Find the real num

Let us assume, z = (x – iy) (3 + 5i)

z = 3x + 5xi – 3yi – 5yi²

= 2x + 5xi – 3yi + 5y

= (3x + 5y) + i (5x – 3y)

∴ = (3x + 5y) – i (5x – 3y)

It is given in the question that, = - 6 – 24i

Now by equating the real and imaginary parts of the equation, we get:

3x + 5y = - 6 (i)

5x – 3y = 24 (ii)

Now, by multiply equation (i) by 3 and equation (ii) by 5 and then by adding them both we get:

34x = 102

∴

= 3

Now putting the value of x in (i), we get:

3 × 3 + 5y = - 6

5y = - 15

∴

Hence, x = 3 and y = - 3