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Does the principal branch square root of z have a Laurent series expansion C-{0}?

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Does the principal branch square root of z have a Laurent series expansion in the domain C-{0}?

    3. The attempt at a solution

    Well I'm not really sure what a principal branch is? I believe that there is a Laurent series expansion for z^(1/2) in C-{0} because originally our only problem is that when we take derivative of z^1/2 we get 1/z^[(2n+1)/2] and this is not defined at 0, but is everywhere else... so I think the answer is yes to this, but again I'm unsure of the details of principal branch?
     
  2. jcsd
  3. Oct 26, 2011 #2
    No. The square root is a multifunction and these functions do not have Laurent series about their branch-points because they're not fully analytic in a punctured disc surrounding the branch-point, specifically not so along their branch-cuts
     
  4. Oct 27, 2011 #3
    But does taking the principal branch of square root z not deal with that? Does the principal branch mean we only take the principal roots of z?
     
  5. Oct 27, 2011 #4
    The principal branch is analytic except along it's branch-cut which extends out from the origin so that we do not have an analytic domain in a punctured disk surrounding the origin and yes, the principal branch is the principal root with arg between -pi and pi
     
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