Question:
Let $z=x+$ iy be a non-zero complex number such that $z^{2}=i|z|^{2}$, where $i=\sqrt{-1}$, then $z$ lies on the :
Correct Option: , 3
Solution:
$z=x+i y$
$z^{2}=i|z|^{2}$
$(x+i y)^{2}=i\left(x^{2}+y^{2}\right)$
$\left(x^{2}-y^{2}\right)-i\left(x^{2}+y^{2}-2 x y\right)=0$
$(x-y)(x+y)-i(x-y)^{2}=0$
$(x-y)((x+y)-i(x-y))=0$
$\Rightarrow x=y$
$z$ lies on $y=x$