The value of 1-21-2-41-4-81-8-16 is- where - is the cube root of unity

(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, nbsp; =( ω ω)( ω^2 ω^2)( ω ω)( ω^2 ω^2) =( 2ω)( 2ω ...

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Standard XII

Mathematics

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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Similar questions

Q. If

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}) =

Q. If 1,

ω

and

ω^{2}

are the cube roots of unity, evaluate

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

Q. lf

1, ω, ω^{2}

are cube roots of unity then the value

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

(upto 2n factors) is?

Q. If

1, ω, ω^{2}

are the three cube roots of unity,

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

...to 2n factors

=

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Cube Root of a Complex Number

Standard XII Mathematics

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The value of (1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8)(1−ω8+ω16) is, where ω is the cube root of unity

(1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8)(1−ω8+ω16) [∵ω3=1] =(1−ω+ω2)(1−ω2+ω)(1−ω+ω2)(1−ω2+ω) Using 1+ω+ω2=0, =(−ω−ω)(−ω2−ω2)(−ω−ω)(−ω2−ω2) =(−2ω)(−2ω2)(−2ω)(−2ω2) =16ω6=16 [∵ω3=1]

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