JEE Advanced 2022 — Paper 1 — Question 5

Let $z=x+i y$

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

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

$(1)+(2)$

$\quad 2 x=-4 x y$

$\Rightarrow x=-2 x y \Rightarrow x(1+2 y)=0 \Rightarrow x=0$ or $y=-\frac{1}{2}$

Put $x=0$ in (1) or (2) we get

$y^{2}=y \Rightarrow y=0,1$

$\therefore 2$ complex numbers are possible $0+0 \mathrm{i}$ and $0+\mathrm{i}$

put $y=-\frac{1}{2}$ in (1) or (2)

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

$\Rightarrow x^{2}=\frac{3}{4} \Rightarrow x= \pm \frac{\sqrt{3}}{2}$

$\therefore \frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}$ and $-\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}$ are possible

$\therefore 4$ solutions are possible.