Roots of Unity MCQ Quiz - Objective Question with Answer for Roots of Unity - Download Free PDF

Concept

If ω  is an imaginary cube root of unity, we have the following properties: 

• ω³  = 1
• 1 + ω + ω² = 0 ⇒ ω² = -1 - ω  

Calculation

Given that (1 + ω - ω2)5

Now, let's simplify the expression (1 + ω - ω²)

⇒ 1 + ω - (-1 - ω) = 1 + ω + 1 + ω  = 2 + 2ω  = 2(1 + ω) 

Next, we need to evaluate (1 + ω - ω²)⁵

(1 + ω - ω²)⁵ = (2(1 + ω))⁵ 

We know that 1 + ω + ω² = 0,

So 1 + ω = -ω²

Substitute back into the expression

⇒ 

Since 

∴ -32ω¹⁰ = -32ω 

∴ Option 3 is correct 

Explanation:

(1+ω +ω²)¹⁰⁰+ (1-ω +ω2)100

= (– ω² – ω² )¹⁰⁰ + (–ω – ω)¹⁰⁰  {∵ 1 + ω  + ω² = 0}

= 2¹⁰⁰(ω²⁰⁰ + ω¹⁰⁰)

= 2¹⁰⁰(ω² + ω)     (∵ ω³ = 1)

= –2¹⁰⁰ 

∴ Option (d) is correct

a, b ∈ I, –3 ≤ a, b ≤ 3, a + b ≠ 0

|z – a| = |z + b|

⇒ z³ = 1

⇒ z = ω , ω², 1

Now

|1 - a| = |1 + b|

⇒ 10 pair

Concept

If ω  is an imaginary cube root of unity, we have the following properties: 

• ω³  = 1
• 1 + ω + ω² = 0 ⇒ ω² = -1 - ω  

Calculation

Given that (1 + ω - ω2)5

Now, let's simplify the expression (1 + ω - ω²)

⇒ 1 + ω - (-1 - ω) = 1 + ω + 1 + ω  = 2 + 2ω  = 2(1 + ω) 

Next, we need to evaluate (1 + ω - ω²)⁵

(1 + ω - ω²)⁵ = (2(1 + ω))⁵ 

We know that 1 + ω + ω² = 0,

So 1 + ω = -ω²

Substitute back into the expression

⇒ 

Since 

∴ -32ω¹⁰ = -32ω 

∴ Option 3 is correct 

Concept:

Cube roots of Unity

Cube roots of unity are given by 1, ω, ω², where

ω = , ω² = 

Some Results Involving Complex Cube Root of Unity (ω)

(i) ω³ = 1

(ii) 1 + ω + ω² = 0

(iii) ω2 = 

(iv) ω̅  = ω2

Calculation:

Statement I: roots are -2, 1 - 3ω, 1 - 3ω2

(z -1)³ = - 27

⇒ (z -1) = (- 27)^1/3 

⇒ (z -1) = - 3 ⋅ (1)^1/3 

⇒ z  =  - 3 ⋅ (1) ^1/3  + 1

⇒ z  =  1 - 3 ⋅ (1) ^1/3 

Cube root of unity are 1, ω, ω²

For 1, z  =  1 - 3 ⋅ 1 ⇒ z  =  1 - 3 =  - 2

For ω, z  =  1 - 3 ⋅ ω ⇒ z  =  1 - 3ω 

For ω², z  =  1 - 3 ⋅ ω² ⇒ z  =  1 - 3ω² 

Statement I is correct.

Statement II: Sum of roots is 0

Roots of the equations (z -1)3 + 27 = 0 are - 2, 1 - 3ω, 1 - 3ω2 

Sum of roots = (- 2) + (1 - 3ω) + (1 - 3ω2) 

 = - 3ω - 3ω2 ⇒ - 3(ω + ω2) = -3(-1) = 3

Statement II is incorrect.

Statement III: Product of roots is 1

Product of roots = (-2) (1 - 3ω)(1 - 3ω2)

 = (-2) (1 - 3ω² - 3ω + 9ω³)

 = (-2)(10 - 3(ω2 + ω))

 = (-2)(10 + 3)

 = -26

Statement III is correct.

∴ I and III are correct.

