Find the real values of x and y for which:
(1 + i) y2 + (6 + i) = (2 + i)x
Given: (1 + i) y2 + (6 + i) = (2 + i)x
Consider, (1 + i) y2 + (6 + i) = (2 + i)x
⇒ y2 + iy2 + 6 + i = 2x + ix
⇒ (y2 + 6) + i(y2 + 1) = 2x + ix
Comparing the real parts, we get
y2 + 6 = 2x
⇒ 2x – y2 – 6 = 0 …(i)
Comparing the imaginary parts, we get
y2 + 1 = x
⇒ x – y2 – 1 = 0 …(ii)
Subtracting the eq. (ii) from (i), we get
2x – y2 – 6 – (x – y2 – 1) = 0
⇒ 2x – y2 – 6 – x + y2 + 1 = 0
⇒ x – 5 = 0
⇒ x = 5
Putting the value of x = 5 in eq. (i), we get
2(5) – y2 – 6 = 0
⇒ 10 – y2 – 6 = 0
⇒ -y2 + 4 = 0
⇒ - y2 = -4
⇒ y2 = 4
⇒ y = √4
⇒ y = ± 2
Hence, the value of x = 5 and y = ± 2