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De-Moivre's Theorem

If 1,ω ,ω 2...

Question

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

A

7

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B

49

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C

64

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D

9

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Solution

The correct option is A 49
Let $ω$ being imaginary cube root of unity,
Given,
$ω^{3} = 1 a n d 1 + ω + ω^{2} = 0$
The problem given is,
$= (2 - w) (2 - w^{2}) (2 - w^{10}) (2 - w^{11})$
$= (2 - w) (2 - w^{2}) (2 - {(w^{3})}^{3} \cdot w) (2 - {(w^{3})}^{3} \cdot w^{2})$
$= (2 - w) (2 - w^{2}) (2 - w) (2 - w^{2})$
$= {(4 - 2 w - 2 w^{2} + w^{3})}^{2}$
$= {(4 - 2 (w + w^{2}) + 1)}^{2}$
$= {(4 - 2 ((- 1)) + 1)}^{2}$
$= 7^{2}$
Hence, the given equation will be equal to $= 49$

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