv}\sqrt[4]{1-\frac{1}{z^{4}}}$

**chisigma** · May 2nd 2009, 06:04 AM

Remembering the binomial series expansion...

$(1+x)^{\alpha}= \sum_{k=0}^{\infty} \frac{\alpha\cdot (\alpha-1)\dots (\alpha-k+1)}{k!}\cdot x^{k}$

... for $x=-\frac{1}{z^{4}}$ and $\alpha=\frac{1}{4}$ we have...

$(1-\frac{1}{z^{4}})^{\frac{1}{4}}= 1 - \frac{1}{4}\cdot z^{-4} - \frac{3}{4\cdot 4\cdot 2!}\cdot z^{-8} - \frac{3\cdot 7}{4\cdot 4\cdot 4\cdot 3!}\cdot z^{-12} + \dots$

... and then...

$f(z)= -i\cdot z \cdot (1-\frac{1}{z^{4}})^{\frac{1}{4}}= -i\cdot (z - \frac{1}{4}\cdot z^{-3} - \frac{3}{4\cdot 4\cdot 2!}\cdot z^{-7} - \frac{3\cdot 7}{4\cdot 4\cdot 4\cdot 3!}\cdot z^{-11} + \dots)$

Kind regards

$\chi$ $\sigma$

**bernardbb** · May 2nd 2009, 06:46 AM

Thanks, that seems helpful.
I think you forgot the (k!)?

**bernardbb** · May 3rd 2009, 08:53 AM

Edit: Nevermind, figured it all out.

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