<s>[INST] 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?
 [/INST] $$z = 1-\cos 2\theta -i\sin 2 \theta$$
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.
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