Finding the modulus of a complex number $1-e^{i\theta }$ 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 :)
 A: HINT:
Now $2-2\cos\theta=4\sin^2\dfrac\theta2$ using $\cos2A=1-2\sin^2A$
Finally for real $u,\sqrt{u^2}=|u|=\begin{cases} +u &\mbox{if } u\ge0 \\ 
-u & \mbox{if } u<0 \end{cases} $
Alternatively,
$$1-e^{i\theta}=-e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)=-2ie^{i\theta/2}\sin\dfrac\theta2$$
Alternatively, $$1-\cos\theta-i\sin\theta=2\sin^2\dfrac\theta2--2i\cos\dfrac\theta2\sin\dfrac\theta2=-2i\sin\dfrac\theta2\left(\cos\dfrac\theta2+i\sin\dfrac\theta2\right)$$
A: 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
$$
4\sin^2\alpha=4\sin^2\frac{\theta}{2}=4\frac{1-\cos\theta}{2}=
2(1-\cos\theta)
$$
A: Just take $2$ common from the final result and apply the formulae $$1-\cos\theta=2\sin^2\frac{\theta}{2}$$ Your answer will be simplified.
