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Laurent series of Sqrt(z)

  1. Feb 1, 2009 #1

    My mathematics professor said that it is possible to construct a Laurent series of sqrt(z) about zero by integrating over a keyhole contour and then taking the limit R --> 0 where R radius of the inner circle. But I think he is mistaken. I dont understand how it is possible to have a Laurent series about zero, as it is a branch point.

    Can someone please clarify this point, and tell me what the series is if such a series exists.

    Also, then is it possible to have a laurent series for any function about its branch point by considering a similar contour.

    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2


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    I think you're right; Laurent series converge on an annulus, and square root cannot be defined* on an annulus about the origin.

    Square root can be expressed by a (rather boring) Puiseux series, but I'm not sure how well that works complex analytically.

    *: I mean in a continuous way, of course.
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