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

Concept :

Cube Roots of unity are 1, ω and ω2

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

Property of cube roots of unity:

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

Calculations :

Given that

 \(\left(\frac{i + \sqrt{3}}{-i + \sqrt{3}}\right)^{200}+ \left(\frac{i - \sqrt{3}}{-i + \sqrt{3}}\right)^{200}\)

Consider the first part of the given equation

⇒ \((\frac{i +√{3}}{-i+√{3}})\times(\frac{i +√{3}}{i+√{3}})\)

⇒ \(\frac{(i+√{3})^2}{(√{3}-i)(√{3}+i)}\)

We know that (a  - b)(a + b) = a2 - b2 and i² = √-1

⇒ \(\frac{2+2√{3}i}{4}\)    

⇒ \(\frac{1+√{3}}{2} = -(\frac{-1-√{3}}{2})=-ω^2\)       ----- (1)

Consider the second part of the given equation

⇒ \((\frac{i -√{3}}{i+√{3}})\times(\frac{i -√{3}}{i-√{3}})\)

⇒ \(\frac{(i-√{3})^2}{(i^2-(√{3})^2)} = \frac{-1+i√{3}}{2}=ω\)         ----- (2)

Using equations (1) and (2) in the given equation, we get

⇒ \((-ω^2)^{200} + ω^{200} +1\)

⇒ \(ω^{400}+ω^{200}+1\)

⇒ \(ω^{3\times133+1} + ω^{3\times66 +2}+1\)

We know that \(\omega^3 =1\)

⇒ \(\omega^2+\omega +1 = 0\)

∴ The value of \(\left(\frac{i + √{3}}{-1 + √{3}}\right)^{200}+ \left(\frac{i - √{3}}{-1 + √{3}}\right)^{200}+1\) is 0.

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

\(1 + {ω ^n} + {ω ^{2n}} = \;\left\{ {\begin{array}{*{20}{c}} {0,\;if\;n\;is\;not\;multiple\;of\;3}\\ {3,\;if\;n\;is\;multiple\;of\;3} \end{array}} \right.\)

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = \(\begin{bmatrix} 1 &ω & ω^{2n} \\ ω^2 & ω^{2n} & 1 \\ ω^{2n}& 1 & ω^{n} \end{bmatrix} \)

on solving determinant we get,

Δ = 1 (ω²ⁿ ωⁿ - 1) - ω (ω2 ωn - ω2n) + ω2n (ω2 - ω2n ω²n)

Δ = (ω³n - 1) - (ω3 ωn - ω2n ω) + (ω2 ω2n - ω⁶n )

Since ω3 = 1, ω3n = 1, ω6n = 1

Δ = 0 - ωn + ω2n ω + ω2 ω2n - 1

Δ = -1 - ωn +ω2n (ω + ω2)

Since 1 + ω + ω2 = 0 ⇒ ω + ω2 = - 1

Δ = -1 - ωn - ω2n 

If n is not a multiple of 3 then:

Δ = -1 - ωn - ω2n  = 0

Concept:

ω³ = 1 where ω is the cube root of unity.

ω³ - 1³ = 0

(ω - 1)(1 + ω + ω²) = 0

∴ ω = 1 and 1 + ω + ω² = 0

Calculation:

Given:

\(1\;+\;ω\;+\frac{1}{ω}\)

We know that;

1 + ω + ω2 = 0

\(1\;+\;ω\;+\frac{ω^2}{1}=0\)

\(1\;+\;ω\;+\frac{ω^2}{ω^3}=0\)      [∵ ω³ = 1]

\(\therefore 1\;+\;ω\;+\frac{1}{ω}=0\)

We know that \(1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0{\rm{\;and\;}}{{\rm{\omega }}^3} = 1\) then

\(\begin{array}{l} \left( {1 - {{\rm{\omega }}^8}} \right)\left( {1 - {{\rm{\omega }}^4}} \right)\left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right) = \left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right)\left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right)\\ = \left( {1 - {{\rm{\omega }}^2} - {\rm{\omega }} + {{\rm{\omega }}^3}{\rm{\;}}} \right)\left( {1 - {{\rm{\omega }}^2} - {\rm{\omega }} + {{\rm{\omega }}^3}} \right)\\ = \left( {2 - \left( {{\rm{\omega }} + {{\rm{\omega }}^2}} \right)} \right)\left( {2 - \left( {{\rm{\omega }} + {{\rm{\omega }}^2}} \right)} \right)\\ = \left( {2 - \left( { - 1} \right)} \right)\left( {2 - \left( { - 1} \right)} \right) = 9 \end{array}\)

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

\(1 + {ω ^n} + {ω ^{2n}} = \;\left\{ {\begin{array}{*{20}{c}} {0,\;if\;n\;is\;not\;multiple\;of\;3}\\ {3,\;if\;n\;is\;multiple\;of\;3} \end{array}} \right.\)

