<s>[INST] 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: $\pi - \tan^{-1}(\frac{b}{a})$?
 [/INST] 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 $\frac{7\pi}{12}$.
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