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I have been trying to write $\omega^2$ in Euler form, first $\omega^2 = \frac{-1}{2}-\frac{\sqrt{3}i}{2} ,$ hence $|\omega^2| = 1$ and $\arg(\omega^2) = -\pi + \frac{\pi}{3}$ as $\omega ^2$ lies in the fourth quadrant which gives $\omega^2 = e^{\frac{-2i\pi}{3}}$.

But using de Moivre's formula, I derived that the cube roots for unity is $1$, $\exp \left(\frac{2i\pi}{3} \right)$ and $\exp \left( \frac{4i\pi}{3} \right)$. I am getting the value of $\omega = e^{\frac{2i\pi}{3}} $ but I am not sure how $e^{\frac{4i\pi}{3}}=e^{\frac{-2i\pi}{3}}$? Please explain your answer.

Thanks,

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    @user6312:Yeps that's an easy fix but Julián Aguirre's answer is the right way to deduce it and works for other cases too :-)2011-04-12

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For any complex number $z\ne0$ there is an infinity of angles $\theta$ such that $ z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}. $ The difference between any two such angles is an integer multiple of $2\pi$. Thus, since $ \frac{4\pi}{3}=-\frac{2\pi}{3}+2\pi, $ you see that $ e^{\frac{4\pi}{3}}=e^{-\frac{2\pi}{3}}. $

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    Yes I meant the same thing :-)2011-04-17
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Observe that $\large e^{\frac{4i\pi}{3}}=e^{\frac{-2i\pi}{3}}$

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    $A$ha got it :)2011-04-12