Show $|1-e^{ix}|^2=2(1-\cos x)$ Show $|1-e^{ix}|^2=2(1-\cos x)$
$$|1-e^{ix}||1-e^{ix}|=1-2e^{ix}+e^{2ix}=e^{ix}(e^{-ix}-2+e^{ix})=e^{ix}(2\cos x-2)=-2e^{ix}(1-\cos x)$$
Not sure how they got rid of the $-e^{ix}$ factor. Did I expand the absolute values wrong? thank you

$$ |1-e^{ix}|^{2} = |1-\cos x - i \sin x|^{2} = (1-\cos x)^{2} + (-\sin x)^{2} = 1 - 2 \cos x + \cos^{2} x + \sin^{2} x $$
$$= 2 - 2 \cos x$$