Concept:

Cube Roots of unity are 1, ω and ω2

Here, ω =  and ω2 = 

Property of cube roots of unity

• ω3 = 1
• 1 + ω + ω2 = 0
• ω3n = 1

Calculation:

ω6 +  ω7 + ω⁵

= ω⁵ (ω + ω² + 1)

= ω⁵ × (1 + ω + ω2)

= ω5 × 0

= 0

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω =   and ω² = 

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ω
• ω³ⁿ = 1

Calculation:

We have to find the value of ω3n + ω3n+1 + ω³ⁿ⁺²

⇒ ω3n + ω3n+1 + ω³ⁿ⁺² 

=  ω3n (1 + ω + ω²)           (∵ 1 + ω + ω2 = 0)

= 1 × 0 = 0

Concept 

Cube Roots of unity are 1, ω and ω2

Here, ω =  and ω2 = 

Property of cube roots of unity:

• ω3 = 1
• 1 + ω + ω2 = 0
• ω = 1 / ω 2 and ω2 = 1 / ω
• ω3n = 1

Calculation:

Given that,

(x - 1)3 + 8 = 0

⇒ (x - 1)³ = (-2)³

⇒ (x - 1) = -2(1)^1/3

(x - 1) = -2(1, ω,  ω2)

⇒ x = -1, 1 - 2ω, 1 - 2ω²  

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω =  and ω² = 

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ω
• ω³ⁿ = 1

Calculation:

We have to find the value of ω¹⁰⁰ + ω²⁰⁰ + ω³⁰⁰

⇒ ω¹⁰⁰ + ω²⁰⁰ + ω³⁰⁰

= ω⁹⁹ ω + ω¹⁹⁸ ω² + (ω³)¹⁰⁰

= (ω³)³³ ω + (ω³)⁶⁶ ω² + 1     (∵ ω³ = 1)

= ω + ω² + 1

= 0                       (∵ 1 + ω + ω² = 0)

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω =   and ω² = 

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ωsum
• ω³ⁿ = 1

Calculation:

We have to find the value of 

As we know that 1 + ω + ω² = 0

⇒ 1 + ω² = - ω and 1 + ω = - ω²

Now,

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω =    and ω² = 

Some properties of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0

Calculation:

Given expression is  ( -1 + i√3 )48 

Since,  ω = 

∴ 2ω = -1 + i√3 

∴ (2ω)⁴⁸ = (-1 + i√3 )⁴⁸ 

∴ (-1 + i√3 )⁴⁸ = 2⁴⁸(ω³)¹⁶

∴ (-1 + i√3 )⁴⁸ = 2⁴⁸, as  ω³ = 1

Concept:

1 + ω + ω² = 0

 ω3 = 1

Calculation:

 (1 - ω + ω2)2 + (1 - ω2 + ω)2

⇒ {(1 + ω2) - ω}² + {(1+ ω) - ω²}²

⇒ (- ω - ω)² + (-ω² - ω2 )² ,since 1 + ω + ω2 = 0 

⇒ 4ω² + 4ω⁴

⇒ 4(ω² + ω) since, ω³ = 1

⇒ - 4

Concept 

Cube Roots of unity are 1, ω and ω²

Here, ω =  and ω² = 

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ω
• ω³ⁿ = 1

Calculation:

Given: α, β, γ are the cube roots

Let z³ = p (p

Let p  = - q where q > 0

⇒  z3 = -q 

So, the roots of the above equation will be, 

Assume α = q, β = qω, γ = qω²

(Because the magnitude of each root will be q)

Now,

            (∵ ω³ = 1)

First of all, we should not think that because of the 4 roots, the degree of the polynomial will be 4.

We need to remember that complex roots always occur in pairs.

Now,

Now, to find the value of the given pair roots,

1) 2ω² = -1 - √3 i

∴ -1 + √3i will also be a root.

2) 3 + 4ω = 3 - 2 + 2√3i

= 1 + 2√3i

∴ 1 - 2√3i will also be a root.

3) 3 + 4ω² = 3 - 2 - 2√3i

= 1 - 2√3i

i.e, 3 + 4ω and 3 + 4ω² are conjugate pairs.

= 6

i.e. 6 is a root.

Total we have 5 roots.

The minimum degree of the polynomial is 5.

Concept:

• ω =  and ω² =  where ω is cube root of unity.

Calculation:

Given that,

 where n is not a multiple of 3.

Here n is not a multiple of 3 so,

Put n = 3k + 1 where k is integer k =1, 2, 3 …..

⇒ -1