I have an answer for this question, but I'm not very confident with it:
$\left | 1-e^{i\theta } \right |^{2}$ i.e. find the square of the modulus of 1-$e^{i\theta }$
$e^{i\theta } = R(cos\theta +isin\theta )$, R=1
= $cos\theta +isin\theta$
In cartesian form:
$a = Rcos\theta \Rightarrow cos\theta$
$b = Rsin\theta \Rightarrow sin\theta$
$\therefore 1-e^{i\Theta } = 1-cos\theta-isin\theta$
$\Rightarrow \left | 1-e^i\theta \right | = \left | 1-cos\theta-isin\theta \right |$
Where the real part, a, = $1-cos\theta$
and the imaginary, b, = $-sin\theta$
The modulus of a complex number being $\sqrt{a^{2} +b^{2}}$
$\therefore \left | 1-cos\theta-isin\theta \right | = \sqrt{(1-cos\theta)^{2} + (-sin\theta)^{2}}$
$\Rightarrow \left | 1-cos\theta-isin\theta \right |^{2} = (1-cos\theta)^{2} + (-sin\theta)^{2} \Rightarrow 1-2cos\theta +cos^{2}\theta +sin^{2}\theta \Rightarrow (\because cos^{2}\theta +sin^{2}\theta = 1) = 2-2cos\theta$
Is this correct? Or is my approach incorrect? Thanks in advance for any feedback/help 🙂
Best Answer
Your argument is lengthy, but sound.
If just the square of the modulus is needed, you can consider that $|z|^2=z\bar{z}$, so $$ |1-e^{i\theta}|^2=(1-e^{i\theta})(1-e^{-i\theta})= 1-(e^{i\theta}+e^{-i\theta})+1=2-2\cos\theta $$
If also the argument is needed, the trick with $1-e^{i\theta}$ (but also $1+e^{i\theta}$) is to set $$ \theta=2\alpha $$ and rewrite $$ 1-e^{i\theta}=1-e^{2i\alpha}= -e^{i\alpha}(e^{i\alpha}-e^{-i\alpha})=-2i(\sin\alpha)e^{i\alpha} $$ Since $0\le\theta<2\pi$, we have $0\le\alpha<\pi$, so $\sin\alpha\ge0$. Hence the modulus is $2\sin\alpha$ and, from $-i=e^{3i\pi/2}$, the argument is $$ \alpha+\frac{3\pi}{2} $$ (up to an integer multiple of $2\pi$).
Thus the square of the modulus is $$ 4\sin^2\alpha=4\sin^2\frac{\theta}{2}=4\frac{1-\cos\theta}{2}= 2(1-\cos\theta) $$