# Find the real values of x and y for which the complex number

Question:

Find the real values of $x$ and $y$ for which the complex number $\left(-3+i y x^{2}\right)$ and $\left(x^{2}+y+4 i\right)$ are conjugates of each other.

Solution:

Let $z_{1}=-3+i y x^{2}$

So, the conjugate of $Z_{1}$ is

$\overline{z_{1}}=-3-i y x^{2}$

And $z 2=x^{2}+y+4 i$

So, the conjugate of z2 is

$\bar{z}_{2}=x^{2}+y-4 i$

Given that: $\overline{z_{1}}=z_{2} \& z_{1}=\bar{z}_{2}$

Firstly, consider $\overline{z_{1}}=z_{2}$

$-3-i y x^{2}=x^{2}+y+4 i$

$\Rightarrow x^{2}+y+4 i+i y x^{2}=-3$

$\Rightarrow x^{2}+y+i\left(4+y x^{2}\right)=-3+0 i$

Comparing the real parts, we get

$x^{2}+y=-3 \ldots$ (i)

Comparing the imaginary parts, we get

$4+y x^{2}=0$

$\Rightarrow x^{2} y=-4 \ldots$ (ii)

Now, consider ${ }^{Z}_{1}=\bar{Z}_{2}$

$-3+i y x^{2}=x^{2}+y-4 i$

$\Rightarrow x^{2}+y-4 i-i y x^{2}=-3$

$\Rightarrow x^{2}+y+i\left(-4 i-y x^{2}\right)=-3+0 i$

Comparing the real parts, we get

$x^{2}+y=-3$

Comparing the imaginary parts, we get

$-4-y x^{2}=0$

$\Rightarrow x^{2} y=-4$

Now, we will solve the equations to find the value of x and y

From eq. (i), we get

$x^{2}=-3-y$

Putting the value of $x^{2}$ in eq. (ii), we get

$(-3-y)(y)=-4$

$\Rightarrow-3 y-y^{2}=-4$

$\Rightarrow y^{2}+3 y=4$

$\Rightarrow y^{2}+3 y-4=0$

$\Rightarrow y^{2}+4 y-y-4=0$

$\Rightarrow y(y+4)-1(y+4)=0$

$\Rightarrow(y-1)(y+4)=0$

$\Rightarrow y-1=0$ or $y+4=0$

$\Rightarrow y=1$ or $y=-4$

When $y=1$, then

$x^{2}=-3-1$

$=-4$ [It is not possible]

When $y=-4$, then

$x^{2}=-3-(-4)$

$=-3+4$

$\Rightarrow x^{2}=1$

$\Rightarrow x=\sqrt{1}$

$\Rightarrow x=\pm 1$

Hence, the values of $x=\pm 1$ and $y=-4$