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Why does $|e^{ix}|^2 = 1$?

The book said $e^{ix} = \cos x + i\sin x$, and square it, then $|e^{ix}|^2 = \cos^2x + \sin^2x = 1$.

But, when I calculated it, $ |e^{ix}|^2 = \left|\cos x + i\sin x\right|^2 = \cos^2x - \sin^2x + 2i\sin x\cos x$.

I can't make it to be equal $1.$ How can I do it?

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    $\begingroup$ $|z|^2 = z\bar z$, not $z^2$. $\endgroup$ – Ted Shifrin Aug 3 '18 at 4:01
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If $a,b$ are real then $\displaystyle \left| a+bi \right| = \sqrt{a^2+b^2\,\,} = \sqrt{(a+bi)(a-bi)\,}.$

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If $x \in \mathbb{R}$, $$\cos^2x - \sin^2 x= \cos(2x)$$

$$2 \sin x \cos x =\sin(2x)$$

$$|(\cos x + i \sin x)^2|=\cos^2(2x)+\sin^2(2x)=1$$

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  • $\begingroup$ up to the point .. +1 $\endgroup$ – Ahmad Bazzi Aug 3 '18 at 4:12
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you can first apply the modulus $$|e^{ix}|=\sqrt{\cos^2 x + \sin^2 x}$$ then square it whole you will get $$|e^{ix}|^2={\cos^2 x + \sin^2 x}=1$$

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$|z|^2=z\bar{z}$

But the complex conjugate of ${e^{xi}}$ is $e^{-xi}$.

$|e^{xi}|^2=e^{xi}e^{-xi}=1$

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