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Partial Fractions Complex Decomposition

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Sorry I don't have equation editor working

    1/(z+1)(z2 + 2z + 2)


    2. Relevant equations



    3. The attempt at a solution

    (z2 + 2z + 2)

    z2 + 1) can be factor as (z - i)(z + i) However, I'm having trouble seeing the pattern on what (z2 + 2z + 2) would become, I tried guessing and checking but haven't found anything that works... I can't find a value for z that would make that equation 0.
     
  2. jcsd
  3. Dec 10, 2009 #2

    Dick

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    You don't have to guess the roots. You could just use the quadratic equation to find them.
     
  4. Dec 10, 2009 #3
    haha, okay, couldn't I just do

    1/(z+1)(z2 + 2z + 2)

    then

    1/(z+1)(z2 + 2z + 2) = A/(z2 + 2z + 2) + B/(z+1)

    Then 1 = A(z+1) + B(z2 + 2z + 2)

    First let z = -1 solve for B, so I found B to be 1. Knowing that couldn't I let z = anything, and solve for A?
     
  5. Dec 10, 2009 #4

    Dick

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    You could try. But you'd find what you get for A is dependent on what you pick for z. There's no A and B that will work for all z. (z^2+2z+2) is the product of two linear factors. You have to split it up completely to make partial fractions work.
     
  6. Dec 10, 2009 #5
    Got it, thanks a million
     
  7. Dec 10, 2009 #6

    Dick

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    Or you could try writing it as (A*z+B)/(z^2+2z+2)+C/(z+1) if you want a real decomposition. Forgot to mention that possibility.
     
  8. Dec 10, 2009 #7
    You can also obtain the partial fraction decomposition by computing the residues at the poles.
     
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