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

$\mathrm{\Delta }(x)=\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0.$

Step 1 — Column operation:  Replace the first column by $C_1-C_2$:

$\mathrm{\Delta }(x)=\left|\begin{array}{ccc}{x}^2-1& 2k+1& 1\\ k-1& k+2& 1\\ 0& 3& 1\end{array}\right|.$

Step 2 — Expansion along the third row:

$\mathrm{\Delta }(x)=-3\left|\begin{array}{cc}{k}^2-1& 1\\ k-1& 1\end{array}\right|+\left|\begin{array}{cc}{k}^2-1& 2k+1\\ k-1& k+2\end{array}\right|,$

which simplifies to

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

Step 3 — Equate to zero:

$\mathrm{\Delta }(x)=0⟹x^3=0⟹x=0.$

∴ 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

⇒ $ei-fh=(0)(1)-(i)(-i)=0-(-1)=1$

⇒ $di-fg=(3-i)(1)-(i)(-1)=3-i+i=3$

⇒ $dh-eg=(3-i)(-i)-(0)(-1)=-3i+i2=-3i-1=-1-3i$

⇒ Δ = $2(1)-(3+i)(3)+(-1)(-1-3i)$

⇒ Δ = $2-9-3i+1+3i$

⇒ $\Delta =-6+0i$

Since we are given that $\Delta =A+iB$ 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  $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$ i.e. (Taking determinant about first row), We get,

 $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$ 

⇒ 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:

$A(\theta )=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$

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

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

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:

$A(\theta {)}^{-1}=\frac{1}{\text{det}(A(\theta ))}\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

Since the determinant is 1, we have:

$A(\theta {)}^{-1}=\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

Now, let’s compute A(-θ) :

$A(-\theta )=\left[\begin{array}{cc}\sin (-\theta )& i\cos (-\theta )\\ i\cos (-\theta )& \sin (-\theta )\end{array}\right]$

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

$A(-\theta )=\left[\begin{array}{cc}-\sin \theta & i\cos \theta \\ i\cos \theta & -\sin \theta \end{array}\right]$

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

Therefore, Statement II is not correct.

Hence, the correct answer is: Option (1)

Calculation:

We are given two matrices:

$A(\theta )=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$

$B(\theta )=A(\frac{\pi }{2}-\theta )=\left[\begin{array}{cc}\sin (\frac{\pi }{2}-\theta )& i\cos (\frac{\pi }{2}-\theta )\\ i\cos (\frac{\pi }{2}-\theta )& \sin (\frac{\pi }{2}-\theta )\end{array}\right]$

Using the trigonometric identities:

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

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

The matrix $B(\theta )$ becomes:

$B(\theta )=\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$

Now, compute the matrix product AB :

$AB=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$

Performing the matrix multiplication:

$AB=\left[\begin{array}{cc}\sin \theta \cos \theta +i\cos \theta i\sin \theta & \sin \theta i\sin \theta +i\cos \theta \cos \theta \\ i\cos \theta \cos \theta +\sin \theta i\sin \theta & i\cos \theta i\sin \theta +\sin \theta \cos \theta \end{array}\right]$

Simplifying the terms:

$AB=\left[\begin{array}{cc}i{\cos }^2\theta +{\sin }^2\theta & i\sin \theta \cos \theta +\sin \theta \cos \theta \\ i\sin \theta \cos \theta +\cos \theta \sin \theta & i{\cos }^2\theta +{\sin }^2\theta \end{array}\right]$

Now, simplify further:

$AB=\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$

Hence, the correct answer is Option (1).

