TITLE: Continuity of $\arg (z)$
QUESTION [6 upvotes]: Let $\mathrm{arg}:\mathbb{C}\setminus \{0\} \to [0,2\pi) $ be the function with $\mathrm{arg}(re^{i\alpha})= \alpha$ and $\alpha \in [0,2\pi)$. How can we prove that $\mathrm{arg}: \mathbb{C}\setminus A \to \mathbb{R}$ is continuous when $A= \{z\in \mathbb{C} | x \geq 0, y=0\}$?

REPLY [3 votes]: Note that
$$
\begin{align}
\frac{\sin(\theta)}{1-\cos(\theta)}
&=\frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)}\\[6pt]
&=\cot(\theta/2)\\[12pt]
&=\tan((\pi-\theta)/2)\tag{1}
\end{align}
$$
Using $(1)$, $\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$, and $\cos(\theta)=\frac{x}{\sqrt{x^2+y^2}}$, we get that
$$
\begin{align}
\arg(x+iy)
&=\theta\\
&=\pi-2\arctan\left(\frac{y}{\sqrt{x^2+y^2}-x}\right)\tag{2}
\end{align}
$$
which is continuous for all $\mathbb{C}$ away from the positive real axis (the positive real axis is precisely where $\sqrt{x^2+y^2}-x=0$).