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A problem about residues(is matlab wrong?)

  1. Dec 12, 2007 #1
    I have written this simple code fragment to the matlab for finding the residue(s) of the function 1/(z-i/9)^3;

    b=[1];
    a=[1 -i/3 -1/27 i/(729) ];
    [r p k]=residue(b,a)

    and get the following result;
    r =

    0
    0
    1


    p =

    0.0000 + 0.1111i
    0.0000 + 0.1111i
    0.0000 + 0.1111i


    k =

    []

    The poles are true. Function has a pole of order three at z=i/9 . However, there are two different values for the residue: 0 and 1. I could not get the meaning behind that? How can a function have different residue at the same point? Additionally if we assume matlab is right then the result of the contour integral should be 2pi*i at the unit circle. However, we can easily show that this contour integral is zero. So, is something wrong with the residue algorithm of matlab or there is something I do not consider?
     
  2. jcsd
  3. Dec 12, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    The residue of a pole in a function f(z) at z=a is the coefficient of 1/(z-a) in the Laurent expansion of f(z). In your case, a=i/9 and the coefficeint of 1/(z-i/9) is zero. So the residue (there is only one residue!) is zero.
     
  4. Dec 13, 2007 #3
    Yes, logically it should have only one residue which is zero. However, what I did not understand is why has matlab find also a residue which is 1 at the pol z=i/9 ?
     
  5. Dec 13, 2007 #4

    Avodyne

    User Avatar
    Science Advisor

    I don't use matlab so I can't help.
     
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