# Does exists a constant $C>0$ such that $|\theta_1 - \theta_2| \leq C|e^{i\theta_1}- e^{i\theta_2}|$?

Consider $S^1 = \{ z \in \mathbb C: |z| = 1 \}$. Given $z_1 = e^{i\theta_1}$ and $z_2 = e^{i\theta_2}$, we have that $|e^{i\theta_1}- e^{i\theta_2}| \leq |\theta_1 - \theta_2|$.

I'm trying to see if exists a constant $C>0$ such that $|\theta_1 - \theta_2| \leq C|e^{i\theta_1}- e^{i\theta_2}|.$

Desenvolving $|e^{i\theta_1}- e^{i\theta_2}|$

$$|e^{i\theta_1}- e^{i\theta_2}|^2 = |(\cos \theta_1 - \cos \theta_2) + i(\sin \theta_1 - \sin \theta_2)|^2 = 2(1 - \sin{(\theta_1 + \theta_2)}).$$

How ever, I don't which constant work.

Help?

Following tristan's remark one can reduce the difference modulo $2 \pi$ such that $$\theta_1 - \theta_2 = \delta + 2 \pi k \quad \text{with }|\delta| \le \pi , k \in \Bbb Z \, .$$ Then $$|e^{i\theta_1} - e^{i\theta_2}| = 2|\sin \frac{\theta_1 - \theta_2}2| = 2 |\sin \frac \delta 2| \ge \frac 2 \pi |\delta|$$ where the first identity is shown in Prove that $|e^{i\theta_1}-e^{i\theta_2}| \leq |\theta_1 - \theta_2|$ and the final estimate follows from $1)$ For $0\le \theta \le\frac{\pi}{2}$, show that $\sin \theta \ge \frac{2}{\pi} \theta$.
If $\theta_1 - \theta_2 = \pi$ then equality holds, so the estimate is best possible.
This is false. Indeed, let $\theta_1=0$ and $\theta_2=2\pi$. The argument is not even a continuous function on $\mathbb{S}^1$.
• It depends on how you define the argument :) One could argue that the argument of a complex number belongs to $\mathbb{R}/2\pi\mathbb{Z}$. Aug 19 '17 at 18:58