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Partial fraction decomposition

  1. Sep 24, 2011 #1
    Yes, another of these. How do you decompose (z3+1)/(z(1-z)2) ?

    I've tried A/z + B/(1-z) + (Cz+D)/(1-z)2 and a slew of others that don't work.

    Thanks a bunch!
     
    Last edited: Sep 24, 2011
  2. jcsd
  3. Sep 24, 2011 #2
    (z3+1)/z(1-z2)

    Do you mean:

    (z3+1)/(z(1-z2))

    or

    ((z3+1)/z)(1-z2)
     
  4. Sep 24, 2011 #3
    First of all, the power of the numerator is the same as the power of the denominator. You have to make long division of polynomials:

    [tex]
    \frac{z^{3} + 1}{z(1 - z^{2})} = \frac{z^{3} + 1}{-z^{3} + z} = \frac{z^{3} - z + z + 1}{-z^{3} + z} = -1 - \frac{z + 1}{z^{3} - z}
    [/tex]

    Next, the denominator factors into linear factors:
    [tex]
    z^{3} - z = z (z^{2} - 1) = z (z - 1) (z + 1)
    [/tex]
    so, you have:
    [tex]
    \frac{z + 1}{z^{3} - z} = \frac{z + 1}{z ( z - 1) (z + 1)} = \frac{A}{z} + \frac{B}{z - 1} + \frac{C}{z + 1}
    [/tex]
    You find A, B and C.
     
  5. Sep 24, 2011 #4
    Hey, I made a quick edit in the problem. I meant the quantity squared. Sorry for the confusion.
     
  6. Sep 24, 2011 #5
    Again, your numerator and denominator have the same power. Do the long division:

    [tex]
    (z^3 + 1) \colon (z^3 - 2 z^2 + z) = ?
    [/tex]

    first. What is the quotient and what is the remainder?
     
  7. Sep 24, 2011 #6
    It comes out to ((z+1)(z2-z+1))/(z(z-1)2)
     
  8. Sep 24, 2011 #7
    no, this is not long division.
     
  9. Sep 24, 2011 #8
    Right. It is 1 + (2z2-z+1)/(z3-2z2+z)
     
    Last edited: Sep 24, 2011
  10. Sep 24, 2011 #9
    How?

    [tex]
    1 \cdot (z^{3} - 2 z^{2} + z) + (2 z^{2} - z) = z^{3} - 2 z^{2} + z + 2 z^{2} - z = z^{3} \neq z^{3} + 1
    [/tex]
     
  11. Sep 24, 2011 #10
    right again. I've edited my last reply. I forgot the +1 in the numerator.
     
  12. Sep 24, 2011 #11
    Ok, now look at the normal fraction (power of numerator is lower than power of denominator). The partional fraction decomposition is:

    [tex]
    \frac{2 z^{2} - z + 1}{z ( z - 1)^{2}} = \frac{A}{z} + \frac{B}{z - 1} + \frac{C}{(z - 1)^{2}}
    [/tex]

    If you multiply by [itex]z (z - 1)^{2}[/itex], and compare coefficients in front of like powers of z, you will get 3 equations for A, B and C. Solve them and you are done.
     
  13. Sep 24, 2011 #12
    Great, appreciate the patience. Imagine, I've made it to a graduate-level complex analysis class without ever doing partial fractions...

    Loads of help, thanks again.
     
  14. Sep 24, 2011 #13
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