If |z^{2} -1| = |z|^{2}+1, then z lies on?

Answers (1)

Let Z = x +iy

|(x+iy)^{2}-1| = |x+iy|^{2}+1

\Rightarrow |(x^{2}-y^{2}+i2xy-1| = \left ( \sqrt{x^{2}+y^{2}} \right )^{2}+1

\Rightarrow \sqrt{(x^2 -y^{2}-1)^2+4x^{2}y^{2}} = {x^{2}+y^{2}}+1

\Rightarrow (x^2 -y^{2}-1)^2+4x^{2}y^{2} = \left ( {x^{2}+y^{2}}+1 \right )^{2}

\Rightarrow 4x^{2}y^{2} = 4x^{2}(y^{2}+1)

\Rightarrow x = 0

Hence z will lie on imaginary axis for every x

 

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