AP EAMCET · Maths · Complex Number
If \(\omega\) is a complex cube root of unity, then \(\sum_{k=1}^6\left(\omega^k+\frac{1}{\omega^k}\right)^2=\)
- A 6
- B 8
- C 12
- D 24
Answer & Solution
Correct Answer
(C) 12
Step-by-step Solution
Detailed explanation
If \( k \equiv 0 \pmod 3 \), then \(\omega^k=1\). So, \(\left(\omega^k+\frac{1}{\omega^k}\right)^2 = \left(1+1\right)^2 = 4\). If \( k \not\equiv 0 \pmod 3 \), then \(\omega^k+\frac{1}{\omega^k} = \omega^k+\omega^{3-k} = \omega+\omega^2 = -1\). So,…
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