- #1
Millertron
- 4
- 0
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!
[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!