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Hi I have trouble doing this exercise:

Let $\omega=e^{2i\pi/5} $ Explain why $\omega^1+\omega^2+\omega^3+\omega^4=−1$ and show that $\omega+\omega^4= 2\cos(2\pi/5)$ and $\omega^2+\omega^3=2\cos(4\pi/5)$

So far I have, $w^1=e^{2iπ/5}, w^2=e^{-2iπ/5}, w^3=e^{4iπ/5}, w^4=e^{-4iπ/5}$

Thank you very much for help!

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  • $\begingroup$ Please share your effort. Observe that $$w\cdot w^4=w^2\cdot w^3=w^5=1$$ $\endgroup$ May 14 '17 at 18:14
  • $\begingroup$ For a start, $\omega+\cdots+\omega^4$ is the sum of a finite geometric progression. $\endgroup$ May 14 '17 at 18:18
  • $\begingroup$ So far I have, $w^1=e^{2iπ/5}, w^2=e^{-2iπ/5}, w^3=e^{4iπ/5}, w^4=e^{-4iπ/5}$ $\endgroup$
    – Markowska
    May 14 '17 at 18:19
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Note that $\omega$ satisfies $\omega^5-1=0$
But $(\omega^5-1)=(\omega-1)(\omega^4 + \omega^3 + \omega^2 + \omega + 1)$
And $\omega \neq 1$.
That's the first bit done.

Then note that $\omega^n + \omega^{5-n} = \omega^n + \omega^{-n} = (\cos(\frac{2\pi n}5) + i\sin(\frac{2\pi n}5)) + \cos(\frac{-2\pi n}5) + i\sin(\frac{-2\pi n}5) = 2\cos(\frac{2\pi n}5) $ Where this last equality is from the fact that cosine is even and sine is odd.

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