Cube roots of Unity MCQ Quiz in বাংলা - Objective Question with Answer for Cube roots of Unity - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Cube Roots of unity are 1, ω and ω²

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

Calculation:

Cube roots of unity are 1, \(\frac{{ - 1 + \sqrt 3 {\rm{i}}}}{2}{\rm{\;}}\& {\rm{\;}}\frac{{ - 1 - \sqrt 3 {\rm{i}}}}{2}\)

So, \({{\rm{\omega }}_1} = \frac{{ - 1 + \sqrt 3 {\rm{i}}}}{2}\:and\:{{\rm{\omega }}_2} = \frac{{ - 1 - \sqrt 3 {\rm{i}}}}{2}\)

\({\left( {{{\rm{\omega }}_1} - {{\rm{\omega }}_2}} \right)^2} = {\left( {\frac{{ - 1 + \sqrt 3 {\rm{i}}}}{2} - \frac{{ - 1 - \sqrt 3 {\rm{i}}}}{2}} \right)^2}\)

\(= {\left( {\sqrt 3 {\rm{i}}} \right)^2}\)

= 3(i)²

= -3

Concept:

ω is a cube root of unity.

Property of cube root of unity:

ω3 = 1, and 1 + ω + ω2  = 0

Calculation:

Let,

Z = (1 - ω + ω²) (1 + ω - ω²)      ----(1)

We know that,

1 + ω + ω2  = 0

⇒ 1  + ω2  = - ω and 1 + ω = - ω2  

⇒ Z = (-2ω) (-2ω2)

⇒ Z = 4ω³ = 4    (∵ ω3 = 1)

Hence, option 4 is correct.

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:

Let the required value be x.

⇒ x = (1 - ω + ω2)5  + (1 + ω - ω2)5 - 32

⇒ x = (- ω - ω)5  + (- ω² - ω2)5 - 32

⇒  x = (-2ω)5  + (-2ω2)5 - 32

⇒  x = - 32ω5  + (-32ω¹⁰) - 32

⇒  x = - 32(ω²  + ω) - 32       (∵  ω³ = 1)

⇒ x = - 32 (-1) - 32           (∵ ω2  + ω = -1)

⇒  x =  32 - 32     

⇒ x = 0

∴ (1 - ω + ω2)5  + (1 + ω - ω2)5 - 32 = 0

Concept:

Properties of cube roots of unity:

1 + ω + ω² = 0

ω³ = 1

Law's of exponent:

(a^m)ⁿ = a^mn

a^m+n = a^maⁿ

Calculation: 

To find: (1 + ω6)(ω + ω4)(1 + ω⁷)

ω⁶ = (ω³)² = 1, ω⁴ = ωω³ = ω, and ω⁷ = (ω3)² ω = ω

So,(1 + ω6)(ω + ω4)(1 + ω7)

= (1 + 1)(ω + ω)(1 + ω)

= 4ω(- ω²)                                    (∵ 1 + ω + ω2 = 0, ∴  1 + ω  = -ω2)

= -4ω³

= -4(1)                                          (∵ ω³ = 1)

= -4

Concept:

1. Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Where a, b and c are constants with a ≠ 0

• \({\rm{x}} = {\rm{\;}}\frac{{ - {\rm{b\;}} \pm {\rm{\;}}\sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}\)

2. x³ – y³ = (x – y) (x² + y² + xy)

3. Cube root of unity:

• \({\rm{\omega }} = {\rm{\;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{\;}}\)
• \({{\rm{\omega }}^2} = \frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\)
• ω³ = 1

Calculation:

We know that x³ – y³ = (x – y) (x² + xy + y²)

∴ (x³ – 1) = (x – 1) (x² + x + 1)       …. (1)

Now, Factor of (x² + x + 1)

\({\rm{x}} = {\rm{\;}}\frac{{ - 1{\rm{\;}} \pm {\rm{\;}}\sqrt {{1^2} - 4{\rm{\;}} \times 1{\rm{\;}} \times 1} }}{{2{\rm{\;}} \times 1}} = {\rm{\;}}\frac{{ - 1{\rm{\;}} \pm {\rm{i}}\sqrt 3 }}{2}\)

\(\Rightarrow {\rm{x}} = {\rm{\;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{\;or\;\;}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\)

⇒ x = ω or ω²

∴ (x² + x + 1) = (x – ω) (x – ω²)

Putting the value in equation 1,

(x³ – 1) = (x – 1) (x² + x + 1)

⇒ (x³ – 1) = (x – 1) (x – ω) (x – ω²)

Concept:

The cube-root of unity 'ω' follows:

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

Calculation:

S = ω¹⁷

S = ω^{3 × 5} .ω²

S = 1 × ω² = ω2

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:

ω15 + ω20+ ω25

= ω¹5 (1+ ω⁵ +ω¹⁰)

= ω¹5 × (1 + ω³.ω² +ω9.ω)

= (ω³)⁵ × (1 + ω² + ω)

= 1 × (1 + ω + ω2)

= 1× 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
• ω3n = 1

Calculation:

We have to find the value of ω3n + 4

ω3n + 4 = ω3n + 3 × ω 

= ω3(n + 1) × ω 

= (1)^{(n + 1)} × ω 

⇒ 1 × ω 

⇒ ω 

Concept:

Properties of cube root of unity:

• 1 + ω + ω² = 0
• ω³ = 1
• ω3n = 1
•  

Calculation:

\(\rm \frac{{a + bω + c{ω ^2}}}{{c + aω + b{ω ^2}}} + \frac{{a + bω + c{ω ^2}}}{{b + cω + a{ω ^2}}}\)

\(\rm = \left( {\frac{{a + bω + c{ω ^2}}}{{c + aω + b{ω ^2}}} \times \frac{{{ω ^2}}}{{{ω ^2}}}} \right) + \left( {\frac{{a + bω + c{ω ^2}}}{{b + cω + a{ω ^2}}} \times \frac{ω }{ω }} \right)\)

\(\rm = {ω ^2}\left( {\frac{{a + bω + c{ω ^2}}}{{c{ω ^2} + a{ω ^3} + b{ω ^4}}}} \right) + ω \left( {\frac{{a + bω + c{ω ^2}}}{{bω + c{ω ^2} + a{ω ^3}}}} \right)\)           (∵ ω3 = 1 and ω⁴ = ω)

\(\rm = {ω ^2}\left( {\frac{{a + bω + c{ω ^2}}}{{a + bω + c{ω ^2}}}} \right) + ω \left( {\frac{{a + bω + c{ω ^2}}}{{a + bω + c{ω ^2}}}} \right)\)

= ω² + ω

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.