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