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Partial Fraction From Hell!

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data

    I know how to use the method of partial fractions in most circumstances, but I'm working on a problem that has gotten the best of me. How do I get from the left side of the following identity to the right side????

    \ = \
    \ 2 \ + \
    \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega -
    \frac{-\sqrt{2}+i\sqrt{2}}{2}} \ + \
    \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega -

    3. The attempt at a solution

    I was able to factor the denominator and write the following equation:

    A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \
    B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \

    but couldn't get much further because A and B don't have [itex]\omega^2[/itex] multiples to match up with the [itex]-2\omega^2[/itex] on the right side of the equation.

    How do I handle this monster?

    Note: [itex]i[/itex] is the imaginary unit.

  2. jcsd
  3. Aug 30, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I would like you to show how you arrived at what you give since that is not what I get.

    The first thing I did was multiply both numerator and denominator by -1 to get
    Then the denominator factors as
    [tex]\omega^2- \sqrt{2}i\omega- 1= (\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})[/tex]
    Putting those factors into the denominators and multiplying through by them gives
    [tex]A(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})+ B(\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})= 2+ 2\omega^2[/tex]

    Now set
    [tex]\omega= \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/tex]
    [tex]\omega= \frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/tex]
    and it should be easy.
  4. Aug 30, 2007 #3
    Thanks for your help!

    Your denominator factors are not in the form I need them to be. I didn't check it, but I assume your work is correct. The problem is I need the denominator factors to be in the [itex](a +i\omega)[/itex] (where [itex]a[/itex] is complex or real) form to take advantage of a known Fourier transform.

    Here is how I arrived at my equation:


    [tex](A+i\omega)(B+i\omega) = 1 + \sqrt{2}i\omega-\omega^2[/tex]

    Then we want A and B to satisfy the following conditions:

    [tex]A+B = \sqrt{2}[/tex]
    [tex]AB = 1[/tex]

    Solving that system we get:

    [tex]A= \frac{\sqrt{2}}{2}(1-i)[/tex]
    [tex]B= \frac{\sqrt{2}}{2}(1+i)[/tex]


    [tex]1+\sqrt{2}i\omega - \omega^2 = \left(\frac{\sqrt{2}}{2}(1-i)+i\omega\right)

    And from that I was able to get the equation I gave before:

    A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \
    B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \

    And that's where I was stuck.
    Last edited: Aug 30, 2007
  5. Aug 31, 2007 #4

    I got it now. I found out that I have to first do long division to get the numerator to be 1 degree less than the denominator, and then use the method of partial fractions.

    Thanks for your help!
  6. Aug 31, 2007 #5


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    Staff Emeritus
    Science Advisor

    Dang, didn't even occur to me!
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