Concept:

$\mathrm{i}=\sqrt{-1}$

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

Calculations: 

Given determinant is $\left|\begin{array}{ccc}\mathrm{i}& {\mathrm{i}}^2& {\mathrm{i}}^3\\ {\mathrm{i}}^4& {\mathrm{i}}^6& {\mathrm{i}}^8\\ {\mathrm{i}}^9& {\mathrm{i}}^{12}& {\mathrm{i}}^{15}\end{array}\right|$

Since, we have, 

$\mathrm{i}=\sqrt{-1}$

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

=$\left|\begin{array}{ccc}\mathrm{i}& -1& -\mathrm{i}\\ 1& -1& 1\\ \mathrm{i}& 1& -\mathrm{i}\end{array}\right|$

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

= i² - i - 2i - i - i²

= - 4i

Concept:

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

Calculation:

Let, $\mathrm{A}=\left|\begin{array}{cc}2+i& 2-i\\ 1+i& i-1\end{array}\right|$

⇒ |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 $\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right]$ 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:

$\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& 0\\ 0& \mathrm{x}& \mathrm{y}\\ \mathrm{y}& 0& \mathrm{x}\end{array}\right|=0$

Expanding along R₁, we get

⇒ x (x² – 0) – y (0 – y²) + 0 = 0

⇒ x³ + y³ = 0

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

$\therefore \left(\frac{{\mathrm{x}}^3}{{\mathrm{y}}^3}+1\right)=0\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{y}}^3\ne 0$

$\Rightarrow \frac{{\mathrm{x}}^3}{{\mathrm{y}}^3}=-1$

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

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 = $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$

R₃ = R₃ + R₁ + R₂

D = $\left|\begin{array}{ccc}1& \omega & 1+\omega +{\omega }^2\\ \omega & {\omega }^2& 1+\omega +{\omega }^2\\ {\omega }^2& 1& 1+\omega +{\omega }^2\end{array}\right|$

D = $\left|\begin{array}{ccc}1& \omega & 0\\ \omega & {\omega }^2& 0\\ {\omega }^2& 1& 0\end{array}\right|$ = 0

Concept:

i2 = - 1

Calculations:

Given, $x+iy=\left|\begin{array}{ccc}6i& -3i& 1\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|$

⇒ 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

 $\left|\begin{array}{ccc}1+\omega & {\omega }^2& -\omega \\ 1+{\omega }^2& \omega & -{\omega }^2\\ {\omega }^2+\omega & \omega & -{\omega }^2\end{array}\right|$

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

= $-{\omega }^2\left|\begin{array}{ccc}1+\omega & \omega & 1\\ 1+{\omega }^2& 1& \omega \\ {\omega }^2+\omega & 1& \omega \end{array}\right|$

=  $-{\omega }^2[(1+\omega )(\omega -\omega )-\omega (\omega +1-1-{\omega }^2)+(1+{\omega }^2-{\omega }^2-\omega )]$

= $-{\omega }^2[-{\omega }^2+1+1-\omega ]$

= $\omega -2{\omega }^2+1$

= $1+\omega +{\omega }^2-3{\omega }^2$

= 0 - 3ω2

= - 3ω2

Hence the required value is -3ω2

Concept:

(i)² = 1

(-i)⁴ = 1

Calculation:

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

On rationalizing,

${(\frac{1-\mathrm{i}}{1+\mathrm{i}}\times \frac{1-\mathrm{i}}{1-\mathrm{i}})}^{{\mathrm{n}}^2}$= 1

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

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

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

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

${(-\mathrm{i})}^{{\mathrm{n}}^2}$ = 1

if we put n = 2 then, 

${(-\mathrm{i})}^{{\mathrm{n}}^2}$ = 1

Satisfy the equation.

Hence, Option 1 is correct.

Calculation:

Given: $\left|\begin{array}{ccc}x& -3i& 1\\ y& 1& i\\ 0& 2i& -i\end{array}\right|=6+11i$

Let $A=\left|\begin{array}{ccc}x& -3i& 1\\ y& 1& i\\ 0& 2i& -i\end{array}\right|$

⇒ 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 = $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$

R₃ = R₃ + R₁ + R₂

D = $\left|\begin{array}{ccc}1& \omega & 1+\omega +{\omega }^2\\ \omega & {\omega }^2& 1+\omega +{\omega }^2\\ {\omega }^2& 1& 1+\omega +{\omega }^2\end{array}\right|$

D = $\left|\begin{array}{ccc}1& \omega & 0\\ \omega & {\omega }^2& 0\\ {\omega }^2& 1& 0\end{array}\right|$ = 0