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Cube Root of a Complex Number

The value of ...

Question

The value of
$(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16})$ is, where $ω$ is the cube root of unity

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Solution

$(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16})$
$[∵ ω^{3} = 1]$
$= (1 - ω + ω^{2}) (1 - ω^{2} + ω) (1 - ω + ω^{2}) (1 - ω^{2} + ω)$
Using $1 + ω + ω^{2} = 0,$
$= (- ω - ω) (- ω^{2} - ω^{2}) (- ω - ω) (- ω^{2} - ω^{2})$
$= (- 2 ω) (- 2 ω^{2}) (- 2 ω) (- 2 ω^{2})$
$= 16 ω^{6} = 16 [∵ ω^{3} = 1]$

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ω

is a complex cube root of unity then

(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16})

(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16}) =

ω

ω^{2}

are the cube roots of unity, evaluate

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are cube roots of unity then the value

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(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8})

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=

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Cube Root of a Complex Number

Standard XII Mathematics

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