Convert $1-\sqrt{3}i$ to polar coordinates $(r,\varphi)$.
I started by computing $r=|1-\sqrt{3}i|=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2$. When I tried to compute the angle I did something like
$\varphi=\arctan\left|\frac{y}{x}\right|=\arctan\left|\frac{-\sqrt{3}}{1}\right|=\arctan\sqrt{3}=\frac{\pi}{3}.$
Although this answer seems plausible to me, I am unsure, because the angle should be $-\frac{\pi}{3}$ otherwise the resulting coordinates would be the first quadrant rather than in the fourth. How do I have to compute $\varphi$ to match the right quadrant?