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = \(\begin{bmatrix} 1 &ω & ω^{2n} \\ ω^2 & ω^{2n} & 1 \\ ω^{2n}& 1 & ω^{n} \end{bmatrix} \)

on solving determinant we get,

Δ = 1 (ω²ⁿ ωⁿ - 1) - ω (ω2 ωn - ω2n) + ω2n (ω2 - ω2n ω²n)

Δ = (ω³n - 1) - (ω3 ωn - ω2n ω) + (ω2 ω2n - ω⁶n )

Since ω3 = 1, ω3n = 1, ω6n = 1

Δ = 0 - ωn + ω2n ω + ω2 ω2n - 1

Δ = -1 - ωn +ω2n (ω + ω2)

Since 1 + ω + ω2 = 0 ⇒ ω + ω2 = - 1

Δ = -1 - ωn - ω2n 

If n is not a multiple of 3 then:

Δ = -1 - ωn - ω2n  = 0

We know that \(1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0{\rm{\;and\;}}{{\rm{\omega }}^3} = 1\) then

\(\begin{array}{l} \left( {1 - {{\rm{\omega }}^8}} \right)\left( {1 - {{\rm{\omega }}^4}} \right)\left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right) = \left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right)\left( {1 - {{\rm{\omega }}^2}} \right)\left( {1 - {\rm{\omega }}} \right)\\ = \left( {1 - {{\rm{\omega }}^2} - {\rm{\omega }} + {{\rm{\omega }}^3}{\rm{\;}}} \right)\left( {1 - {{\rm{\omega }}^2} - {\rm{\omega }} + {{\rm{\omega }}^3}} \right)\\ = \left( {2 - \left( {{\rm{\omega }} + {{\rm{\omega }}^2}} \right)} \right)\left( {2 - \left( {{\rm{\omega }} + {{\rm{\omega }}^2}} \right)} \right)\\ = \left( {2 - \left( { - 1} \right)} \right)\left( {2 - \left( { - 1} \right)} \right) = 9 \end{array}\)

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

\(1 + {ω ^n} + {ω ^{2n}} = \;\left\{ {\begin{array}{*{20}{c}} {0,\;if\;n\;is\;not\;multiple\;of\;3}\\ {3,\;if\;n\;is\;multiple\;of\;3} \end{array}} \right.\)

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = \(\begin{bmatrix} 1 &ω & ω^{2n} \\ ω^2 & ω^{2n} & 1 \\ ω^{2n}& 1 & ω^{n} \end{bmatrix} \)

on solving determinant we get,

Δ = 1 (ω²ⁿ ωⁿ - 1) - ω (ω2 ωn - ω2n) + ω2n (ω2 - ω2n ω²n)

Δ = (ω³n - 1) - (ω3 ωn - ω2n ω) + (ω2 ω2n - ω⁶n )

Since ω3 = 1, ω3n = 1, ω6n = 1

Δ = 0 - ωn + ω2n ω + ω2 ω2n - 1

Δ = -1 - ωn +ω2n (ω + ω2)

Since 1 + ω + ω2 = 0 ⇒ ω + ω2 = - 1

Δ = -1 - ωn - ω2n 

If n is not a multiple of 3 then:

Δ = -1 - ωn - ω2n  = 0

Concept:

ω³ = 1 where ω is the cube root of unity.

ω³ - 1³ = 0

(ω - 1)(1 + ω + ω²) = 0

∴ ω = 1 and 1 + ω + ω² = 0

Calculation:

Given:

\(1\;+\;ω\;+\frac{1}{ω}\)

We know that;

1 + ω + ω2 = 0

\(1\;+\;ω\;+\frac{ω^2}{1}=0\)

\(1\;+\;ω\;+\frac{ω^2}{ω^3}=0\)      [∵ ω³ = 1]

\(\therefore 1\;+\;ω\;+\frac{1}{ω}=0\)

Concept :

Cube Roots of unity are 1, ω and ω2

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

Property of cube roots of unity:

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

Calculations :

Given that

 \(\left(\frac{i + \sqrt{3}}{-i + \sqrt{3}}\right)^{200}+ \left(\frac{i - \sqrt{3}}{-i + \sqrt{3}}\right)^{200}\)

Consider the first part of the given equation

⇒ \((\frac{i +√{3}}{-i+√{3}})\times(\frac{i +√{3}}{i+√{3}})\)

⇒ \(\frac{(i+√{3})^2}{(√{3}-i)(√{3}+i)}\)

We know that (a  - b)(a + b) = a2 - b2 and i² = √-1

⇒ \(\frac{2+2√{3}i}{4}\)    

⇒ \(\frac{1+√{3}}{2} = -(\frac{-1-√{3}}{2})=-ω^2\)       ----- (1)

Consider the second part of the given equation

⇒ \((\frac{i -√{3}}{i+√{3}})\times(\frac{i -√{3}}{i-√{3}})\)

⇒ \(\frac{(i-√{3})^2}{(i^2-(√{3})^2)} = \frac{-1+i√{3}}{2}=ω\)         ----- (2)

Using equations (1) and (2) in the given equation, we get

⇒ \((-ω^2)^{200} + ω^{200} +1\)

⇒ \(ω^{400}+ω^{200}+1\)

⇒ \(ω^{3\times133+1} + ω^{3\times66 +2}+1\)

We know that \(\omega^3 =1\)

⇒ \(\omega^2+\omega +1 = 0\)

∴ The value of \(\left(\frac{i + √{3}}{-1 + √{3}}\right)^{200}+ \left(\frac{i - √{3}}{-1 + √{3}}\right)^{200}+1\) is 0.

