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TS EAMCET · Maths · Complex Number

If \(\omega\) is a complex cube root of unity, then \(\begin{aligned} & \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right)=\end{aligned}\)

  1. A \(-2\)
  2. B \(2\)
  3. C \(-1\)
  4. D \(0\)
Verified Solution

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 \(\frac{1+\sqrt{3} i}{2}=-\left[\frac{-1-\sqrt{3} i}{2}\right]=-\omega^2\) \(\therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026}\)…