Question

Find the value of expression ${\omega ^{99}} + {\omega ^{100}} + {\omega ^{101}}$ if $\omega $ is a complex cube root of unity ?
A) 1
B) -1
C) 3
D) 0

Answer 1

Hint: To obtain the answer we need to understand the definition of the complex cube root of unity, which is defined as the numbers which when raised to the power of 3 gives the result as 1 . After taking out the common part from the question , use ${\omega ^2} + \omega + 1 = 0$ to get the desired answer.

Complete step-by-step answer:
Since $\omega $ is the cube root of unity
$ \Rightarrow {\omega ^3} = 1$
$ \Rightarrow {\omega ^3} - 1 = 0$
$ \Rightarrow \left( {\omega - 1} \right)\left( {{\omega ^2} + \omega + 1} \right) = 0$ ( Since ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$)
$\therefore \omega - 1 = 0,{\omega ^2} + \omega + 1 = 0$
Consider ${\omega ^{99}} + {\omega ^{100}} + {\omega ^{101}}$=${\omega ^{99}}\left( {1 + \omega + {\omega ^2}} \right)$
$ = {1^{99}}\left( {1 + \omega + {\omega ^2}} \right) = 1 \times 0$ ( from above )
$ = 0$

Note: In such types of questions students can do two mistakes first is wrong substitution in the formula ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ and second is substitution of value $\omega = 1$ in expression.