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nth Root of a Complex Number

If 1, ω, ω2...

Question

If $1, ω, ω^{2}$ are the cube roots of unity, then prove that $(2 - ω) (2 - ω^{2}) (2 - ω^{10}) (2 - ω^{11}) = 49$

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Solution

Given $1, ω, ω^{2}$ are the cube roots of unity
$⟹ 1 + ω + ω^{2} = 0, ω^{3} = 1$
$L H S = (2 - ω) (2 - ω^{2}) (2 - ω^{10}) (2 - ω^{11}) = (2 - ω) (2 - ω^{2}) (2 - (ω^{3})^{3} ω) (2 - (ω^{3})^{3} ω^{2}) = ((2 - ω) (2 - ω)^{2})^{2}$
$= (2^{2} - 2 (ω + ω^{2}) + ω^{3})^{2} = (4 + 2 + 1)^{2} = 7^{2} = 49 = R H S$
Hence proved

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