Correct Answer - Option 2 : -1
Concept:
Properties of cube root of unity:
ω3 = 1 and (ω2 + ω + 1) = 0
Calculations:
Consider, ω10 + ω-10
= ω10 + \(\dfrac {1}{ω^{10}}\)
= \(\dfrac {ω^{20}+1}{ω^{10}}\)
= \(\dfrac {ω^{18}.ω^{2}+1}{ω^{9.}ω}\)
= \(\dfrac {{(ω^3)^6}.ω^{2}+1}{(ω^3)^3ω}\)
= \(\dfrac {{(1)^6}.ω^{2}+1}{(1)^3ω}\) (∵ ω3 = 1)
= \(\dfrac {ω^{2}+1}{ω}\)
From equation (1), we have
= \(\dfrac {-ω}{ω}\) (∵ 1 + ω2 = -ω)
= - 1
Hence, if ω is a complex cube root of unity, then ω10 + ω-10 equal to -1.
