Question

The value of ${\left[\frac{-1+i\sqrt{3}}{2}\right]}^{100}+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^{100}$ is

• A. $2$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

The correct option is C $-1$

Fact: If $\omega =\frac{-1+i\sqrt{3}}{2}$ then ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$, where $\omega$ is a cube root of unity.

$\therefore {\left[\frac{-1+i\sqrt{3}}{2}\right]}^{100}+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^{100}={\omega }^{100}+({\omega }^2{)}^{100}$

$={\omega }^{100}+{\omega }^{200}={\omega }^{99}\cdot \omega +{\omega }^{198}\cdot {\omega }^2=1\cdot \omega +1\cdot {\omega }^2=\omega +{\omega }^2=-1$

Note: ${\omega }^{3n}=1,n\in N$ and $1+\omega +{\omega }^2=0$