I appreciate that if one does a formal solution (such as the arctangent) for $\arg(z)$ and $\arg(-z)$ you'll get the results that you presented. However, when you get right down to it, there is no difference between $\pm\pi$. I mean exactly this:
$$z=|z|e^{i(\theta+\pi)}=|z|e^{i(\theta-\pi)}$$
Therefore you can say that
$$\arg(-z)-\arg(z)=\pm\pi$$
and use either $+\pi$ or $-\pi$. If I had to make a choice I would choose $+\pi$ because that would mean that I was always rotating $z$ anticlockwise to create $-z$.