# Show $|\left(e^{-ixt}-1 \right)/t| \le |x|$, $x ,t\ne 0$

Show $$\left|\frac{e^{-ixt}-1}{t}\right| \le |x|$$ for all real $x \ne 0$ and all real $t \ne 0$.

I tried taylor theorem but as the reminder is complex, I don't see how to get the result.

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First step: denote $\alpha=xt$. You need to prove that $\left|e^{-i\alpha}-1\right|\leq|\alpha|$ for $\alpha\neq0$ –  Dennis Gulko Oct 29 '12 at 19:40

As explained by @Dennis Gulko in a comment, the task is to prove that $|\mathrm e^{-\mathrm is}-1|\leqslant|s|$ for every real number $s$. But the function $u:\mathbb R\to\mathbb C$, $t\mapsto\mathrm e^{-\mathrm it}$, has derivative $u'(t)=-\mathrm i\mathrm e^{-\mathrm it}$, whose modulus is (at most) $1$ everywhere, hence $$|u(s)-u(0)|\leqslant|s-0|\cdot\sup\{|u'(t)|\mid0\leqslant t\leqslant s\}=|s|.$$
\begin{align} \exp(-ixt) - 1 &= \cos(xt) - i \sin(xt) - 1\\ & = -\left(2 \sin^2(xt/2) + 2i \cos(xt/2) \sin(xt/2)\right)\\ &= -2i \sin(xt/2) \left(\cos(xt/2) - i \sin(xt/2)\right)\\ & = -2i \sin(xt/2) \exp(-ixt/2) \end{align} Hence, $$\vert \exp(-ixt) - 1 \vert = \vert -2i \sin(xt/2) \exp(-ixt/2) \vert = 2 \vert \sin(xt/2) \vert$$