Q50 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

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

  • A
    23
  • B
    -1
  • C
    1
  • Correct answer
    D
    0

Solution

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

Hence, the correct answer is option: 'D'.

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