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: $\pi - \tan^{-1}(\frac{b}{a})$?

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 $\theta =\left(\dfrac{7 \pi}{12}\right)$
