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

Calculation:

Given,

Let ω be a non-real cube root of unity, so \( \omega^{3}=1 \) and \( 1+\omega+\omega^{2}=0 \).

Consider the determinant

\( \Delta(x)= \begin{vmatrix} x+1 & \omega & \omega^{2}\\ \omega & x+\omega^{2} & 1\\ \omega^{2} & 1 & x+\omega \end{vmatrix}=0. \)

Step 1 — Column operation:  Replace the first column by \(C_{1}-C_{2}\):

\( \Delta(x)= \begin{vmatrix} x^{2}-1 & \,2\!k\!+\!1\, & 1\\ k-1 & k+2 & 1\\ 0 & 3 & 1 \end{vmatrix}. \)

Step 2 — Expansion along the third row:

\( \Delta(x)= -3\! \begin{vmatrix} k^{2}-1 & 1\\ k-1 & 1 \end{vmatrix} + \begin{vmatrix} k^{2}-1 & 2k+1\\ k-1 & k+2 \end{vmatrix}, \)

which simplifies to

\( \Delta(x)=x(x^{2}-1)-x\bigl(\omega+\omega^{2}\bigr) =x(x^{2}-1)+x =x^{3}. \)

Step 3 — Equate to zero:

\( \Delta(x)=0 \;\Longrightarrow\; x^{3}=0 \;\Longrightarrow\; x=0. \)

∴ The root of the equation is  \( x = 0 \).

Hence, the correct answer is Option 1.

Calculation:

Determinant Δ = \(a(ei−fh)−b(di−fg)+c(dh−eg)\)

Now, For our matrix, 

\(a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1\)

calculate the subdeterminants

⇒ \( ei−fh=(0)(1)−(i)(−i)=0−(−1)=1\)

⇒ \(di−fg=(3−i)(1)−(i)(−1)=3−i+i=3\)

⇒ \(dh−eg=(3−i)(−i)−(0)(−1)=−3i+i 2=−3i−1=−1−3i\)

⇒ Δ = \(2(1)−(3+i)(3)+(−1)(−1−3i)\)

⇒ Δ = \(2−9−3i+1+3i\)

⇒ \(Δ=−6+0i\)

Since we are given that $\backslash (\Delta =A+iB\backslash )$ comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

Concept:

Since ω is the cube root of unity,

Thus, ω³ = 1 and ω⁴ = ω.

Explanation:

We are given  \(\begin{vmatrix} 1&\omega&\omega^2 \\\ \omega&\omega^2&1 \\\ \omega^2&1&\omega \end{vmatrix}\) i.e. (Taking determinant about first row), We get,

 \(\begin{vmatrix} 1&\omega&\omega^2 \\\ \omega&\omega^2&1 \\\ \omega^2&1&\omega \end{vmatrix}\) 

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

⇒ 1(1 - 1) - (0) + ω²(ω - ω)

⇒ 0

Calculation:

We are given the following matrix:

\( A(θ) = \begin{bmatrix} \sin θ & i \cos θ \\ i \cos θ & \sin θ \end{bmatrix} \)

The determinant of \(A(θ) \) is calculated as follows:

\( \text{det}(A(θ)) = \sin^2 θ - (i \cos θ)(i \cos θ) = \sin^2 θ + \cos^2 θ = 1 \)

Since the determinant is 1 (non-zero), matrix Aθ  is invertible for all θ in \(\mathbb{R} \)

Therefore, Statement I is correct.

We first compute the inverse of A(θ)^-1 using the formula for the inverse of a 2x2 matrix:

\( A(θ)^{-1} = \frac{1}{\text{det}(A(θ))} \begin{bmatrix} \sin θ & -i \cos θ \\ -i \cos θ & \sin θ \end{bmatrix} \)

Since the determinant is 1, we have:

\( A(θ)^{-1} = \begin{bmatrix} \sin θ & -i \cos θ \\ -i \cos θ & \sin θ \end{bmatrix} \)

Now, let’s compute A(-θ) :

\( A(-θ) = \begin{bmatrix} \sin(-θ) & i \cos(-θ) \\ i \cos(-θ) & \sin(-θ) \end{bmatrix} \)

Using the trigonometric identities\(\sin(-θ) = -\sin(θ) \) and \(\cos(-θ) = \cos(θ) \) we get:

\( A(-θ) = \begin{bmatrix} -\sin θ & i \cos θ \\ i \cos θ & -\sin θ \end{bmatrix} \)

This is not equal to\(A(θ)^{-1} \), because the signs are different.

Therefore, Statement II is not correct.

Hence, the correct answer is: Option (1)

Calculation:

We are given two matrices:

\( A(\theta) = \begin{bmatrix} \sin \theta & i \cos \theta \\ i \cos \theta & \sin \theta \end{bmatrix} \)

\( B(\theta) = A\left( \frac{\pi}{2} - \theta \right) = \begin{bmatrix} \sin\left( \frac{\pi}{2} - \theta \right) & i \cos\left( \frac{\pi}{2} - \theta \right) \\ i \cos\left( \frac{\pi}{2} - \theta \right) & \sin\left( \frac{\pi}{2} - \theta \right) \end{bmatrix} \)

Using the trigonometric identities:

\( \sin\left( \frac{\pi}{2} - \theta \right) = \cos(\theta) \)

\( \cos\left( \frac{\pi}{2} - \theta \right) = \sin(\theta) \)

The matrix \(B(\theta) \) becomes:

\( B(\theta) = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix} \)

Now, compute the matrix product AB :

\( AB = \begin{bmatrix} \sin \theta & i \cos \theta \\ i \cos \theta & \sin \theta \end{bmatrix} \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix} \)

Performing the matrix multiplication:

\( AB = \begin{bmatrix} \sin \theta \cos \theta + i \cos \theta i \sin \theta & \sin \theta i \sin \theta + i \cos \theta \cos \theta \\ i \cos \theta \cos \theta + \sin \theta i \sin \theta & i \cos \theta i \sin \theta + \sin \theta \cos \theta \end{bmatrix} \)

Simplifying the terms:

\( AB = \begin{bmatrix} i \cos^2 \theta + \sin^2 \theta & i \sin \theta \cos \theta + \sin \theta \cos \theta \\ i \sin \theta \cos \theta + \cos \theta \sin \theta & i \cos^2 \theta + \sin^2 \theta \end{bmatrix} \)

Now, simplify further:

\( AB = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \)

Hence, the correct answer is Option (1).

