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Anyone familiar with residue theory

  1. Dec 11, 2005 #1
    from my understanding we use residue theory when we have poles.
    The question i have is
    if f(z) = 1/(1-Z^2) has two poles at 1, -1 each of order 1
    then does
    Res[f(z),-1] = lim as z -> -1 of (z+1)(f(z)) = -1/2

    if we have a pole of order 1 then

    Res[f(z),z0] = lim as z -> z0 of (z - z0)f(z)

    or does
    Res[f(z),-1] = lim as z -> -1 of (z+2)(f(z)) = undefined
     
    Last edited: Dec 11, 2005
  2. jcsd
  3. Dec 11, 2005 #2

    shmoe

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    This is correct. So for the residue of a pole of order one at z0=-1 you look at the limit of (z+1)f(z) as z->-1, which is what your professor appears to have done, though check the sign carefully.

    Where did z+2 come from?
     
  4. Dec 11, 2005 #3
    from the formula in the text it states (z-z0)(f(x)) so if zo = -1 and z = z+ 1 would that not = z+2
     
  5. Dec 11, 2005 #4

    shmoe

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    z=z+1? That ain't true. If z0=-1 then z-z0=z-(-1)=z+1.

    remember the reside at z0 is the coefficient of 1/(z-z0) in the Laurent series at z0. A simple pole means

    [tex]f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1 (z-z_0)^1+a_2 (z-z_0)^2+\ldots[/tex]

    It's [tex]a_{-1}[/tex] you're after. Multiply f(z) by (z-z0) and let z->z0, poof the other terms vanish.
     
  6. Dec 11, 2005 #5
    yes i just caught my mistake z is actually just a variable where z0 is where the pole is at
     
  7. Dec 11, 2005 #6
    correct me if im wrong but with residues at poles you dont actually have to use laurent series you could just use residue at poles formulas. I have not really gone over laurent series but it looks confusing.
     
  8. Dec 11, 2005 #7
    sorry thanks for the help
     
  9. Dec 11, 2005 #8

    shmoe

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    That's correct, but understanding where these formulas come from is a good thing.

    happy to help.
     
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