TS EAMCET · Maths · Complex Number
If \(\omega\) is a complex cube root of unity, then \(\begin{aligned} & \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right)=\end{aligned}\)
- A \(-2\)
- B \(2\)
- C \(-1\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(-2\)
Step-by-step Solution
Detailed explanation
We have, \(\frac{1-\sqrt{3} i}{2}=-\left[\frac{-1+\sqrt{3} i}{2}\right]=-\omega\) and \(\frac{1+\sqrt{3} i}{2}=-\left[\frac{-1-\sqrt{3} i}{2}\right]=-\omega^2\) \(\therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026}\)…
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