Proving by calculation that $\arg(-2) = \pi$ The fact that it is true, seems very obvious, if one draws the complex number $z = (-2 + 0i)$ on the complex plane. The angle is certainly 180 degrees, or pi radians.
But how can this be proven by calculation? Using $\arg(z)=\arctan(b/a)$ or even the "extended" version $\arg(z) = \arctan(b/a) + \text{sign}(b)(1-\text{sign}(a))$ gives $0$.

More complicated than $\arctan(y/x)$, yet valid for all $z\ne0$ is
$$
\arg(x+iy)=\left\{\begin{array}{cl}
2\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)&\text{if }y\ne0\text{ or }x\gt0\\[6pt]
\pi&\text{otherwise}
\end{array}\right.
$$
which is based on the identity
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}
$$