Concept:

\(\rm i = \sqrt {-1}\)

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

Calculations: 

Given determinant is \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\)

Since, we have, 

\(\rm i = \sqrt {-1}\)

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

=\(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{-1}}}}&{{{\rm{-i}}}}\\ {{{\rm{1}}}}&{{{\rm{-1}}}}&{{{\rm{1}}}}\\ {{{\rm{i}}}}&{{{\rm{1}}}}&{{{\rm{-i}}}} \end{array}} \right|\)

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

= i² - i - 2i - i - i²

= - 4i

Concept:

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

Calculation:

Let, \({\rm{A}} = \left| {\begin{array}{*{20}{c}} {2 + i}&{2 - i}\\ {1 + i}&{i - 1} \end{array}} \right|\)

⇒ |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 \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) 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:

\(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\)

Expanding along R₁, we get

⇒ x (x² – 0) – y (0 – y²) + 0 = 0

⇒ x³ + y³ = 0

\(\Rightarrow {\rm{\;}}{{\rm{y}}^3}{\rm{\;}}\left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right){\rm{\;}} = {\rm{\;}}0{\rm{\;}}\)

\(\therefore \left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right) = {\rm{\;}}0{\rm{\;and\;}}{{\rm{y}}^3} \ne 0\)

\(\Rightarrow \frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} = {\rm{}} - 1\)

\(\Rightarrow \frac{{\rm{x}}}{{\rm{y}}} = {\left( { - 1} \right)^{\frac{1}{3}}}\)

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 = \(\rm\begin{vmatrix} 1 & ω &ω^2 \\ ω & ω^2 &1 \\ ω^2& 1 & ω \end{vmatrix}\)

R₃ = R₃ + R₁ + R₂

D = \(\rm\begin{vmatrix} 1 & ω &1+ω+ω^2 \\ ω & ω^2 &1+ω+ω^2 \\ ω^2& 1 & 1+ω+ω^2 \end{vmatrix}\)

D = \(\rm\begin{vmatrix} 1 & ω &0 \\ ω & ω^2 &0 \\ ω^2& 1 & 0 \end{vmatrix}\) = 0

Concept:

i2 = - 1

Calculations:

Given, \(x+iy=\begin{vmatrix}6i & -3i & 1 \\\ 4 & 3i & -1 \\\ 20 & 3 & i \end{vmatrix}\)

⇒ 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

 \(\rm \begin{vmatrix} 1 + ω & ω^2 & -ω \\\ 1 + ω^2 & ω & -ω^2 \\\ ω^2 + ω & ω & -ω^2 \end{vmatrix}\)

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

= \(\rm-ω^2 \begin{vmatrix} 1 + ω & ω & 1\\\ 1 + ω^2 &1 & ω \\\ ω^2 + ω & 1& ω \end{vmatrix}\)

=  \(\rm -ω^2 [(1+ω)(ω-ω)-ω(ω +1-1-ω^2)+(1+ω^2-ω^2-ω)]\)

= \(\rm -ω^2 [-ω^2+1+1-ω]\)

= \(\rm ω -2ω^2+ 1\)

= \(\rm 1+ ω + ω^2-3ω^2\)

= 0 - 3ω2

= - 3ω2

Hence the required value is -3ω2

Concept:

(i)² = 1

(-i)⁴ = 1

Calculation:

\(\rm \left(\frac{1-i}{1+i}\right)^{n^2}=1\)

On rationalizing,

\(\rm \left (\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \right )^{n^{2}}\)= 1

\(\rm \left (\frac{(1 - i)^{2}}{1 - i^{2}} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{1 + i^{2} - 2i}{1 - (-1)} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{1 - 1 - 2i}{1 + 1} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{- 2i}{2} \right )^{n^{2}}\)= 1

\(\rm \left (- i \right )^{n^{2}} \) = 1

if we put n = 2 then, 

\(\rm \left (- i \right )^{n^{2}} \) = 1

Satisfy the equation.

Hence, Option 1 is correct.

Calculation:

Given: \(\left| {\begin{array}{*{20}{c}} x&-3i&1\\ y&1&{i}\\ 0&2i&-i \end{array}} \right|=6+11i\)

Let \(A=\left| {\begin{array}{*{20}{c}} x&-3i&1\\ y&1&{i}\\ 0&2i&-i \end{array}} \right|\)

⇒ 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

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 = \(\rm\begin{vmatrix} 1 & ω &ω^2 \\ ω & ω^2 &1 \\ ω^2& 1 & ω \end{vmatrix}\)

R₃ = R₃ + R₁ + R₂

D = \(\rm\begin{vmatrix} 1 & ω &1+ω+ω^2 \\ ω & ω^2 &1+ω+ω^2 \\ ω^2& 1 & 1+ω+ω^2 \end{vmatrix}\)

D = \(\rm\begin{vmatrix} 1 & ω &0 \\ ω & ω^2 &0 \\ ω^2& 1 & 0 \end{vmatrix}\) = 0