Question

If $w$ is a complex cube root of unity, then the value of $w^{99}+w^{100}+w^{101}$ is :

• A. $1$
• B. $-1$
• C. $3$
• D. $0$

Answer 1

The correct option is D $0$

$\omega$ is a complex cube root of unity.

$\Rightarrow \omega =\sqrt[3]{31}$

$\Rightarrow {\omega }^3=1$

$\Rightarrow {\omega }^3-1=0$

$\Rightarrow (\omega -1)({\omega }^2+\omega +1)=0$

$\therefore \omega -1=0,{\omega }^2+\omega +1=0$ .......$\left(1\right)$

Consider ${\omega }^{99}+{\omega }^{100}+{\omega }^{101}$

$={\omega }^{99}(1+\omega +{\omega }^2)$

$=1^{99}\times 0$ from $\left(1\right)$

$\therefore {\omega }^{99}+{\omega }^{100}+{\omega }^{101}=0$