Question

For what real value of $x$ and $y$ are the number $-3+i{x}^{2}y$ and $x^2+y+4i$ conjugate complex?

Answer 1

Given two complex number

$z_1=-3+i{x}^{2}y{z}_2=x^2+y+4i$

For $z_1$ &$z_2$ to be complex conjugate of each other i.e.,$z_1=\overline{z_2}$ or vice versa.

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

Comparing both sides for real and imaginary part

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

Using $\left(ii\right)$ in $\left(i\right)$ we get

$y-\frac{4}{y}=-3y^2+3y-4=0\Rightarrow y^2+4y-y-4=0\Rightarrow (y+4)(y-1)=0\therefore y=4or-1$

Now,$x^2=\frac{-4}{y}>0$(Since Square of a number is always positive)

$\therefore$ we can chose only $y=-1$

$\therefore x=\pm 2$

Hence $x$ can assume $\pm 2$ and $y$ can assume $-1$.