Let z₁ = -3 + iyx²

So, the conjugate of z₁ is

and z₂ = x² + y + 4i

So, the conjugate of z₂ is

Given that:

Firstly, consider

- 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

-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