Prove that $|e^{i\theta_1}-e^{i\theta_2}| \leq |\theta_1 - \theta_2|$

Question

I'm trying to prove the inequality

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

I have tried to use Taylor's formula and got this

$$|e^{i\theta_1}-e^{i\theta_2}| = |(1+i\theta_1 - \frac{\theta_1^2}{2} + \ldots) -(1 +i\theta_2 - \frac{\theta_2^2}{2} + \ldots)| = |i(\theta_1-\theta_2) + \frac{\theta_2^2-\theta_1^2}{2}+\ldots|$$

The first term looks right, but how do I proceed?

Answer 1

$$|e^{i\theta_1}-e^{i\theta_2}|= |(e^{i\theta_1}-1)-(e^{i\theta_2}-1)|= |\int^{\theta_1}_0ie^{it}dt-\int^{\theta_2}_0ie^{it}dt | = |\int^{\theta_1}_{\theta_2}ie^{it}dt |\leq |\theta_1 - \theta_2|.$$

All credits go to @1015 and his great answer in here.

Answer 2

$$\begin{align} |e^{i\theta_1} - e^{i\theta_2}| &= |e^{i\theta_2}||e^{i(\theta_1 - \theta_2)} - 1|\\ &= |e^{i(\theta_1 - \theta_2)} - 1| \\ &= |e^{i(\theta_1 - \theta_2)/2}||e^{i(\theta_1 - \theta_1)/2} - e^{-i(\theta_1 - \theta_2)/2}| \\ &= |e^{i(\theta_1 - \theta_2)/2} - e^{-i(\theta_1 - \theta_2)/2}| \\ &= |2i \sin((\theta_1 - \theta_2)/2)| \\ &= 2|\sin((\theta_1 - \theta_2)/2)| \\ &\leq 2|(\theta_1 - \theta_2)/2| \\ &= |\theta_1 - \theta_2| \end{align}$$ where the inequality is due to the fact that $|\sin(x)| \leq |x|$ for all real $x$.

Answer 3

Think geometrically. We can represent complex exponential on the unit circle. The quantity $|e^{i\theta_1} - e^{i\theta_2}|$ is the distance between the two points on the circle, the blue line. Geometrically $\theta_1,\theta_2$ are the arc lengths and so $|\theta_1 - \theta_2|$ is the length of the arc that bounds the blue line. So we see that $|\theta_1 - \theta_2|$ is longer - since the line is the shortest distance between two points.