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Let $f(z) = \sqrt{z(z-1)}$. The branch cut is the real interval $[0,1]$, and $f(z)>0$ for real $z$ that are greater than 1. I need to find the first few terms of the Laurent expansion of $f(z)$ for $\left|z\right| > 1$ (centered at zero). I also need the radius of convergence.

I don't really know where to start for this one. I tried rewriting as $f(z) = z\sqrt{1 - \frac{1}{z}}$, as some have suggested, but this doesn't appear that enlightening.

This is a study question when reviewing for a qualifying exam.

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    $\begingroup$ Now use the binomial theorem $(1-z^{-1})^{1/2}=\sum_{k=0}^{\infty}\binom{1/2}{k}(-1)^kz^{-z}$. $\endgroup$
    – Hellen
    Commented Jul 21, 2017 at 4:36
  • $\begingroup$ This works! Thanks! But how do I find the radius of convergence? Is it as simple as $\left|z\right| > 1$? $\endgroup$
    – cgmil
    Commented Jul 22, 2017 at 1:46
  • $\begingroup$ I used the information from this Wikipedia article to answer the second part of the question. I may write up an answer to my own question if I get the time. $\endgroup$
    – cgmil
    Commented Jul 22, 2017 at 4:36

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