Question

Find how many nonzero complex numbers $z$ satisfy $z=i{z}^2$.

Answer 1

$z=i{z}^2$

Let $z=x+iy$

$z^2=(x+iy{)}^2$

$=x^2-y^2+2ixy$

$i{z}^2=i(x^2-y^2)-2xy$

$x+2xy=0$

$x(1+2y)=0$

$⟹x=0,y=-\frac{1}{2}$

or, $x^2-y^2=y$

When $x=0$

$y(y+1)=0$

$y=0,y=-1$

When $y=-\frac{1}{2}$

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

$x^2=-\frac{1}{4}$(not possible)

So, answer is $2$