MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

49

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $= 49$

Suggest Corrections

thumbs-up

0

Similar questions

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})

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

1, ω, ω^{2}

are the cube roots of unity then

(2 - ω) (2 - ω^{2}) (2 - ω^{13}) (2 - ω^{17})

1, ω, ω^{2}

are the cube roots of unity, then the value of

(3 + ω) (3 + ω^{2}) (3 + ω^{10}) (3 + ω^{11})

1, ω, ω^{2}

are three cube roots of unity, then

(1 - ω + ω^{2}) (1 + ω - ω^{2})

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

De-Moivre's Theorem

MATHEMATICS

Watch in App

explore_more_icon

Explore more

De-Moivre's Theorem

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon