Prove that the principal argument $\rm{Arg}:S^1\setminus\{-1\}\to (-\pi,\pi)$ is continuous using $\epsilon$-$\delta$ definition. How to prove that $\rm{Arg}:S^1\setminus\{-1\}\to (-\pi,\pi)$ is continuous using $\epsilon$-$\delta$ definition.
The argument function is defined as $\rm{Arg}(e^{i\theta})= \theta$, where $\theta\in(-\pi,\pi)$.
What I tried is given $\theta,\theta_0\in (-\pi,\pi)$, I compute
$$|e^{i\theta}-e^{i\theta_o}|= 2\sin\left(\dfrac{|\theta-\theta_0|}{2}\right).$$ Then I made a guess that for any given any $\epsilon>0$, choose $\delta=2\sin\left(\frac{\epsilon}{2}\right)$. Then
$$|e^{i\theta}-e^{i\theta_o}|<\delta \implies 2\sin\left(\dfrac{|\theta-\theta_0|}{2}\right)<2\sin\left(\dfrac{\epsilon}{2}\right).$$ From here how do I conclude that $|\theta-\theta_0|<\epsilon?$. My problems here are:
(i) The chosen $\delta=\sin(\frac{\epsilon}{2})$ may not be positive always, and
(ii) $\sin:(-\pi,\pi)\to (-1,1)$ is srictly increasing only on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ and strictly decreasing on the rest. So from here it follows that $|\theta-\theta_0|<\epsilon$  when $|\theta-\theta_0|\in [-\frac{\pi}{2},\frac{\pi}{2}]$, and when $|\theta-\theta_0|\in(-\frac{\pi}{2},0)\cup(0, \frac{\pi}{2})$, I got the inequality $|\theta-\theta_0|>\epsilon$.
So my question is what will be the right choice of $\delta$? Any help will be appreciated.
 A: Your idea cannot work. Given $\epsilon >0$, your attempt is $\delta = 2 \sin (\frac \epsilon 2)$ which is independent of $\theta_0$. But no choice of $\delta$ which only depends on $\epsilon$ can do because this would mean that $\operatorname{Arg}$ is uniformly continuous which is not true. In fact, let $\theta^n_0 = \pi -\frac{1}{n}$ and $\theta^n = -\pi + \frac{1}{n}$. Then $e^{i\theta^n} - e^{i\theta^n_0} \to 0$, but $\theta^n - \theta^n_0 \to 2\pi$. Therefore the choice of $\delta$ must depend on $\theta_0$.
Let us analyze what happens with your choice. Of course it suffices to consider $\epsilon \le \pi$. In this case $\frac \epsilon 2 \in (0, \frac \pi 2]$ and thus $\sin (\frac \epsilon 2) \in (0,1]$.  Hence $\delta = 2 \sin (\frac \epsilon 2) >0$. Therefore $\lvert e^{i\theta} - e^{i\theta_0} \rvert <\delta$ means
$$2\sin\left(\dfrac{|\theta - \theta_0|}{2}\right)<2\sin\left(\dfrac{\epsilon}{2}\right) .$$
If $|\theta - \theta_0| \le \pi$, we can indeed conclude that $|\theta - \theta_0| < \epsilon$. But in general $|\theta - \theta_0| > \pi$ and the argement breaks down. However, for $\theta_0 = 0$ we get a correct proof that $|\theta - 0| < \epsilon$.
Let us now consider $\theta_0 \ne 0$. We only consider the case that $\theta_0 \in (0,\pi)$, the case $\theta_0 \in (-\pi,0)$ can be treated similarly. Let $e^{i\theta_0} = a_0 + ib_0$. Since $\theta_0 \in (0,\pi)$, we have $b_0 > 0$. Now consider $e^{i\theta} =a + ib$. Then $\lvert e^{i\theta} - e^{i\theta_0} \rvert < b_0$ implies  $b_0 > \lvert (a +ib) - (a_0 + ib_0) \rvert \ge \lvert b - b_0 \rvert$. This is possible only if $b > 0$ which means $\theta \in (0,\pi)$ so that $|\theta - \theta_0| \le \pi$. Now take
$$\delta = \min\left(b_0, 2\sin (\frac{\epsilon}{2})\right). $$
