Question

The value of ${\left(\frac{-1+i\sqrt{3}}{2}\right)}^{12}+{\left(\frac{-1-i\sqrt{3}}{2}\right)}^{12}$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (C)

$2$

We know that $\omega =\frac{-1+\sqrt{3}i}{2}$ and ${\omega }^2=\frac{-1-\sqrt{3}i}{2}$ are complex cube roots of unity
$\therefore$ the expression becomes
$(w{)}^{12}+(w^2{)}^{12}$
$=w^{12}+w^{24}$
$=1+1$
$=2$