Question

${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{-3}}{2}\right)}^{100}$ is equal to

• A. 2
• B. zero
• C. -1
• D. 1

Answer 1

Answer: (C)

Let $\frac{-1+i\sqrt{-3}}{2}=\omega$ then $\frac{-1-\sqrt{3}i}{2}={\omega }^2$
where $\omega =$ cube root of unity
Consider ${\left(\frac{-1+\sqrt{-3}}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{-3}}{2}\right)}^{100}$
$={\left(\frac{-1+\sqrt{3}i}{2}\right)}^{100}+{\left(\frac{-1-\sqrt{3}i}{2}\right)}^{100}$
$={\omega }^{100}+{\omega }^{200}=\omega +{\omega }^2=-1$
($\therefore {\omega }^3=1$ and $1+\omega +{\omega }^2=0$)