Question

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

• A. -2
• B. 2
• C. -1
• D. 0

Answer 1

Answer: (A)

We have
$\frac{1-\sqrt{3} i}{2}=-\left[\frac{-1+\sqrt{3} i}{2}\right]=-\omega $
and $ \frac{1+\sqrt{3} i}{2} \frac{1+\sqrt{3} i}{2}=-\omega^{2}$
$\therefore \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} $
$=(-\omega)^{2020}+\left(-\omega^{2}\right)^{2026} $
$=\omega^{2020}+\omega^{4052}=\omega+\omega^{2}=-1$
Now, $\displaystyle\sum_{j=1}^{6}(j+\omega)\left(j+\omega^{2}\right)$
$=\displaystyle\sum_{j=1}^{6}\left(j^{2}+j \omega^{2}+j \omega+\omega^{3}\right)$
$=\displaystyle\sum_{j=1}^{6}\left(j^{2}-j+1\right)=\displaystyle\sum_{j=1}^{6} j^{2}-\displaystyle\sum_{j=1}^{6} j+\sum_{j=1}^{6} 1$
$=\frac{6(6+1)(12+1)}{6}-\frac{6(6+1)}{2}+6$
$=7 \times 13-3 \times 7+6=91-21+6=76$
$\therefore \sin \left(\displaystyle\sum_{j=1}^{6}(j+\omega)\left(j+\omega^{2}\right) \frac{3 \pi}{152}\right)=\sin \left(\frac{76 \times 3 \pi}{152}\right)$
$=\sin \frac{3 \pi}{2}=-1$
$\therefore \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026}+\sin \left(\displaystyle\sum_{j=1}^{6}(j+\omega)\left(j+\omega^{2}\right) \frac{3 \pi}{512}\right)$
$=-1-1=-2$