Question

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

• 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, $\sum _{j=1}^6(j+\omega )\left(j+{\omega }^2\right)$
$=\sum _{j=1}^6\left(j^2+j{\omega }^2+j\omega +{\omega }^3\right)$
$=\sum _{j=1}^6\left(j^2-j+1\right)=\sum _{j=1}^6{j}^2-\sum _{j=1}^{6}j+\sum _{j=1}^{6}1$
$=\frac{6(6+1)(12+1)}{6}-\frac{6(6+1)}{2}+6$
$=7\times 13-3\times 7+6=91-21+6=76$
$\therefore \sin \left(\sum _{j=1}^6(j+\omega )\left(j+{\omega }^2\right)\frac{3\pi }{152}\right)=\sin \left(\frac{76\times 3\pi }{152}\right)$
$=\sin \frac{3\pi }{2}=-1$
$\therefore {\left(\frac{1-\sqrt{3}i}{2}\right)}^{2020}+{\left(\frac{1+\sqrt{3}i}{2}\right)}^{2026}+\sin \left(\sum _{j=1}^6(j+\omega )\left(j+{\omega }^2\right)\frac{3\pi }{512}\right)$
$=-1-1=-2$