Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{3}}}\right)$. I can put the squareroot sign over the whole fraction, but that still doesn't really help me get an actual number for theta. 
ALSO: It is a rule that I add $\pi$ to theta if $a < 0$. Why? Looking at the diagram of the triangle form by the complex vector $z$ and the real axis, the 'triangle' is in the second quadrant. Hence, the theta we find with $\theta = \tan^{-1}(\frac{b}{a})$ is the angle closes to the real axis on the left-side. However, we measure angle going counter-clockwise. Shouldn't we do the following computation to get the angle going clock-wise: $\pi - \tan^{-1}(\frac{b}{a})$?
 A: Note
$$\frac{1}{2}\sqrt{2 - \sqrt{3}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1 - \cos \frac{\pi}{6}}{2}} = \sin \frac{\pi}{12} $$
and similarly $$\frac{1}{2}\sqrt{2 + \sqrt{3}} = \cos \frac{\pi}{12}.$$ Therefore $$z = 2\left(-\sin \frac{\pi}{12} + i\cos \frac{\pi}{12}\right) = 2e^{i(\frac{\pi}{2} + \frac{\pi}{12})} = 2e^{i\frac{7\pi}{12}}.$$
Since $-\pi < \frac{7\pi}{12} \le \pi$, the principal argument of $z$ is $\frac{7\pi}{12}$.
A: Note that:
$$
\dfrac{\sqrt{2+\sqrt{3}}}{-\sqrt{2-\sqrt{3}}}=-\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=-\sqrt{\dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}=-(2+\sqrt{3})
$$
since $2+\sqrt{3}=\tan \left(\dfrac{5 \pi}{12}\right)$, and $\theta$ must be in the second quadrant we have $\theta =\left(\dfrac{7 \pi}{12}\right)$
A: These numbers may look pretty complicated but notice that if we square both sides a lot can be simplified.
$$ z^2 = \big((2 - \sqrt{3}) - (2 + \sqrt{3})\big) - 2\sqrt{(2 - \sqrt{3})(2 + \sqrt{3})}\,i
= -2\sqrt{3} - 2i = 4 \left(\cos \left(\frac{7\pi}{6}\right) + i\sin \left(\frac{7\pi}{6}\right) \right) $$
So now we know the argument of $z^2$ is $\frac{7\pi}{6}$, we can relate this to the argument of $z$ according to de Moivre's theorem
$$ (\cos \theta + i\sin \theta)^2 = \cos (2\theta + 2n\pi) + i\sin (2\theta + 2n\pi) $$
If $\theta$ is the argument of $z$, the argument of $z^2$ is $$2\theta + 2n\pi =  \frac{7\pi}{6}$$
Solving that, we get 
$$\theta = \frac{7\pi}{12} + n\pi = \frac{7\pi}{12}, \frac{19\pi}{12} $$
We got two answers, because one is the argument of $z$, and the other is the argument of $-z$. 
Since $\cos \theta < 0$ and $\sin \theta > 0$, $\theta$ must lie on the second quadrant (between $\pi/2$ and $\pi$), therefore $ \theta = \frac{7\pi}{12}$
