Proving the function $f(z)=\arg(z)$ is continuous on a particular set Prove that $f: \{z \in \mathbb C : |z|=1\} \setminus \{-1\} \to \mathbb R$, $f(z)=\arg(z)$, is continuous.
I am lost with this exercise, I don't know which identity to use in order to show that if $z \to z_0$ in this set, then $\arg(z) \to \arg(z_0)$. 
Sorry if I am asking something trivial, but I don't see why the point $-1$ is excluded from this set. And, considering the general case of the complex plane as the domain: if a sequence of complex numbers $\{z_n\}_{n \in \mathbb N}$ is converging to some number $z$, wouldn't this mean that the argument of $z_n$ is getting closer and closer to the argument of $z$? I can't see why this is not true. 
Could someone please enlighten me?
 A: The function $\arg\colon \mathbb C^* \to \mathbb R/(2\pi)$ is one of the most important functions in analysis and gives the polar angle of a point $z \in \mathbb C^*\,$ "up to a multiple of $2\pi$". 
Its principal value $\DeclareMathOperator{\Arg}{Arg}\Arg(z)$ is real-valued and is defined on the set
$$\mathbb C^{-\cdot}:=\mathbb C\setminus \{z = x + iy \mid x\leq 0,\, y=0 \}.$$
The principal value $\Arg(z)$ is the representive of $\arg(z)$ lying in the interval $\, ]-\pi,\pi[\,$, and can be expressed in terms of the $\arctan$ function familiar from calculus as follows:
$$\Arg(z)=\cases{
-{\pi\over2}+\arctan{x\over -y} & $(y<0)$ \cr
\arctan{y\over x} & $(x>0)$ \cr
{\pi\over2}-\arctan{x\over y} & $(y>0)$\cr}
\tag{1}$$
It is easy to check that $\Arg$ is well defined on $\mathbb C^{-\cdot}$ by $(1)$. As any $z\in \mathbb C^{-\cdot}$ has a neighborhood in which a single alternative of $(1)$ applies it follows that $\Arg$ is continuous (even $C^\infty$) in all of $\mathbb C^{-\cdot}$, and a fortiori $\Arg$ is continuous on $\mathbb C^{-\cdot}\cap S^1$.
A: See any point on the set $A = \{z \in \mathbb{C}: |z| = 1\}$ as $z = e^{i \theta}$ where $-\pi \le \theta < \pi$. $e^{- i \pi} = -1$
$arg (z) = \theta $ and $\log(z) = -i arg(z)$ where $z \in A$
On the set $A - \{-1\}$ the complex logarithm $\log(z)$ is single valued and continuous.
Take any point $z \in A$ $z \neq -1$. Consider the sequence $\{z_n\}$ in $A$ converges to $z$. We shall see the sequence $\{\log(z_n)\}$ will converge to $Arg(z)$ because of the continuity of the function $\log(z)$
For $z = -1$ the function $\log(z)$ is multivalued and not continuous. So $arg(z)$ will not also be continuous at $z = -1$.
A: To answer your second question, consider the sequence $x_n=\left(\frac{-1} 2\right)^n i$. This clearly converges to $0$, but the argument oscillates (does not converge).
