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I know branch points are the point where $z^{-1/2} =0$ or $z^{-1/2} =\infty$, so for this I think $z$ should be $\infty$ for $z^{-1/2}$ to go $\infty$?? However for $z^{-1/2}$ to be equal $0$, I couldn't find a point where $z^{-1/2}$ goes to zero because when I put $z=0$, $z^{-1/2}$ is undefined. What should I do in this case to find $z^{-1/2}=0$??

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  • $\begingroup$ sorry I think I realized z^(-1/2) is zero when z goes to inf but if I put (z=-inf), I get z^(-1/2) becomes undefined... $\endgroup$
    – kevin
    Commented Oct 6, 2017 at 3:41
  • $\begingroup$ so the question is does z^(-1/2) ever goes to infinity?? $\endgroup$
    – kevin
    Commented Oct 6, 2017 at 3:43

1 Answer 1

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The branch point you've already found is the only branch point, as $z^\frac{-1}2=0$ has no solutions.

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    $\begingroup$ Correct me if I'm wrong, but $\infty$ is considered a branch point as well. So the function has two branch points $\endgroup$
    – Dylan
    Commented Dec 6, 2017 at 0:47

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