is argument function continuous? Is argument function defined as $f:\mathbb{C} \rightarrow \mathbb{R}$ $f(z)=Arg z$ continuous?
And what about the function $\iota f$?
Any hints are appreciated.
 A: I'm assuming here that $$ f :\mathbb{C}^* \to [0, 2\pi[$$ with $$f(\rho e^{i\theta}) = \theta - 2k\pi$$ with $k\in\mathbb{Z}$ so as to have $\theta - 2k\pi \in [0, 2\pi[$.
If we consider the sequence $$\forall n\in\mathbb{N}^*, \qquad z_n = \exp(i(2\pi - \frac{1}{n}))$$
Then we have $$\forall n\in\mathbb{N}^*, \qquad f(z_n) = 2\pi - \frac{1}{n} \to 2\pi$$ but $$z_n \to \exp(2i\pi) = 1$$ and $$\arg 1 = 0$$
Therefore, because of the sequencial characterization of continuity, $f$ is not continuous.
A: The argument function has domain $ℂ\setminus\{0\}$ and values in $ℝ/2\piℝ$ (a space homeomorphic to a circle) and as such, it is continuous. It's not well-defined as a function to $ℝ$.
EDIT (clarification):
Without more specification, the function $Arg$ is not defined from $ℂ\setminus\{0\}$ to $ℝ$. It is not a function at all.
There is a function $Arg: ℂ\setminus\{0\} \to S_1$ and another function $\mod_{2\pi}: ℝ \to S_1$ but none of them is injective and they have different kernels (in a generic sense), so you can't compose one with the inverse of the other.
If you specify what branch of the inverse of $\mod_{2\pi}$ you use, then you can compose, and obtain a function $ArgR:ℂ\setminus\{0\} \to ℝ$ which is not continuous for any choice of the branch.
