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Let $\omega=\mathrm{e}^{\mathrm{i} \frac{2 \pi}{7}}$. Calculate $$ A=\omega+\omega^{2}+\omega^{4} $$

Since $\omega=e^{i\frac{2\pi}{7}}$ it mean that $\omega$ is the 7th roots of unity. So it have known that $1+\omega+\omega^2+\cdots+\omega^6=0$. But the induction want to calculate $A=\omega+\omega^2+\omega^4$. I'm not quite sure how to do this. Please kindly give a help for me or something recommendations for me. Thank is advanced!

user326159
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Matin
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  • Most of it can be found in this related question https://math.stackexchange.com/q/2408982/399263 and linked questions. – zwim Mar 01 '21 at 16:52

1 Answers1

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On the one hand, observe that $$A = \omega+\omega^{2}+\omega^{4} \quad \text{and} \quad B = \omega^3+\omega^{5}+\omega^{6}$$ are conjugate numbers since $AB = 2$. On the other hand, as you noticed, $A + B = -1$. Therefore, $A$ and $B$ are solutions of the polynomial $x^2 + x + 2 =0$.

Then, you can solve the equation and find $A$ and $B$. To know which solution correspond to $A$, notice that $\text{Im}(A) > 0$.

user326159
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  • Can you give a little bit why $Im(A)>0$? Please kindly explain me – Matin Mar 01 '21 at 17:32
  • Observe that $\text{Im}(\omega)=\sin\left(\tfrac{2\pi}{7}\right) > \sin\left(\tfrac{\pi}{7}\right) = - \sin\left(\tfrac{8\pi}{7}\right) = - \text{Im}(\omega^4)$ and $\text{Im}(\omega^2)=\sin\left(\tfrac{4\pi}{7}\right) > 0$, so $\text{Im}(\omega) + \text{Im}(\omega^2) + \text{Im}(\omega^4) > 0$ – user326159 Mar 01 '21 at 17:57