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

⇒ \((2 \cdot (-ω^2))^5 =(-2ω^2)^5 =((-2)^5 \cdot (ω^2)^5) = -32 \cdot ω^{10} \)

Since \(ω^3 = 1 \implies ω^9 = 1\implies ω^{10} = ω \)

∴ -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|

\(\left|\begin{array}{ccc} z+1 & \omega & \omega^{2} \\ \omega & z+\omega^{2} & 1 \\ \omega^{2} & 1 & z+\omega \end{array}\right|=1 \)

\(\Rightarrow\left|\begin{array}{ccc} z & z & z \\ \omega & z+\omega^{2} & 1 \\ \omega^{2} & 1 & z+\omega \end{array}\right|=1\)

\( \Rightarrow z\left|\begin{array}{ccc} 1 & 1 & 1 \\ \omega & z+\omega^{2} & 1 \\ \omega^{2} & 1 & z+\omega \end{array}\right|=1\)

\(\Rightarrow z\left|\begin{array}{ccc} 1 & 0 & 0 \\ \omega & z+\omega^{2}-\omega & 1-\omega \\ \omega^{2} & 1-\omega^{2} & z+\omega-\omega^{2} \end{array}\right|=1\)

⇒ 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

⇒ \((2 \cdot (-ω^2))^5 =(-2ω^2)^5 =((-2)^5 \cdot (ω^2)^5) = -32 \cdot ω^{10} \)

Since \(ω^3 = 1 \implies ω^9 = 1\implies ω^{10} = ω \)

∴ -32ω¹⁰ = -32ω 

∴ Option 3 is correct 

Concept:

Cube roots of Unity

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

ω = \(\frac{-1+\sqrt3}{2}\), ω² = \(\frac{-1-\sqrt3}{2}\)

Some Results Involving Complex Cube Root of Unity (ω)

(i) ω³ = 1

(ii) 1 + ω + ω² = 0

(iii) ω2 = \(\frac{1}{ω}\)

(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 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;i}}\sqrt 3 }}{2}\) and ω2 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;i}}\sqrt 3 }}{2}\)

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 =  \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\) and ω² = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\)

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 = \(\frac{{ - \;1\; + \;i\sqrt 3 }}{2}\) and ω2 = \(\frac{{ - \;1\; - \;i\sqrt 3 }}{2}\)

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 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\) and ω² = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\)

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 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\)  and ω² = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\)

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 \(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}}\)

As we know that 1 + ω + ω² = 0

⇒ 1 + ω² = - ω and 1 + ω = - ω²

Now,

\(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} = {\rm{\;}}\sqrt {\frac{{ - {\rm{\omega }}}}{{ - {{\rm{\omega }}^2}}}} = {\rm{\;}}\sqrt {\frac{1}{{\rm{\omega }}} \times \frac{{{{\rm{\omega }}^2}}}{{{{\rm{\omega }}^2}}}} = \sqrt {\frac{{{{\rm{\omega }}^2}}}{{{{\rm{\omega }}^3}}}} = \sqrt {{{\rm{\omega }}^2}} = {\rm{\;}}{{\rm{\omega }}}\)

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω = \(\rm(- 1 + i√3)\over2\)   and ω² = \(\rm(- 1 - i√3)\over2\)

Some properties of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0

Calculation:

Given expression is  ( -1 + i√3 )48 

Since,  ω = \(\rm(- 1 + i√3)\over2\)

∴ 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 = \(\frac{{ - \;1\; + \;i\sqrt 3 }}{2}\) and ω² = \(\frac{{ - \;1\; - \;i\sqrt 3 }}{2}\)

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 < 0)

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,

\(\rm \frac {x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha}\)

\(=\rm \frac {p(x + yω+ zω^2)}{p(xω + yω^2 + z)}\)

\(=\rm \frac {x + yω+ zω^2}{xω + yω^2 + z} \times \frac{ω}{ω}\\\)

\(=\rm \frac{1}{ω} \times \frac {xω + yω^2 + zω^3}{xω + yω^2 + z}\)

\(=\rm \frac{1}{ω} \times \frac {xω + yω^2 + z}{xω + yω^2 + z}\)            (∵ ω³ = 1)

\(=\rm \frac{1}{\omega} \\ = \rm \omega^2\)

\(=\frac{{ - \;1\; - \;i\sqrt 3 }}{2}\)

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,

\(ω = {\cos \frac{2\pi}{3}}+{i \sin \frac{2\pi}{3}}\)

\(= - \frac{1}{2} + \frac{√ 3}{2} i\)

\(ω^2 = \cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}\)

\(= - \frac{1}{2} - \frac{√ 3}{2} i\)

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.

\(4) ~5 - \omega - \omega^2 = 5 + \frac{1}{2} - \frac{\sqrt 3}{2}i + \frac{1}{2} + \frac{\sqrt 3}{2}i\)

= 6

i.e. 6 is a root.

Total we have 5 roots.

The minimum degree of the polynomial is 5.

Concept:

• ω = \(\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}\) and ω² = \(\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\) where ω is cube root of unity.
• \({{\rm{\omega }}^{3{\rm{k}}}} = 1,{\rm{\;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 1} \right)}} = {\rm{\;\omega \;and\;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 2} \right)}} = {\rm{\;}}{{\rm{\omega }}^2}\)
• \(1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0\)

Calculation:

Given that,

\({\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}}\) where n is not a multiple of 3.

\(\Rightarrow {{\rm{\omega }}^{\rm{n}}} + {\rm{}}{{\rm{\omega }}^{2{\rm{n}}}}{\rm{}}\left[ {{\rm{}}{\rm{\omega }} = {\rm{}}\frac{{ - 1 + {\rm{i}}\sqrt {3,} }}{2},{\rm{\;}}{{\rm{\omega }}^2}{\rm{}} = {\rm{}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}{\rm{}}} \right]\)

Here n is not a multiple of 3 so,

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

\(\Rightarrow {{\rm{\omega }}^{3{\rm{k}} + 1}} + {{\rm{\omega }}^{6{\rm{k}} + 2}}\)

\(\Rightarrow {{\rm{\omega }}^{3{\rm{k}}}}.{{\rm{\omega }}^1} + {{\rm{\omega }}^{6{\rm{k}}}}.{{\rm{\omega }}^2}\left[ {{{\because\rm{\omega }}^{3{\rm{k}}}} = 1} \right]\)

\( \Rightarrow {\rm{\omega }} + {{\rm{\omega }}^2}\left[ {\because 1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0} \right]\)