Find the real values of x and y for which:(1 + i) y2 + (6 + i) = (2 +

1.	Find the real values of x and y for which:(1 + i) y2 + (6 + i) = (2 + i)x
Answer» 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

Find the real values of x and y for which:(1 + i) y2 + (6 + i) = (2 + i)x

Answer»

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