Question

ii) If \( \omega \) be the imaginary cube root of unity then the value of \( \omega^{3 n}+\omega^{3 n+1}+\omega^{3 n+2} \) (where \( n \in N \) ) is a) 0 b) \( -1 \) d) \( \omega \).

Answer 1

\(\omega^3\) = 1

(∵ ω = 1^1/3 & ω be imaginary (given))

It means ω is a root of equation x³ - 1 = 0.

It is cleared that 1 is also a root of equation x³ - 1 = 0.

Now,

(ω²)³-1 = ω⁶ - 1 

= (ω³)² -1

= 1² - 1

= 1-1 

= 0

Hence,

I, ω & ω² are roots of equation x³ - 1 = 0.

∴ Sum of roots = 1 + ω + ω² 

= \(\frac{-b}{a}\) = \(\frac{-0}{1}\) 

= 0.

i.e., 1 + ω + ω² = 0

Now,

ω³ⁿ + ω³ⁿ⁺¹ + ω³ⁿ⁺²

= (ω³)ⁿ + (ω³)ⁿ.ω + (ω³)ⁿ.ω²

= 1ⁿ + 1ⁿ.ω + 1ⁿ ω²

= 1 + ω + ω² 

= 0.

Hence,

ω³ⁿ  + ω³ⁿ⁺¹ + ω³ⁿ⁺² = 0