Quick Complex Numbers Q (1 Viewer)

gurmies · Dec 13, 2009

@ cyl, you can get to the final result by factoring out -i from the numerator

An alternative method below:

$\frac{1-\cos\theta-i\sin\theta}{1+\cos\theta+i\sin\theta}\times \frac{1+\cos\theta-i\sin\theta}{1+\cos\theta-i\sin\theta} = \frac{1+\cos\theta-i\sin\theta-\cos\theta-\cos^{2}\theta+i\sin\theta\cos\theta-i\sin\theta-i\sin\theta\cos\theta-\sin^{2}\theta}{1+2\cos\theta+\cos^{2}\theta+\sin^{2}\theta} \\\\ = \frac{-2i\sin\theta}{2(1+\cos\theta)} \\\\ = \frac{-i\sin\theta}{1+\cos\theta} \\\\ = \frac{-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}} \\\\ = -i\tan\frac{\theta}{2}$

@ Luky, initially you are dealing with @, however the answer requires @/2, so trigonometry knowledge needs to be utilised =]

Quick Complex Numbers Q (1 Viewer)

gurmies

Drover

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