Is it possible to say that $\arg(0)= \infty $ since it's not undefined or it is $0$ since $0|0$?

Question

let $z$ be a complex variable and $z=0+0i$ , it is clear at all if want to compute $\arg(z)$ then should be check the value of $\tan^{-1}(y/x)$ then $\tan^{-1}(0/0)$ could be $0$ since $0 |0 $ using the divisibility notion or Is it possible to say that $\arg(0)= \infty $ since it's not undefined ? Where is the approach definition ?

Answer 1

If you avoid talking about branches and multivalued functions as much as possible, then you fix some half-open interval $I$ of length $2\pi$. Given $I$, the definition of $\mathrm{arg}(z)$ is a real number $\theta$ in $I$ such that $z=re^{i\theta}$ for some $r \geq 0$. Such a $\theta$ exists and is unique if $z \neq 0$.

But when $z=0$, $r$ is necessarily zero. At this point, $\theta$ can be any number in $I$, thus $\mathrm{arg}(z)$ is not uniquely defined by the properties that we want it to satisfy.