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.

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