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We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?

$$|1-e^{i\theta}|\le|\theta|$$

I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.

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    $\begingroup$ Draw the unit circle and the two points $1$ and $e^{i\theta}$. Draw the line segment between these points. Compare the length of this segment with the arc joining the two points. $\endgroup$ Commented May 18, 2013 at 19:22

3 Answers 3

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Hint: $$ \mathrm e^{\mathrm i\theta}-1=\int_0^\theta\mathrm i\mathrm e^{\mathrm ix}\mathrm dx,\qquad|\mathrm i\mathrm e^{\mathrm ix}|\leqslant1\ (x\in\mathbb R). $$

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  • $\begingroup$ +1: IIRC, the OP's problem appears in "Berkeley Problems in Mathematics" and your hint appears as their solution. $\endgroup$
    – JavaMan
    Commented May 18, 2013 at 19:41
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    $\begingroup$ @JavaMan Thanks. One thing is sure, which is that neither the authors of BPM nor I invented this. $\endgroup$
    – Did
    Commented May 18, 2013 at 20:10
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Geometically, $|1-e^{i\theta}|$ can be viewed as the distance between $1$ and $e^{i\theta}$, which is smaller than the arc from $1$ to $e^{i\theta}$(portion of the unit circle).

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  • $\begingroup$ thank you Chung for the geometric interpretation ;) $\endgroup$
    – mounaim
    Commented May 18, 2013 at 19:43
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$$ \left|1 - e^{i\theta}\right| = \left|\left(1 - \cos \theta\right) - i\sin\theta\right| = \sqrt{(1 - \cos \theta)^2 + \sin^2 \theta} = \sqrt{2 - 2 \cos \theta} = 2\left|\sin\frac{\theta}{2}\right| $$ Because $ \sin \theta $ is an odd function and $ \sin x < x\ \forall x \in \mathbb{R}^+ $, the result follows.

Here is a proof that $ \sin x < x $ using pre-calculus topics. It's a bit lengthy; the overall idea is using a geometrical interpretation.

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  • $\begingroup$ Is it significantly simpler to show that $\sin x\lt x$ than the inequality asked in the question? $\endgroup$
    – Did
    Commented May 18, 2013 at 20:11
  • $\begingroup$ I would say it is neither simpler not more complex. It's just a well known result which is perfectly applicable. In my opinion, $ \sin x < x $ is a more elementary result than the integral you used, as that inequality can be proven using pre-calculus topics. $\endgroup$
    – Jon Claus
    Commented May 18, 2013 at 20:17
  • $\begingroup$ "Neither simpler no[r] more complex" but "more elementary"? You lost me... :-) // OK, if indeed sin x < x has a pre-calculus proof, your answer could include this proof. $\endgroup$
    – Did
    Commented May 18, 2013 at 20:21
  • $\begingroup$ Sorry, I didn't really understand what you meant because in my opinion it is simple to show that $ \sin x < x $ than the result in the problem because one relies on the other. As for the proof, it's in my edit. $\endgroup$
    – Jon Claus
    Commented May 18, 2013 at 20:41
  • $\begingroup$ OK. The video in the link proves this for $0\lt x\lt\pi/2$, comparing the areas of a triangle and a sector. Since quite similar arguments (comparing the lengths of an arc and a segment) show the result asked by the OP, the answer to the question in my first comment seems to be "No". $\endgroup$
    – Did
    Commented May 19, 2013 at 6:35

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