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

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:

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

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 \(\left| {\begin{array}{*{20}{c}}1&ω^3&ω^2\\ ω^3&1&ω\\ ω^2&ω&1 \end{array}} \right|\) = \(\left| {\begin{array}{*{20}{c}}1&1&ω^2\\ 1&1&ω\\ ω^2&ω&1 \end{array}} \right|\)   [ ∵ ω.ω² = ω³ = 1]

Applying R₁ → R₁ + R₂ + R₃ , we get

⇒ \(\left| {\begin{array}{*{20}{c}}2+ω^2&2+ω&0\\ 1&1&ω\\ ω^2&ω&1 \end{array}} \right|\)     [∵ 1 + ω + ω² = 0]

Applying C₁ → C1 + C₂ + C_3, we get

⇒ \(\left| {\begin{array}{*{20}{c}}4+ω+ω^2&2+ω&0\\ 2+ω&1&ω\\ 0&ω&1 \end{array}} \right|\)

⇒ \(\left| {\begin{array}{*{20}{c}}3&2+ω&0\\ 2+ω&1&ω\\ 0&ω&1 \end{array}} \right|\)   

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:

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

Applying R₃ → R3 + R2 

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

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

Applying R₃ → R₃ - R₁

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

\(\begin{vmatrix}1-i&ω^2&-ω\\\ ω^2+i&ω&-i\\\ 0&0&0\end{vmatrix}\) = 0

∴ \(\begin{vmatrix}1-i&ω^2&-ω\\\ ω^2+i&ω&-i\\\ 1-2i-ω^2&ω^2-ω&i-ω\end{vmatrix}\) = 0 

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.

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