complex numbers - Polar coordinates - issue with direction denoted by angle

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?

Answer

Why did you do

$$\arg z=\arctan\left|\frac{y}{z}\right|??$$

It should be

$$\arg z=\arctan\frac{y}{z}=\arctan-\sqrt 3=-\frac{\pi}{3}\,,\,\frac{2\pi}{3}$$

Since in $\,z=1-\sqrt 3i\,\;$ the real part is positive and the imaginary part is negative, the vector(=the complex number) is in the fourth quadrant, so the answer must be $\,-\dfrac{\pi}{3}\,$ , or if a positive number is wanted, $\,\dfrac{5\pi}{3}\,$