Question

Find real values of $x$ and $y$ for which the complex numbers $-3+i{x}^{2}y$ and $x^2+y-4i$, where $i=\sqrt{-1}$, are conjugate to each other.

Answer 1

Given:Conjugate of $-3+i{x}^{2}y=x^2+y-4i$

Comparing real and imaginary parts, we get

$x^2+y=-3$

$x^{2}y=-4$

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

$\Rightarrow x^2+y=-3$

$\Rightarrow \frac{-4}{y}+y=-3$

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

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

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

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

$\Rightarrow y=-4,1$

$\Rightarrow x^2=\frac{-4}{-4}=1,x^2=\frac{-4}{1}=-4$

$\Rightarrow x=\pm 1,x=\pm 2i$

$x=\pm 2i$ is impossible since $x$ is real.

$\therefore x=\pm 1,y=-4$