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## Main Question or Discussion Point

I wonder why z^(-1/2) cannot be expanded in Laurent series with center z=0. Anyone knows?

- Thread starter vinovinovino
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I wonder why z^(-1/2) cannot be expanded in Laurent series with center z=0. Anyone knows?

- #2

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To see this, we use the definition of complex exponentiation

[tex]z^{-1/2}=e^{-\frac{1}{2}Log(z)}[/tex]

but the complex logarithm isn't holomorphic, so the composition isn't either.

- #3

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That's not quite right in my opinion micromass. You're looking at the square root in terms of some single-valued determination of it when you say it's not continuous on some line through the origin. Any line you choose, we can just move the line and then the square root becomes not only continuous over the old line but analytic there as well except at the origin. Rather, the reason we can't expand roots at the origin in terms of Laurent series is that they have non-issolated singular points there and a Laurent series is an expansion around issolated singular points. Also, the function doesn't need to be holomorphic on [itex]\mathbb{C}[/itex] to have a Laurent series. It need only have to be single-valued and analytic in some punctured disc surrounding the expansion center. But since roots cannot have such analytic punctured discs centered at the origin being they have non-issolated singular points there, they cannot have Laurent expansions centered at the origin.

To see this, we use the definition of complex exponentiation

[tex]z^{-1/2}=e^{-\frac{1}{2}Log(z)}[/tex]

but the complex logarithm isn't holomorphic, so the composition isn't either.

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