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Residue of cos(z)/z

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    So the problem at hand is to calculate the contour integral [tex]\oint[/tex] cos(z)/z around the circle abs(z)=1.5 .


    2. Relevant equations
    The integral is going to follow from the Cauchy-Integral Formula and the Residue theorem. The problem I am having is figuring out what the residue is going to be.


    3. The attempt at a solution

    So I know the pole is at z = 0, which lies inside of the contour. So the integral reduces to I=2*pi*i*residue @ 0. What I can't figure out is how to determine the residue. If I use maple, I know that the residue is 1, but I want to figure out where it comes from it. Any help?
     
  2. jcsd
  3. Oct 19, 2009 #2

    Dick

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    If f(z) has a simple pole at z=c then the residue is lim z->c of (z-c)*f(z), isn't it? What does that give you?
     
  4. Oct 19, 2009 #3
    Ah, thank you. I just needed someone to write it out clearly for me.

    So the formal answer =2*Pi*i
     
    Last edited: Oct 19, 2009
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