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# 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.

### 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 = 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

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