Splitting fractional expression into real/imaginary parts

[Millertron]

Hi guys, I'm having a bit of trouble splitting the RHS of the following expression into real and imaginary parts:

$(χ'+iχ")/A = \frac{1}{ω-ω_{0}-iγ/2}$

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

I've defined a = $ω-ω_{0}$ 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

$= \frac{1}{a-ib}$

$= \frac{a-ib}{(a-ib)^{2}}$

$= \frac{a}{(a-ib)^{2}}$ - $\frac{ib}{(a-ib)^{2}}$

only when I expand the squared brackets in each denominator I get $(a-ib)^{2}=a^{2}-b^{2}-2iab$, 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!

 

[lanedance]

the trick is to multiply by the complex conjugate to get a real denominator
$$\frac{1}{a-ib}= \frac{1}{a-ib}\frac{a+ib}{a+ib}= \frac{a+ib}{a^2+b^2}$$

 

[SammyS]

the trick is to multiply by the complex conjugate to get a real denominator
$$\frac{1}{a-ib}= \frac{1}{a-ib}\frac{a+ib}{a+ib}= \frac{a+ib}{a^2+b^2}$$

$\displaystyle=\frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i$

 

[Millertron]

Of course! Definitely should've known this, thanks a lot guys!

 

[lanedance]

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