NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise

Question 1. Evaluate 

Solution:

= [-1 – i]³

= (-1)³ [1 + i]³

= -[1³ + i³ + 3 × 1 × i (1 + i)]

= -[1 + i³ + 3i + 3i²]

= -[1 – i + 3i – 3]

= -[2 + 2i]

= 2 – 2i 

Question 2. For any two complex numbers z₁ and z₂, prove that, Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Solution:

Let’s assume z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ as two complex numbers

Product of these complex numbers, z₁z₂

z₁z₂ = (x₁ + iy₁)(x₂ + iy₂)

= x₁(x₂ + iy₂) + iy₁(x₂ + iy₂)

= x₁x₂ + ix₁y₂ + iy₁x₂ + i²y₁y₂

= x₁x₂ + ix₁y₂ + iy₁x₂ – y₁y₂             [i² = -1]

= (x₁x₂ – y₁y₂) + i(x₁y₂ + y₁x₂)

Now, 

Re(z₁z₂) = x₁x₂ – y₁y₂

⇒ Re(z₁z₂) = Rez₁Rez₂ – Imz₁Imz₂

Hence, proved.

Question 3. Reduce to the standard form 

Solution:

On multiplying numerator and denominator by (14+5i)

Hence, this is the required standard form.

Question 4. If x – iy =  prove that (x² + y²)²

Solution:

Given:

x – iy = 

On multiplying numerator and denominator by (c+id)

So,

(x – iy)² = 

x² – y² – 2ixy  

On comparing real and imaginary parts, we get

x² – y² = , -2xy =       (1)

(x² + y²)² = (x² – y²)² + 4x²y²

= 

Hence proved

Question 5. If z₁ = 2 – i, z₂ = 1 + i, find 

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

Hence, the value of  is √2

Question 6. If a + ib =  , prove that a² + b² = 

Solution:

Given:

a + ib = 

On comparing the real and imaginary parts, we have

a =  and b = 

Therefore,

a² + b² = 

Hence, proved,

a² + b² = 

Question 7. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) 

(ii) 

Solution:

(i) Given:

z₁ = 2 – i, z₂ = -2 + i

(i) z₁z₂ = (2 – i)(-2 + i) = -4 + 2i + 2i – i² = -4 + 4i – (-1) = -3 + 4i

 = 2 + i

Therefore,

On multiplying numerator and denominator by (2 – i), we get

On comparing the real parts, we have

(ii)

On comparing the imaginary part, we get

 = 0

Question 8. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let us assume z = (x – iy) (3 + 5i)

z = 3x + xi – 3yi – 5yi² = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)

Therefore,

 =(3x + 5y) – i(5x – 3y)

Also given,  = -6 – 24i 

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

After equating real and imaginary parts, we get

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

After doing (i) x 3 + (ii) x 5, we have

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3 respectively.

Question 9. Find the modulus of 

Solution:

Question 10. If (x + iy)³ = u + iv, then show that  = 4(x² – y²)

Solution:

(x + iy)³ = u + iv

x³ + (iy)³  + 3 × x × iy(x + iy) = u + iv

x³ + i³y³ + 3x²yi + 3xy² = u + iv

x³ – iy³ + 3x²yi – 3xy² = u + iv

(x³ – 3xy²) + i(3x²y – y³) = u + iv

On equating real and imaginary parts, we get

u = x³ – 3xy², v = 3x²y – y³

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

Hence proved

Question 11. If α and β are different complex numbers with |β| = 1, then find 

Solution:

Assume α = a + ib and β = x + iy

Given: |β| = 1

So, 

= x2 + y2 = 1            ….(1)

= 1

Question 12. Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x

Solution:

|1 – i|^x = 2^t

x = 2x

2x – x = 0

Thus, ‘0’ is the only integral solution of the given equation.

Therefore, the number of non-zero integral solutions of the given equation is 0.

Question 13. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Solution:

Given:

(a + ib)(c + id)(e + if)(g + ih) = A + iB

Therefore,

|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|

On squaring both sides, we get

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Hence, proved.

Question 14. If, then find the least positive integral value of m.

Solution:

i^m = 1

Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).