If ω is a complex cube root of unity, then ∑ k = 1 6 ( ω k + ω k 1 ) 2 =, 6, 8, 12, 24

If ω is a complex cube root of unity, then ∑ k = 1 6 ( ω k + ω k 1 ) 2 =

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If $\omega$ is a complex cube root of unity, then $\sum_{k=1}^6\left(\omega^k+\frac{1}{\omega^k}\right)^2=$

MathematicsComplex NumberAP EAMCETAP EAMCET 2017 (24 Apr Shift 1)

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1322 Upvotes Verified Answer

The correct answer is: 12

No solution. Refer to answer key.

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