Q. 234.3( 4 Votes )

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

Answer :

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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