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Splitting fractional expression into real/imaginary parts

  1. Jan 18, 2012 #1
    Hi guys, I'm having a bit of trouble splitting the RHS of the following expression into real and imaginary parts:

    [itex](χ'+iχ")/A = \frac{1}{ω-ω_{0}-iγ/2}[/itex]

    (It's to find expressions for absorption coefficient and index of refraction, but that's irrelevant).

    I've defined a = [itex]ω-ω_{0}[/itex] and b = γ/2 for simplicity, and am looking for the form given by Wolfram under 'Alternate form assuming a and b are real', as this has a clear real and imaginary part. So far I've got to

    [itex]= \frac{1}{a-ib}[/itex]

    [itex]= \frac{a-ib}{(a-ib)^{2}}[/itex]

    [itex]= \frac{a}{(a-ib)^{2}}[/itex] - [itex]\frac{ib}{(a-ib)^{2}}[/itex]

    only when I expand the squared brackets in each denominator I get [itex](a-ib)^{2}=a^{2}-b^{2}-2iab[/itex], which is no good as I need to remove the i's in the denominator.

    I know it boils down to a simple algebra/complex nos question but I've been working on this problem for so long that my brain is ceasing to function. Any help is much appreciated!
  2. jcsd
  3. Jan 18, 2012 #2


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    the trick is to multiply by the complex conjugate to get a real denominator
    [tex] \frac{1}{a-ib}= \frac{1}{a-ib}\frac{a+ib}{a+ib}=
  4. Jan 18, 2012 #3


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    [itex]\displaystyle=\frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i [/itex]
  5. Jan 19, 2012 #4
    Of course!!! Definitely should've known this, thanks a lot guys!
  6. Jan 19, 2012 #5


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    no worries, one way to remember is that mulipllying by a complex congujate gives a you a magnitude, which is always real
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