finding the argument of $1-\cos 2\theta -i\sin 2 \theta$ "Let z = $1-\cos 2\theta -i\sin 2 \theta$, $0\leq \theta \leq \pi$. Find the modulus and argument of z in terms of $\theta$ in their simplest forms."
The modulus is pretty easy, I got $2\sin \theta$ which is correct.
The problem is, when finding the argument of z, I got $\tan \alpha = \frac{-1}{\tan \theta}$, where $\alpha$ is the argument.
the correct answer is $\alpha = \theta - \pi/2$, however I am curious why $\alpha = \theta + \pi/2$ doesn't work. This also fits the equation above, but is wrong. Can someone explain?
 A: When you calculate $\tan \alpha$, you are specifying the slope of the line connecting the origin to your point $(1-\cos2\theta,-\sin2\theta)$. What's missing is the direction traveled along this line with the origin as the starting point.
Your two proposed arguments separated by a difference of $\pi$, represent the two opposite directions possible on your line, only one of which is correct for actually getting from the origin to your destination. To orient yourself, multiply the cosine of each possible argument by the modulus you found, $2\sin\theta$, and compare with the intended $x$-coordinate $1-\cos2\theta$. Only the candidate $\alpha=\theta-\pi/2$ points you the right way.
A: $$z = 1-\cos 2\theta -i\sin 2 \theta$$
Modulus
$$|z| = \sqrt{(1-\cos 2 \theta)^2 + (-\sin 2 \theta)^2}  = \\
= \sqrt{1 + \cos^2 2  \theta - 2 \cos 2 \theta + \sin^2 2 \theta} = \\
= \sqrt{2 - 2 \cos 2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2  \sin^2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2  (1-\cos^2 \theta)} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2  -2\cos^2 \theta} = \\
= \sqrt{4 - 4 \cos^2 \theta} = \\
= 2 \sqrt{1 - \cos^2 \theta} = \\
2 |\sin\theta|.$$
Argument
First of all, notice that
$$\eta = \arctan\frac{- \sin 2\theta}{1 - \cos 2\theta} 
=\arctan\frac{- 2 \sin \theta \cos \theta}{1 - \cos^2 \theta + \sin^2 \theta}
=\arctan\frac{- 2 \sin \theta \cos \theta}{2\sin^2 \theta} = \\
= -\arctan\text{cotan} \theta = \theta - \frac{\pi}{2}.
$$
since:
$$\arctan(\text{cotan} \beta) = \frac{\pi}{2} - \beta.$$
Additionally, notice that:
$$\mathcal{Re}[z] = 1-\cos 2 \theta \geq 0 ~\forall \theta \in [0, \pi]$$
Then, in this case, under the assumption that $\arg{z} \in (-\pi, \pi]$:
$$\arg{z} = \eta = \theta - \frac{\pi}{2}.$$
Why does $\alpha = \theta + \frac{\pi}{2}$ not work?
First of all, notice that:
$$\mathcal{Im}[z] = -\sin 2 \theta \geq 0 ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right]$$
Hence, for $\theta \in \left[\frac{\pi}{2}, \pi\right]$, $z$ represents a vector in the first quadrant of the complex plane. This means that its argument must be positive and less that $\frac{\pi}{2}$. Indeed:
$$\eta = \theta - \frac{\pi}{2} \in \left[0, \frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right],$$
while
$$\alpha = \theta + \frac{\pi}{2} \in \left[\pi, 3\frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right].$$
The last shows that $\alpha = \theta + \frac{\pi}{2}$ does not work.
