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) \( \omega \).
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) \( \omega \).
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\(\omega^3\) = 1
(∵ ω = 11/3 & 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
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