Solve this following

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 :

  1. imaginary axis

  2. real axis

  3. line, $y=x$

  4. line, $y=-x$


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 $y=x$

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