Consider $$\operatorname{Arg}:\mathbb{C}^*\rightarrow\mathbb{R}/2\pi\mathbb{Z}$$ $$\operatorname{Arg}(z)=\arg(z)+2\pi\mathbb{Z}$$
where $\arg(z)$ is the principal argument. Show that such function is continuous in the quotient topology. I never encountered such topology before so I don't know how to do that, however I have a hint (that actually seems to me harder):
Show that for every convergent sequence of non-zero complex numbers $z_n\rightarrow z$ and $\theta \in \operatorname{Arg}(z)$ we can choose for every $n\in \mathbb{N}$ a $\theta_n \in \text{Arg}(z_n)$ s.t. $\theta_n\rightarrow \theta$
Best Answer
Define $arg:\Bbb C-\{(0,0)\}\rightarrow \Bbb R$ as follows: $$arg(x,y)=tan^{-1}(|\frac{y}{x}|); \ y≥0,x>0$$ $$arg(0,y)=\frac{\pi}{2};\ y>0$$ $$arg(x,y)=\pi-tan^{-1}(|\frac{y}{x}|);\ y≥0,x<0$$ $$arg(x,y)=\pi+tan^{-1}(|\frac{y}{x}|);\ y≤0,x<0$$ $$arg(0,y)=\frac{3\pi}{2};\ y<0$$ $$arg(x,y)=2\pi-tan^{-1}(|\frac{y}{x}|);\ y<0,x>0$$
Where $tan^{-1}:\Bbb R\rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ is the inverse of $tan_{|(-\frac{\pi}{2},\frac{\pi}{2})}\rightarrow \Bbb R$.
Now $\Bbb C^*$ is a metric space. Therefore to show the continuity of the map $$\operatorname{Arg}:\mathbb{C}^*\rightarrow\mathbb{R}/2\pi\mathbb{Z}$$ $${Arg}(z)=\arg(z)+2\pi\mathbb{Z}$$ we may use sequential criteria. But note that $arg $ is continuous at each point of $\Bbb C-\{(x,0):x≥0\}$ , hence $Arg$ is also continuous on $\Bbb C-\{(x,0):x≥0\}$ as the quotient map $\pi:\Bbb R\rightarrow \frac{\Bbb R}{2\pi \Bbb Z}$ is continuous.
Now to check the continuity of $Arg$ on $\{(x,0)\in \Bbb C:x>0\}$ notice that if a sequence $\{(x_n,y_n)\}_n$ in $\Bbb C ^*$ converges to $(x,0)$ with $x>0$, then either $arg$ converges to $0$ or converges to $2\pi$ i.e. $Arg$ converges to either $0\ +\ 2\pi \Bbb Z$ or $2\pi\ +\ 2\pi \Bbb Z$ but $0\ +\ 2\pi \Bbb Z\ =\ 2\pi\ +\ 2\pi \Bbb Z$. Hence $Arg$ is also continuous.