Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Let Z1 = (x – iy) (3 + 5i) = 3c + 5xi - 3yi - 5yi2 = (3x + 5y) + i ( 5x - 3y)
Let Z2 = -6 -24i
5
It is given that Z1 is conjugate of Z2
Z1 =
(3x + 5y) + i (5x - 3y) = -6 + 24 i
Equating real and imaginary parts, we get
3x + 5y = - 6 ....(i)
and 5x - 3y = 24 ...(ii)
Multiplying (i) by 5 and (ii) by 3, we get
15x + 25y = -30 ...(iii)
15x - 9y = 72 ...(iv)
Subtracting (iv) from (iii), we get
34y = -102 or
Putting y = -3 in (i), we get
3x + 5 (-3) = -6
or 3x = -6 +15 = 9 or x = 3
Hence, x = 3, y = -3