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

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

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

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

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

⇒ $\frac{2+2\surd 3i}{4}$    

⇒ $\frac{1+\surd 3}{2}=-(\frac{-1-\surd 3}{2})=-{\omega }^2$       ----- (1)

Consider the second part of the given equation

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

⇒ $\frac{(i-\surd 3{)}^2}{(i^2-(\surd 3{)}^2)}=\frac{-1+i\surd 3}{2}=\omega$         ----- (2)

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

⇒ $(-{\omega }^2{)}^{200}+{\omega }^{200}+1$

⇒ ${\omega }^{400}+{\omega }^{200}+1$

⇒ ${\omega }^{3\times 133+1}+{\omega }^{3\times 66+2}+1$

We know that ${\omega }^3=1$

⇒ ${\omega }^2+\omega +1=0$

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

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

$1+{\omega }^n+{\omega }^{2n}=\{\begin{array}{c}0,ifnisnotmultipleof3\\ 3,ifnismultipleof3\end{array}$

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = $\left[\begin{array}{ccc}1& \omega & {\omega }^{2n}\\ {\omega }^2& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right]$

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:

$1+\omega +\frac{1}{\omega }$

We know that;

1 + ω + ω2 = 0

$1+\omega +\frac{{\omega }^2}{1}=0$

$1+\omega +\frac{{\omega }^2}{{\omega }^3}=0$      [∵ ω³ = 1]

$\therefore 1+\omega +\frac{1}{\omega }=0$

We know that $1+\omega +{\omega }^2=0\mathrm{a}\mathrm{n}\mathrm{d}{\omega }^3=1$ then

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

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

$1+{\omega }^n+{\omega }^{2n}=\{\begin{array}{c}0,ifnisnotmultipleof3\\ 3,ifnismultipleof3\end{array}$

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = $\left[\begin{array}{ccc}1& \omega & {\omega }^{2n}\\ {\omega }^2& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right]$

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 $1+\omega +{\omega }^2=0\mathrm{a}\mathrm{n}\mathrm{d}{\omega }^3=1$ then

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

Concept:

The cube roots of unity are 1, ω and ω2 

where,

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

$1+{\omega }^n+{\omega }^{2n}=\{\begin{array}{c}0,ifnisnotmultipleof3\\ 3,ifnismultipleof3\end{array}$

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = $\left[\begin{array}{ccc}1& \omega & {\omega }^{2n}\\ {\omega }^2& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right]$

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:

$1+\omega +\frac{1}{\omega }$

We know that;

1 + ω + ω2 = 0

$1+\omega +\frac{{\omega }^2}{1}=0$

$1+\omega +\frac{{\omega }^2}{{\omega }^3}=0$      [∵ ω³ = 1]

$\therefore 1+\omega +\frac{1}{\omega }=0$

Concept :

Cube Roots of unity are 1, ω and ω2

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

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

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

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

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

⇒ $\frac{2+2\surd 3i}{4}$    

⇒ $\frac{1+\surd 3}{2}=-(\frac{-1-\surd 3}{2})=-{\omega }^2$       ----- (1)

Consider the second part of the given equation

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

⇒ $\frac{(i-\surd 3{)}^2}{(i^2-(\surd 3{)}^2)}=\frac{-1+i\surd 3}{2}=\omega$         ----- (2)

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

⇒ $(-{\omega }^2{)}^{200}+{\omega }^{200}+1$

⇒ ${\omega }^{400}+{\omega }^{200}+1$

⇒ ${\omega }^{3\times 133+1}+{\omega }^{3\times 66+2}+1$

We know that ${\omega }^3=1$

⇒ ${\omega }^2+\omega +1=0$

∴ The value of ${\left(\frac{i+\surd 3}{-1+\surd 3}\right)}^{200}+{\left(\frac{i-\surd 3}{-1+\surd 3}\right)}^{200}+1$ 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

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

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

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

$=-\mathrm{i}(\cos 3\theta -\mathrm{i}\sin 3\theta )$

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

⇒ 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.