Complex Number elements MCQ Quiz - Objective Question with Answer for Complex Number elements - Download Free PDF

Concept:

Since ω is the cube root of unity,

Thus, ω³ = 1 and ω⁴ = ω.

Explanation:

We are given   i.e. (Taking determinant about first row), We get,

⇒ 1(ω³ - 1) - ω(ω² - ω²) + ω²(ω - ω⁴) ....(ω3 = 1 and ω4 = ω.)

⇒ 1(1 - 1) - (0) + ω²(ω - ω)

⇒ 0

Concept:

• The cube root of unity has three roots, which are 1, ω, ω².
• Here the roots ω and ω² are imaginary roots and one root is a square of the other root. 
• The product of the imaginary roots of the cube root of unity is equal to 1 (ω.ω² = ω³ = 1), and
• The sum of the cube roots of unity is equal to zero.(1 + ω + ω² = 0).

​Calculations:

Given  =    [ ∵ ω.ω² = ω³ = 1]

Applying R₁ → R₁ + R₂ + R₃ , we get

⇒      [∵ 1 + ω + ω² = 0]

Applying C₁ → C1 + C₂ + C_3, we get

⇒ 

⇒    

On simplification, we get 

⇒ 3 [1 - w²] - (2  + w) [(2  + w) - 0]

⇒ 3 - 3w2 - 4 - w² - 4w

⇒ 3 - 4 - 4w - 4w2 

⇒ 3 - 4(1 + ω + ω²)

⇒ 3 - 0 = 3

Hence, the correct answer is option 4).  

Concept:

If ω is the cube root of unity, then 

 ω³ = 1 and 1 +  ω + ω² = 0

Applying Row operation Rj → Rj + t.Rk (t is a real number) does not change the value of the determinant.

Where Rj denotes the j-th row of the matrix

If any row or column of the matrix has all zero elements, then the determinant of that matrix is 0.

Calculation:

Applying R₃ → R3 + R2 

Applying R₃ → R₃ - R₁

 = 0

∴  = 0 

Concept:

(i)² = 1

(-i)⁴ = 1

Calculation:

On rationalizing,

= 1

 = 1

 = 1

 = 1

= 1

 = 1

if we put n = 2 then, 

 = 1

Satisfy the equation.

Hence, Option 1 is correct.

Concept:

Complex number = a + bi

|z| = 

Calculation:

Given:

z = 

By rationalization,

z = 

z = 

z = 

z = 

z = -1 + 2i

a = -1, b = 2

|z| = 

|z| = 

Concept:

i² = -1 , i³ = - i, i⁴ = 1, i⁶ = - 1 , i⁸ = 1 , i⁹ = i, i ¹² = 1, and i¹⁵ = - i

Calculations: 

Given determinant is 

Since, we have, 

i2 = -1 , i3 = - i, i4 = 1, i6 = - 1 , i8 = 1 , i9 = i, i 12 = 1, and i15 = - i

=

=i(i - 1) + 1(-i - i) - i (1 + i)

= i² - i - 2i - i - i²

= - 4i

Concept:

If  then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

Calculation:

Let, 

⇒ |A| = (2 + i) (i – 1) – (2 – i) (1 + i)

= 2i + i² – 2 – i – (2 – i + 2i – i²)

= i – 1 – 2 – (2 + i + 1)                   (∵ i² = -1)

= i – 3 – 2 – i – 1

= -6

∴ |A| is real number.

Concept:

If  then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

Calculation:

Given:

Expanding along R₁, we get

⇒ x (x² – 0) – y (0 – y²) + 0 = 0

⇒ x³ + y³ = 0

Hence x/y is one of the cube roots of -1

Concept:

If the 1, ω and ω²  are the cube roots of unity, then 1 + ω + ω² = 0

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

Calculation:

D = 

R₃ = R₃ + R₁ + R₂

D = 

D =  = 0

Concept:

i2 = - 1

Calculations:

Given, 

⇒ x + iy = 6i (3i² + 3) + 3i (4i + 20) + 1(12 - 60i)

⇒ x + iy = 6i (3i2 + 3) + 12i2 + 60i + 12 - 60i

We know that i2 = - 1

⇒ x + iy = 6i [3(-1)+3] +12(-1) + 60i + 12 - 60i

⇒ x + iy = 0 

⇒ x + iy = 0 + i 0

⇒ x = 0 and y = 0

Consider,  x - iy

Put the value of x and y in above expression, we get

⇒ x - iy = 0 - i0

⇒ x - iy = 0 

Hint

To find the value of x - iy, find the value of x and y.

To find the value of x and y, solve the given determinant and compare the real and imaginary part. 

Concept:

ω is a cube root of unity.

Property of cube root of unity:

• ω3 = 1
• 1 + ω + ω2  = 0

Calculations:

Given, ω is a cube root of unity.

⇒ω³ = 1, ad 1 + ω + ω²  = 0

Now, consider the determinant

Taking ω and -ω common from C₂ and C₃ respectively,

= 

=  

= 

= 

= 

= 0 - 3ω2

= - 3ω2

Hence the required value is -3ω2

Concept:

(i)² = 1

(-i)⁴ = 1

Calculation:

On rationalizing,

= 1

 = 1

 = 1

 = 1

= 1

 = 1

if we put n = 2 then, 

 = 1

Satisfy the equation.

Hence, Option 1 is correct.

Calculation:

Given:

Let

⇒ det A = x[-i - 2i²] - (-3i)[-yi - 0] + 1[2yi - 0]

As we know that, i2 = -1 

det A = x[-i + 2] + 3i[-yi] + [2yi]

det A = -ix + 2x + 3y + 2yi

det A = (2x + 3y) + (2y - x)i

According to the question, 

det A =  6 + 11i

(2x + 3y) + (2y - x)i = 6 + 11i

⇒ 2x + 3y = 6 and 2y – x = 11

Solving the above equations we get

⇒ x = - 3, y = 4

Concept:

If r(cos θ + i sin θ) is a complex number then:

{r (cos θ + i sin θ)}ⁿ = rⁿ (cos nθ + i sin nθ)

Calculation:

Given:

(cos 5θ – i sin 5θ)² = z

Using identity

Z = {cos (10 θ) – i sin (10 θ)}

z = (cos θ + i sin θ)^{– 10}

Additional Information

DERIVATION:

De Moivre’s theorem

Prove by induction z = r(cos θ + i sin θ)

⇒ zⁿ = rⁿ (cos nθ + i sin nθ)

Step 1:

Show true for n = 1; z¹ = r¹ (cos θ + i sin θ)

Step 2:

Assume true for n = k; z^k = r^k (cos kθ + i sin kθ)

Step 3:

Prove true for n = k + 1

z^{k + 1} = z^k z¹ = r^k (cos kθ + i sin kθ) × r(cos θ + i sin θ)

= r^{k + 1} (cos (kθ + θ) + i sin (kθ + θ))

⇒ z^{k + 1} = r^{k + 1} (cos (k + 1)θ + i sin (k + 1)θ)

Step 4:

Conclusion

By induction, the statement is true  ∀ n ≥ 1, n ϵ N

Concept:

If the 1, ω and ω²  are the cube roots of unity, then 1 + ω + ω² = 0

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

Calculation:

D = 

R₃ = R₃ + R₁ + R₂

D = 

D =  = 0