As part of a larger problem, I wish to find the arguments $\theta$ for the complex numbers $-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$. This is $\theta = \arctan(-\sqrt{3})$, which is $-60^{\circ}=300^{\circ}$ for the positive and $\theta = \arctan(\sqrt{3})=60^{\circ}$ for the negative.
However, it is then given in a solution to the problem that the possible arguments are only $120^{\circ}$ and $240^{\circ}$. I feel as though I am missing something painfully obvious, but I don't understand what I am missing.
The correct formula is $$\tan(\theta) = \frac{y}{x},$$ where $x$ and $y$ are the real and imaginary parts, respectively of a complex number $z$, and $\theta$ is its argument. However, students tend to make the following incorrect simplification: $$\theta = \tan^{-1}\left(\frac{y}{x}\right).$$ The above is wrong, as $\tan^{-1}$ can only produce angles between $-\pi/2$ and $\pi/2$, i.e. complex numbers in the first or fourth quadrants, i.e. complex numbers with positive real components.
Try plotting the number. You'll see immediately that its argument is definitely not $-\pi/3$. Use the fact that $\tan$ has period $\pi$, and you'll see that instead the answer is $-\pi/3 + \pi = 2\pi/3$, or $120^\circ$; this angle produces the same $\tan$ value, and lies in the correct quadrant.