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

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

Since ${\omega }^3=1⟹{\omega }^9=1⟹{\omega }^{10}=\omega$

∴ -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

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

Since ${\omega }^3=1⟹{\omega }^9=1⟹{\omega }^{10}=\omega$

∴ -32ω¹⁰ = -32ω 

∴ Option 3 is correct 

Concept:

Cube roots of Unity

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

ω = $\frac{-1+\sqrt{3}}{2}$, ω² = $\frac{-1-\sqrt{3}}{2}$

Some Results Involving Complex Cube Root of Unity (ω)

(i) ω³ = 1

(ii) 1 + ω + ω² = 0

(iii) ω2 = $\frac{1}{\omega }$

(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 = $\frac{-1+\mathrm{i}\sqrt{3}}{2}$ and ω2 = $\frac{-1-\mathrm{i}\sqrt{3}}{2}$

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 =  $\frac{-1+\mathrm{i}\sqrt{3}}{2}$ and ω² = $\frac{-1-\mathrm{i}\sqrt{3}}{2}$

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 = $\frac{-1+\mathrm{i}\sqrt{3}}{2}$ and ω² = $\frac{-1-\mathrm{i}\sqrt{3}}{2}$

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 = $\frac{-1+\mathrm{i}\sqrt{3}}{2}$  and ω² = $\frac{-1-\mathrm{i}\sqrt{3}}{2}$

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 $\sqrt{\frac{1+{\omega }^2}{1+\omega }}$

As we know that 1 + ω + ω² = 0

⇒ 1 + ω² = - ω and 1 + ω = - ω²

Now,

$\sqrt{\frac{1+{\omega }^2}{1+\omega }}=\sqrt{\frac{-\omega }{-{\omega }^2}}=\sqrt{\frac{1}{\omega }\times \frac{{\omega }^2}{{\omega }^2}}=\sqrt{\frac{{\omega }^2}{{\omega }^3}}=\sqrt{{\omega }^2}=\omega$

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω = $\frac{(-1+\mathrm{i}\surd 3)}{2}$   and ω² = $\frac{(-1-\mathrm{i}\surd 3)}{2}$

Some properties of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0

Calculation:

Given expression is  ( -1 + i√3 )48 

Since,  ω = $\frac{(-1+\mathrm{i}\surd 3)}{2}$

∴ 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,

$\frac{\mathrm{x}\alpha +\mathrm{y}\beta +\mathrm{z}\gamma }{\mathrm{x}\beta +\mathrm{y}\gamma +\mathrm{z}\alpha }$

$=\frac{\mathrm{p}(\mathrm{x}+\mathrm{y}\omega +\mathrm{z}{\omega }^2)}{\mathrm{p}(\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z})}$

$=\frac{\mathrm{x}+\mathrm{y}\omega +\mathrm{z}{\omega }^2}{\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z}}\times \frac{\omega }{\omega }$

$=\frac{1}{\omega }\times \frac{\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z}{\omega }^3}{\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z}}$

$=\frac{1}{\omega }\times \frac{\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z}}{\mathrm{x}\omega +\mathrm{y}{\omega }^2+\mathrm{z}}$            (∵ ω³ = 1)

$$\begin{gathered} =\frac{1}{\omega } \\ ={\omega }^2 \end{gathered}$$

$=\frac{-1-i\sqrt{3}}{2}$

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,

$\omega =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$

$=-\frac{1}{2}+\frac{\surd 3}{2}i$

${\omega }^2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}$

$=-\frac{1}{2}-\frac{\surd 3}{2}i$

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

Calculation:

Given that,

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

$\Rightarrow {\omega }^{\mathrm{n}}+{\omega }^{2\mathrm{n}}\left[\omega =\frac{-1+\mathrm{i}\sqrt{3,}}{2},{\omega }^2=\frac{-1-\mathrm{i}\sqrt{3}}{2}\right]$

Here n is not a multiple of 3 so,

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

$\Rightarrow {\omega }^{3\mathrm{k}+1}+{\omega }^{6\mathrm{k}+2}$

$\Rightarrow {\omega }^{3\mathrm{k}}.{\omega }^1+{\omega }^{6\mathrm{k}}.{\omega }^2\left[{\because \omega }^{3\mathrm{k}}=1\right]$

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