Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Miscellaneous Exercise on Chapter 5 | Set 1

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. Convert the following in the polar form:

(i) 

(ii) 

Solution:

(i) Here, z = 

= 

Multiplying by its conjugate in the numerator and denominator

= -1+i

Let r cos θ = -1 and r sin θ = 1

On squaring and adding, we get

r² (cos²θ + sin²θ) = 1 + 1 = 2

r² = 2           [cos²θ + sin²θ = 1]

r = √2

So, 

√2 cosθ = -1 and √2 sinθ = 1

⇒ cosθ =  and sinθ  = 

Therefore,

θ =            [As θ lies in II quadrant]

Expressing as, z = r cos θ + i r sin θ 

= 

Therefore, this is the required polar form.

(ii) Let, z = 

= -1 + i

Now, Let r cosθ = -1 and r sin θ = 1

On squaring and adding, we get

r²(cos²θ + sin²θ) = 1 + 1

r(cos²θ + sin²θ) = 2

r² = 2          [cos2θ + sin2θ = 1]

= r = √2     [Conventionally, r > 0]

Therefore,

√2 cosθ = -1 and √2 sinθ = 1

cosθ =  and sinθ = 

Therefore,

θ =       [As θ lies in II quadrant]

Expressing as, z = r cosθ + i r sinθ 

z = 

Therefore, this is the required polar form.

Solve each of the equation in Exercises 6 to 9.

Question 6. 3x² – 4x + 20/3 = 0

Solution:

Given quadratic equation, 3x² – 4x + 20/3 = 0

It can be re-written as: 9x² – 12x + 20 = 0

On comparing it with ax² + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b² – 4ac = (–12)² – 4 × 9 × 20 = 144 – 720 = –576

Hence, the required solutions are

X = 

Question 7. x² – 2x + 3/2 = 0

Solution:

Given:

Quadratic equation, x² – 2x + = 0

After re-written 2x² – 4x + 3 = 0

On comparing it with ax² + bx + c = 0, 

We get

a = 2, b = –4, and c = 3

So, the discriminant of the given equation will be

D = b² – 4ac = (–4)² – 4 × 2 × 3 = 16 – 24 = –8

Hence, the required solutions are

x = 

Question 8. 27x² – 10x + 1 = 0

Solution:

Given:

Quadratic equation, 27x² – 10x + 1 = 0

On comparing it with ax² + bx + c = 0, 

We get

a = 27, b = –10, and c = 1

So, the discriminant of the given equation will be

D = b² – 4ac = (–10)² – 4 × 27 × 1 = 100 – 108 = –8

Hence, the required solutions are

Question 9. 21x² – 28x + 10 = 0

Solution:

Given:

Quadratic equation, 21x² – 28x + 10 = 0

On comparing it with ax² + bx + c = 0, 

We have

a = 21, b = –28, and c = 10

So, the discriminant of the given equation will be

D = b² – 4ac = (–28)² – 4 × 21 × 10 = 784 – 840 = –56

Hence, the required solutions are

Question 10. If z₁ = 2 – i, z₂ = 1 + i, find 

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

Hence, the value of  is √2