NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 14 (Mis Ex)

Question 14:

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

Answer

Let z = (x – iy) (3 + 5i)

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

∴ (3x + 5y) – i(5x – 3y) = −6 – 24i

Equating real and imaginary parts, we obtain

3x + 5y = −6                     ….(i)

5x – 3y = 24                      ….(ii)

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 9x + 15y = −18

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

3(3) + 5y = −6

⇒ 5y = −6 – 9 = −15

⇒ y = −3

Thus, the values of x and y are 3 and −3 respectively.

NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 5 (Mis Ex)

Question 5:

Convert the following in the polar form :

Answer

Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ) = 1 + 1

⇒ r2(cos2 θ + sin2 θ) = 1 + 1

⇒ r2(cos2 θ + sin2 θ) = 2

⇒ r2 = 2                      [cos2 θ + sin2 θ = 1]

⇒ r = √2                  [Conventionally, r > 0]

∴ √2 cos θ = −1 and √2 sin θ = 1

Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ)=1 + 1

⇒ r2(cos2 θ + sin2 θ) = 2

⇒ r2 = 2                     [cos2 θ + sin2 θ = 1]

⇒ r = √2                  [Conventionally, r > 0]

∴ √2 cos θ =−1 and √2 sin θ = 1

This is the required polar form.