Concept:

1 + ω + ω² = 0

ω³ = 1

Calculation: 

We can write the given expression as

⇒ \(\frac{\omega( a \ + \ b \omega \ + \ c\omega^2)}{(c\omega \ + \ a \omega^2 \ + \ b)} + \frac{a \ + \ b \omega \ + \ c\omega^2}{b \ + \ c \omega \ + \ a\omega^2} \)

⇒ \(\frac{( a \ + \ b \omega \ + \ c\omega^2) (\omega + 1)}{(c\omega \ + \ a \omega^2 \ + \ b)} \)

⇒ \(\frac{( a \ + \ b \omega \ + \ c\omega^2) (-\omega^2)}{(c\omega \ + \ a \omega^2 \ + \ b)} \)

⇒ \(-\frac{( a\omega^2 \ + \ b \ + \ c\omega)}{(c\omega \ + \ a \omega^2 \ + \ b)} \)

⇒ -1

∴ The required value is -1.

Explanation:

Given:

P(x) = ƒ(x³) + xg(x³)...(1)

Roots of x² + x + 1 are ω and ω² as ω² + ω + 1 = 0 

where ω is the cube root of unity

Now P(x) is divisible by x2 + x + 1 ⇒ P(x) = Q(x)(x2 + x + 1)

So P(ω) = P(ω2) = 0 --(2) ---- (Since roots of x2 + x + 1 are ω and ω2)

Now put x = ω and ω2 in equation (1)

P (ω) = f (1) + ω g (1) = 0 …(3) ------(Since ω³ = 1)

P (ω²) = f (1) + ω² g (1) = 0 …(4)

Subtracting (3) and (4)

ω g (1) - ω2 g (1) = P (ω) - P (ω2)

from (2)

(ω - ω2) g (1) = 0

As ω - ω2 ≠ 0 so  g (1) = 0 -----(5)

Adding (3) and (4)

2 f (1) – g (1) = P(ω) + P(ω2) -----(Since ω2 + ω = -1)

From (2)

g (1) = 2 f (1)

From (5) we get

f (1) = g (1) = 0

Therefore , P (1) = f (1) + g (1)

= 0 + 0

= 0

\(\rm {\left( {\cos 2{\rm{\theta }} + {\rm{i}}\sin 2{\rm{\theta }}} \right)^{ - 3}}{\left( {\cos 3{\rm{\theta }} - {\rm{i}}\sin 3{\rm{\theta }}} \right)^{ - 2}}{\left( {\sin {\rm{\theta }} - {\rm{i}}\cos {\rm{\theta }}} \right)^{ - 3}}\)

\(\rm = {\left( {{{\rm{e}}^{{\rm{i}}2{\rm{\theta }}}}} \right)^{ - 3}}.{\left( {{{\rm{e}}^{ - {\rm{i}}3{\rm{\theta }}}}} \right)^{ - 2}}.{\left( -i\cdot {{{\rm{e}}^{ {\rm{i\theta }}}}} \right)^{ - 3}}\)

\(\rm = {{\rm{e}}^{ - {\rm{i}}6{\rm{\theta }}}}.{{\rm{e}}^{{\rm{i}}6{\rm{\theta }}}}\cdot (-i)\cdot{{\rm{e}}^{{\rm{-i}}3{\rm{\theta }}}} =(- i)\cdot {{\rm{e}}^{{\rm{-i}}3{\rm{\theta }}}}\)

\(\rm =- i\left(\cos 3{\rm{\theta }} - {\rm{i}}\sin 3{\rm{\theta \;}}\right)\)

\(\left| {\frac{{z - 5i}}{{z + 5i}}} \right| = 1 \Rightarrow {\left| {x + iy - 5i} \right|^2} = {\left| {x + iy + 5i} \right|^2}\)

⇒ y = 0            ⇒ x – axis

f = |cos (2n + 1)π + isin nπ|ⁱ

for any n (integer value           cos(2n + 1)π = ±1) and sin nπ = 0

so        f = |±1 + 0|^I = 1ⁱ

^{\( = {1^{\left( {0 + i} \right)}} = {1^{\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)}} = {e^{{e^{i\frac{\pi }{2}}}}}\)}

or we can say for multi values \(f = {e^{{e^{i\left( {4n + 1} \right)\frac{\pi }{2}}}}}\) for any values of n, function will always be real & non-negative.