If $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the 23 rd roots of unity, then $\alpha_1^{47}+\alpha_2^{47}+\ldots .+\alpha_{23}^{47}=$

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Aicharya · Accepted Answer

$\because \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{23}$ are the 23 rd roots of unity. $ 	herefore \alpha^{23}-1=0 \Rightarrow \alpha^{23}=1 $ Now, $\alpha_1^{47}+\alpha_2^{47}+\ldots+\alpha_{23}^{47}$ $ \begin{aligned} & =\alpha_1\left(\alpha_1^{23}ight)^2+\alpha_2\left(\alpha_2^{23}ight)^2+\ldots+\alpha_{23}\left(\alpha_{23}^{23}ight)^2 \ & =\alpha_1+\alpha_2+\ldots+\alpha_{23} \end{aligned} $ Sum of roots $=0$

If $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the 23 rd roots of unity, then…

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