How to prove $|e^{i\theta} - 1| \leq |\theta|$?

Question

Prove $|e^{i\theta} - 1| \leq |\theta|$, for all real numbers $\theta$.

Answer 1

$|e^{i\theta} -1| =|i\theta\int_0^{1} e^{it\theta}\, dt|\leq |\theta| |\int_0^{1} 1\, dt|=|\theta|$.

Answer 2

$|e^{i\theta} - 1| = |\cos\theta + i\sin\theta -1| = |-2\sin^2\frac{\theta}{2} + 2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}| = |2\sin\frac{\theta}{2}||-\sin\frac{\theta}{2} + i\cos\frac{\theta}{2}|$

$|e^{i\theta} - 1| = 2|\sin\frac{\theta}{2}| $

For $\theta$ in radians, $\sin\theta \leq |\theta|$

$\therefore |e^{i\theta} - 1| = 2|\sin\frac{\theta}{2}| \leq 2\cdot|\frac{\theta}{2}| = |\theta|$

$\therefore |e^{i\theta} - 1| \leq |\theta|$

Q.E.D.

Answer 3

Draw the unit circle. Then compare the length of a chord to the arc length.