inequality involving complex exponential Is it true that 
$$|e^{ix}-e^{iy}|\leq |x-y|$$ for $x,y\in\mathbb{R}$? I can't figure it out. I tried looking at the series for exponential but it did not help.
Could someone offer a hint? 
 A: The function $u$ defined on $[0,1]$ by $u(t)=\exp(\mathrm i x+\mathrm it(y-x))$ is such that $u'(t)=\mathrm i(y-x)u(t)$ and $|u(t)|=1$ hence $|u'(t)|=|y-x|$ and $|u(1)-u(0)|\leqslant\sup\{|u'(t)|;t\in[0,1]\}=|y-x|$. Note finally that $u(0)=\exp(\mathrm i x)$ and $u(1)=\exp(\mathrm i y)$.
A: One way is to use
$$
|e^{ix} - e^{iy}| = \left|\int_x^ye^{it}\,dt\right|\leq \int_x^y\,dt = y-x,
$$
assuming $y > x$.
A: Hints: $e^{ix} -e^{iy}= e^{ix}(1 -e^{i(y-x)})$, and $|e^{ix}|=1$. Let $\theta=x-y$.
A: Hint: $$\begin{align}|e^{ix}-e^{iy}|^2 &= (\cos x-\cos y)^2 + (\sin x - \sin y)^2 \\ &= 2-2(\sin x \sin y +\cos x \cos y) = 2-2\cos (x-y)\end{align}$$
For the last step, see ProofWiki. Now use the series expansion of $\cos(x-y)$ and compare to $(x-y)^2$.
A: $$|e^{ix}-e^{iy}|^2=(\cos x-\cos y)^2+(\sin x-\sin y)^2\\=2-2(\cos x\cos y+\sin x\sin y)=2-2\cos(x-y)$$
and we have 
$$1-\cos x\leq \frac{1}{2}x^2\quad \forall x\in\mathbb{R}$$
A: A graphical hint (apparently I need 30 characters... doesn't stackexchange know that a picture is worth a thousand words?).

