# Find the real values of x and y for which the complex number (-3 + iyx2) and (x2 + y + 4i) are conjugates of each other.

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

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