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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?

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6 Answers 6

up vote 27 down vote accepted

One way is to use $$ |e^{ix} - e^{iy}| = \left|\int_x^ye^{it}\,dt\right|\leq \int_x^y\,dt = y-x, $$ assuming $y > x$.

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4  
I shred a tear from this beautiful and clever answer. –  Daniel Montealegre May 5 '13 at 8:49

A graphical hint (apparently I need 30 characters... doesn't stackexchange know that a picture is worth a thousand words?).

A triangle and a circle

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you could still add some words to explain what s,l and theta are, since they are not in the question –  BЈовић Apr 25 '13 at 13:06
3  
I originally did explain how the new variables relate to the old, but since the original poster only asked for a hint, and not a solution, I thought I'd leave it out and let him think it through. –  in_wolframAlpha_we_trust Apr 25 '13 at 13:11

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 $u(1)=\exp(\mathrm i y)$.

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Hints: $e^{ix} -e^{iy}= e^{ix}(1 -e^{i(y-x)})$, and $|e^{ix}|=1$. Let $\theta=x-y$.

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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 $(x-y)^2$.

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$$|e^{ix}-e^{iy}|^2=(\cos x-\cos y)^2+(\sin x-\sin y)^2\\=2-2(\cos x\cos y+\sin x\sin y)=2-2\cos(x-y)$$

and we have $$1-\cos x\leq \frac{1}{2}x^2\quad \forall x\in\mathbb{R}$$

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