Question

Find the value of $x$ and $y$ for which $(2+3i)x^2-(3-2i)y=2x-3y+5i$ where $x,yϵR.$

Answer 1

$(2+3i)x^2-(3-2i)y=2x-3y+5i$
$\Rightarrow$ $2x^2-3y=2x-3y$
$\Rightarrow$ $x^2-x=0$
$\Rightarrow$ $x=0,1$ and $3x^2+2y=5$
$\Rightarrow$ if $x=0,y=\frac{5}{2}$ and if $x=1,y=1$
$\therefore$ $x=0,y=\frac{5}{2}$ and $x=1,y=1$
are two solutions of the given equation which can also be represented as $(0,\frac{5}{2})$ & $(1,1),(0,\frac{5}{2})(1,1)$