When the argument of complex numbers is a well defined real valued function?

Question

I know that the argument $\arg:\Bbb C\setminus\{0\}\to\Bbb R$ is multivalued function and also that if we consider $\arg:\Bbb C\setminus\{0\}\to{\Bbb R}/{2\pi \Bbb Z}$, then it is a well defined function. I am having trouble in understanding

1. when the argument of non-zero complex numbers can be treated as a well defined real valued function and when it will be continuous as a real valued function,

2.If we define $\arg:\Bbb C\setminus\{0\}\to{\Bbb R}/{2\pi \Bbb Z}$, is the argument a continuous function ?

Help is appreciated.

Answer 1

Yes, it is continuous as a map $\mathbb{C}\setminus\{0\}\to \mathbb{R}/2\pi\mathbb{Z}\simeq S^1$. In polar coordinates, it can be identified with the projection $(0,\infty)\times S^1\to S^1$ onto the second coordinate.

But as a function to $\mathbb{R}$, it can only be made continuous if you restrict the domain to $\mathbb{C}\setminus H$ where $H\supseteq \{0\}$ is some set such that each loop in $\mathbb{C}\setminus H$ is contractible in $\mathbb{C}\setminus\{0\}$.

This is not an if and only if statement, it is just necessary (if $\mathbb{C}\setminus H$ contained a loop incontractible in $\mathbb{C}\setminus\{0\}$, then this loop has a nonzero winding number $n$ and the argument would rise by $2\pi n$ when going around this loop). Sufficient is, for example, if $\mathbb{C}\setminus H$ is an open connected and simply connected subset of $\mathbb{C}\setminus\{0\}$.

Answer 2

1. It is multivalued if taking the first definition, and well-defined (i.e. single-valued) if taking the second.

2. Since $\mathbb R/2\pi\mathbb Z$ is isomorphic to $S^1$, it is indeed a continuous function. However if you choose a single representative for each equivalence class (such as $\mathbb R\to [0,2\pi)$), the resulting function is necessary noncontinuous.