Let z1 = - 3 + iyx2
So, the conjugate of z1 is
\(\bar z_1\) = - 3 - iyx2
And z2 = x2 + y + 4i
So, the conjugate of z2 is
\(\bar z_2\) = x2 + y - 4i
Given that: \(\bar z_1\) = z2 and z1 = \(\bar z_2\)
Firstly, consider \(\bar z_1\) = z2
- 3 – iyx2 = x2 + y + 4i
⇒ x2 + y + 4i + iyx2 = -3
⇒ x2 + y + i(4 + yx2) = -3 + 0i
Comparing the real parts, we get
x2 + y = -3 …(i)
Comparing the imaginary parts, we get
4 + yx2 = 0
⇒ x2y = -4 …(ii)
Now, consider z1 = \(\bar z_2\)
-3 + iyx2 = x2 + y – 4i
⇒ x2 + y – 4i – iyx2 = - 3
⇒ x2 + y + i(-4i – yx2) = - 3 + 0i
Comparing the real parts, we get
x2 + y = -3
Comparing the imaginary parts, we get
-4 – yx2 = 0
⇒ x2y = -4
Now, we will solve the equations to find the value of x and y
From eq. (i), we get
x2 = - 3 – y
Putting the value of x2 in eq. (ii), we get
(- 3 – y)(y) = - 4
⇒ - 3y – y2 = - 4
⇒ y2 + 3y = 4
⇒ y2 + 3y – 4 = 0
⇒ y2 + 4y – y – 4 = 0
⇒ y(y + 4) – 1(y + 4) = 0
⇒ (y – 1)(y + 4) = 0
⇒ y – 1 = 0 or y + 4 = 0
⇒ y = 1 or y = -4
When y = 1, then
x2 = - 3 – 1 = - 4 [It is not possible]
When y = - 4, then
x2 = - 3 –(-4) = - 3 + 4
⇒ x2 = 1
⇒ x = √1
⇒ x = ± 1
Hence, the values of x = ± 1 and y = - 4