Question

If ω is a cube root of unity , then the value of ω⁹⁹ +ω¹⁰⁰+ω¹⁰¹ is *

Answer 1

Answer:

The answer is 0

Step-by-step explanation:

Results: If $\omega$ is a complex cube roots of unity, then

                 $\boxed{1+\omega+\omega^2=0 \text{ and } \omega^3=1}$

Therefore

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

                $=\omega^{99}(1+\omega+\omega^2)\\=\omega^{99}(0)\\=0$$\boxed{\text{Using } 1+\omega+\omega^2=0}$