$\omega = \frac{1 + \sqrt3 i}{2}$ , $ \omega^5 = ? $

Question

$\omega = \frac{1 + \sqrt3 i}{2} $, $ \omega^5 = ? $

$\omega^3 = 1$ by definition?
So, $\omega^5 = \omega^2$

But why do i get wrong answer?

Answer 1

Not by definition, just check that $$\omega=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i=(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}))=e^{i\frac{\pi}{3}}$$

If you make $\omega^3$, then you have $e^{i\frac{\pi}{3}\cdot 3}=e^{i\pi}=-1$.

So $\omega^5=\omega^3\omega^2=-\omega^2$ is your mistake

Answer 2

Because $\omega=\cos\left(\frac\pi3\right)+i\sin\left(\frac\pi3\right)$ and therefore it is not true that $\omega^3=1$. Actually, $\omega^3=-1$. And$$\omega^5=-\omega^2=-\cos\left(\frac{2\pi}3\right)-i\sin\left(\frac{2\pi}3\right)=\frac{1-i\sqrt3}2.$$

Answer 3

$$ \omega = \frac{1 + \sqrt3 i}{2} =e^{i\pi/3 }$$

$$ \omega^5 = e^{5i\pi/3 } = \cos (5\pi /3)+i\sin(5\pi /3) = 1/2 -i\sqrt 3 /2$$