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Residue calculus for essential singularities

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Im not sure if i understood correctly how to calculate the residue for functions with essential singularities like:
    f(z)=sin(1/z)
    h(z)=z*sin(1/z)
    j(z)=sin(1/z^2)
    k(z)=z*(1/z^2)



    2. Relevant equations
    So, according to what ive read, when we have a functions with an essential singularity, we expand it in laurent serie and the residue will be the coefficient a-1.
    Im a bit confused, if this is for all kind of functions, because in every book ive read, the exemle is always for 1/z(f(z)=e^(1/z) or sin(1/z) etc) and my question is if its 1/z^2 or 1/z^n the residue is also the a-1 coefficient???


    3. The attempt at a solution
    For all given functions the essential singularity is 0.
    f(z)=1/z-1/(z^3*3!)+1/(z^5*5!)+......
    the a-1 coefficient is 1 so the residue is 1.
    h(z)=z*f(x)=1-1/(z^2*3!)+1/(z^4*5!)+...
    In this case the residue is 0.
    j(z)=1/(z^2)-1/(z^6*3!)+1/(z^10*5!)+..
    For j(z) the residue is 0.
    k(z)=z*j(z)=1/(z)-1/(z^5*3!)+1/(z^9*5!)+...
    for k(z) the residue is 1.
     
  2. jcsd
  3. Apr 7, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, it's just the coefficient of 1/z. I think you've got it. The answers look right.
     
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