If a complex number $z$ satisfies $\left|z^2-1\right|=|z|^2+1$, then $z$ lies on

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Aicharya · Accepted Answer

Given, $\quad\left|z^2-1ight|=|z|^2+1$ Let $ z=x+i y $ $ \begin{aligned} & \Rightarrow \quad\left|(x+i y)^2-1ight|=|x+i y|^2+1 \ & \Rightarrow\left|x^2-y^2+2 i x y-1ight|=\left(x^2+y^2ight)+1 \ & \Rightarrow\left|\left(x^2-y^2-1ight)+2 i x yight|=\left(x^2+y^2+1ight) \ & \Rightarrow \sqrt{\left(x^2-y^2-1ight)^2+4 x^2 y^2}=x^2+y^2+1 \ & \Rightarrow\left(x^2-y^2ight)^2+1-2\left(x^2-y^2ight)+4 x^2 y^2 \ & =\left(x^2+y^2+1ight)^2 \ & =x^4+y^4+2 x^2 y^2+1+2 x^2+2 y^2 \ & \Rightarrow-2 x^2 y^2-2 x^2+4 x^2 y^2=2 x^2 y^2+2 x^2 \ & \Rightarrow \quad-2 x^2=2 x^2 \ & \Rightarrow \quad 4 x^2=0 \Rightarrow x=0 \ & 	herefore \quad z=x+i y=0+i y \ & \Rightarrow \quad z=i y \quad \Rightarrow \quad(x, y)=(0, y) \ & \end{aligned} $ Hence, $z$ lies on the imaginary axis

If a complex number $z$ satisfies $\left|z^2-1\right|=|z|^2+1$, then $z$ lies on

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