How should I prove that $ |e^{ia}-e^{ib}|=2\sin\frac{|a-b|}{2}\leq|a-b| $?

Question

How is it possible to prove that: $$ |e^{ia}-e^{ib}|=2\sin\frac{|a-b|}{2}\leq|a-b|? $$

Specifically, I'm looking for an analytic technique to show that the equality $|e^{ia}-e^{ib}|=2\sin\frac{|a-b|}{2}$ is correct.

Answer 1

$$\|e^{ia}-e^{ib}\|^2=(e^{ia}-e^{ib})(e^{-ia}-e^{-ib}) = 2-2\cos(a-b) = 4\sin^2\left(\frac{a-b}{2}\right)$$

Due to the fact that $\sin$ is a Lipschitz-continuous, odd function with a derivative bounded by $1$, by assuming that $|a-b|$ does not exceed $2\pi$ we have: $$ \|e^{ia}-e^{ib}\|=2\sin\left(\frac{|a-b|}{2}\right) \leq |a-b|.$$

Answer 2

Draw a unit circle in the complex plane, and the points $e^{ia}$ and $e^{ib}$. The inequality $|e^{ia}-e^{ib}|\leq |a-b|$ is true because the first term is the distance between $e^{ia}$ and $e^{ib}$ along a straight line, while $|a-b|$ is the distance between the same two points along the arc of the unit circle, possibly plus some multiple of $2\pi$.

The equality $|e^{ia}-e^{ib}|=2\sin\frac{|a-b|}2$ you get if you draw the four line segments

• From the origin to $e^{ai}$
• From the origin to $e^{bi}$
• From $e^{ai}$ to $e^{bi}$
• From the origin to the middle of the previous line

Now $|e^{ia}-e^{ib}|$ is the length of the third segment, while $\frac{|a-b|}2$ is the angle of one of the two right angled triangles you have just drawn.

Answer 3

A useful trick when you have expressions such as $e^{ia}-e^{ib}$ is to collect $e^{ib}$, so $$ e^{ia}-e^{ib}=e^{ib}(e^{i(a-b)}-1) $$ Now set $a-b=2c$: $$ e^{i(a-b)}-1=e^{ic}(e^{ic}-e^{-ic})=2ie^{ic}\sin c $$ Thus $$ |e^{ia}-e^{ib}|=|e^{ib}|\,|e^{ic}|\,|2ie^{ic}\sin c|=2\lvert\sin c\rvert= 2\sin\lvert c\rvert=2\sin\frac{|a-b|}{2} $$ Next recall that $-x\le \sin x\le x$, for every real $x$.