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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.


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Everything is right! Note that $e^{-2\pi i/3}=e^{4\pi i/3}$. (Divide the second by the first. You get $e^{2\pi i}$, which is $1$.) – André Nicolas Apr 12 '11 at 12:22
@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 :-) – Quixotic Apr 12 '11 at 12:31
up vote 2 down vote accepted

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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Hm I got it thanks,but I think the general value of the argument is $Arg(z)= (2n\pi + \theta)$ where $arg(z) = \theta$ and $n \in \mathbb{N}$ so $\large e^{\frac{4i\pi}{3}}=e^{-\frac{2i\pi}{3}} = e^{\frac{10i\pi}{3}} ... $. – Quixotic Apr 12 '11 at 12:25
The principal value of the argument of $z\ne0$ is usually defined as the only $\theta\in(-\pi,\pi]$ such that $z=|z|e^{i\theta}$, and is represented as $\hbox{arg}(z)$. Sometimes $\hbox{Arg}(z)$ is defined as the set of all arguments: $\hbox{Arg}(z)=\{\hbox{arg}(z)+2k\pi:k\in\mathbb{Z}\}$. In general, $z=|z|e^{(\hbox{arg}(z)+2k\pi)i}$ for all $k\in\mathbb{Z}$. – Julián Aguirre Apr 12 '11 at 13:59
Yes I meant the same thing :-) – Quixotic Apr 17 '11 at 15:13

Observe that $\large e^{\frac{4i\pi}{3}}=e^{\frac{-2i\pi}{3}}$

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But could you hint me how? – Quixotic Apr 12 '11 at 12:16
Since $e^{2i\pi}=1$ – Dennis Gulko Apr 12 '11 at 12:18
Okay but I still don't get how is your hint going to lead me to that thing? – Quixotic Apr 12 '11 at 12:27
Aha got it :) – Quixotic Apr 12 '11 at 12:28

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