Question

If $\omega$ is a non-real complex cube root of unity, then $(\omega +{\omega }^2+{\omega }^3+...+{\omega }^{100}{)}^2+1$ is equal to

• A. ${\omega }^2$
• B. $\omega$
• C. $-\omega$
• D. $-{\omega }^2$

Answer 1

The correct option is D $-{\omega }^2$

According to question

$(\omega +{\omega }^2+{\omega }^3+...+{\omega }^{100}{)}^2+1$

Solving G.P,

$⟹\frac{\omega (\omega -1)}{{\omega }^{100}-1}+1$

$⟹\frac{\omega (\omega -1)}{\omega -1}+1$ ..........as $({\omega }^{99}={\omega }^3=1)$

$⟹\omega +1$ ............. as $({\omega }^2+\omega +1=0)$

$⟹-{\omega }^2$