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

Question

Dr. Abubacker · Accepted Answer

(d)
$\omega^{99} + \omega^{100} + \omega^{101} = \left( \omega^{3} \right)^{33} + \left( \omega^{3} \right)^{33}\omega + \left( \omega^{3} \right)^{33}\omega^{2}$
$= 1 + \omega + \omega^{2} = 0$

Q. If $\omega$ is a complex cube root of unity, then the value of $\omega^{99} + \omega^{100} + \omega^{101}$ is

Explanation

Reviewed by Dr. Abubacker

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