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So, I'm trying to find the regions in which the function may be represented as a Laurent series (expanded about the origin). Then I want to find those expansions:

$$f(z)=(z^2+1)^{-1/2}$$

Now, since there's a singularity at $-i \space \space and \space \space i$, the regions must be 0<|z|<1 and 1<|z|< infinity. Well, I'm lost as to how the laurent series can be calculated. If we expand about a point that isn't the singularity, is that expansion not just the taylor series? In that case, we would form a binomial series, but I believe that only converges for 0<|z|<1. How do we obtain the expansion for 1<|z|< infinity?

Incognito
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  • Why is it not analytic? Could you please explain further? – Incognito Mar 12 '14 at 23:55
  • Just an FYI: This response to my question is incorrect. $f$ may be analytic for |z| > 1 if you choose a branch connecting i and -i (if you go around both singularities, then you end up at the same value. It's only when you go around one pole that another value is generated). So, you can have a Taylor series at |z|<1 and a Laurent series at |z|>1. It depends on how you define your branch cut. – Incognito Mar 14 '14 at 22:41
  • You are correct, of course. If $|z|>1$ then $f(z) = {1 \over z} (1+{1 \over z^2})^{-{1 \over 2}}$ which you can expand using the binomial series expansion. (I was incorrectly thinking of $\log$ in my earlier responses.) – copper.hat Mar 14 '14 at 22:54
  • Exactly, that's what I'm doing. I just wanted to share the information here :) – Incognito Mar 14 '14 at 22:58

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