Cube Roots of Unity MCQ Quiz - Objective Question with Answer for Cube Roots of Unity - Download Free PDF
Last updated on Apr 1, 2025
Latest Cube Roots of Unity MCQ Objective Questions
Cube Roots of Unity Question 1:
What is the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 1 Detailed Solution
Concept :
Cube Roots of unity are 1, ω and ω2
Here, ω =
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
Consider the first part of the given equation
⇒
⇒
We know that (a - b)(a + b) = a2 - b2 and i2 = √-1
⇒
⇒
Consider the second part of the given equation
⇒
⇒
Using equations (1) and (2) in the given equation, we get
⇒
⇒
⇒
We know that
⇒
∴ The value of
Cube Roots of Unity Question 2:
If 1, ω, ω2 are cube roots of unity then the value of Δ =
Answer (Detailed Solution Below)
Cube Roots of Unity Question 2 Detailed Solution
Concept:
The cube roots of unity are 1, ω and ω2
where,
ω3 = 1
1 + ω + ω2 = 0
ω3n = 1
Calculation:
Given:
Δ =
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
Cube Roots of Unity Question 3:
If a complex number ω satisfies the equation ω3 =1, then the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 3 Detailed Solution
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:
We know that;
1 + ω + ω2 = 0
Cube Roots of Unity Question 4:
If ω is a cube root of unity then the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 4 Detailed Solution
We know that
Top Cube Roots of Unity MCQ Objective Questions
If 1, ω, ω2 are cube roots of unity then the value of Δ =
Answer (Detailed Solution Below)
Cube Roots of Unity Question 5 Detailed Solution
Download Solution PDFConcept:
The cube roots of unity are 1, ω and ω2
where,
ω3 = 1
1 + ω + ω2 = 0
ω3n = 1
Calculation:
Given:
Δ =
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
Cube Roots of Unity Question 6:
If ω is a cube root of unity then the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 6 Detailed Solution
We know that
Cube Roots of Unity Question 7:
If 1, ω, ω2 are cube roots of unity then the value of Δ =
Answer (Detailed Solution Below)
Cube Roots of Unity Question 7 Detailed Solution
Concept:
The cube roots of unity are 1, ω and ω2
where,
ω3 = 1
1 + ω + ω2 = 0
ω3n = 1
Calculation:
Given:
Δ =
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
Cube Roots of Unity Question 8:
If a complex number ω satisfies the equation ω3 =1, then the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 8 Detailed Solution
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:
We know that;
1 + ω + ω2 = 0
Cube Roots of Unity Question 9:
What is the value of
Answer (Detailed Solution Below)
Cube Roots of Unity Question 9 Detailed Solution
Concept :
Cube Roots of unity are 1, ω and ω2
Here, ω =
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
Consider the first part of the given equation
⇒
⇒
We know that (a - b)(a + b) = a2 - b2 and i2 = √-1
⇒
⇒
Consider the second part of the given equation
⇒
⇒
Using equations (1) and (2) in the given equation, we get
⇒
⇒
⇒
We know that
⇒
∴ The value of
Cube Roots of Unity Question 10:
+
Answer (Detailed Solution Below)
Cube Roots of Unity Question 10 Detailed Solution
Concept:
1 + ω + ω2 = 0
ω3 = 1
Calculation:
We can write the given expression as
⇒
⇒
⇒
⇒
⇒ -1
∴ The required value is -1.
Cube Roots of Unity Question 11:
If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x3) + x g(x3) is divisible by x2 + x + 1, then P(1) is equal to ________.
Answer (Detailed Solution Below) 0
Cube Roots of Unity Question 11 Detailed Solution
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
Cube Roots of Unity Question 12:
is equal to
Answer (Detailed Solution Below)
Cube Roots of Unity Question 12 Detailed Solution
Cube Roots of Unity Question 13:
The complex number z = x + iy, which satisfy the equation,
Answer (Detailed Solution Below)
X – axis
Cube Roots of Unity Question 13 Detailed Solution
⇒ y = 0 ⇒ x – axis
Cube Roots of Unity Question 14:
Value of complex function
Answer (Detailed Solution Below)
real and non-negative
Cube Roots of Unity Question 14 Detailed Solution
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
or we can say for multi values