If a complex number z satisfies |z^2-1|=|z|^2 | vedclass

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(B) Given,$|z^2-1|=|z|^2+1$. Let $z=x+iy$. Then,$|(x+iy)^2-1|=|x+iy|^2+1$. $|x^2-y^2+2ixy-1| = x^2+y^2+1$. $|(x^2-y^2-1)+i(2xy)| = x^2+y^2+1$. Squarin

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the imaginary axis

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(B) Given,$|z^2-1|=|z|^2+1$. Let $z=x+iy$. Then,$|(x+iy)^2-1|=|x+iy|^2+1$. $|x^2-y^2+2ixy-1| = x^2+y^2+1$. $|(x^2-y^2-1)+i(2xy)| = x^2+y^2+1$. Squaring both sides,we get: $(x^2-y^2-1)^2 + (2xy)^2 = (x^2+y^2+1)^2$. $(x^2-y^2)^2 + 1 - 2(x^2-y^2) + 4x^2y^2 = (x^2+y^2)^2 + 1 + 2(x^2+y^2)$. $x^4+y^4-2x^2y^2 + 1 - 2x^2+2y^2 + 4x^2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. $x^4+y^4+2x^2y^2 + 1 - 2x^2+2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. $-2x^2 = 2x^2$. $4x^2 = 0 \Rightarrow x=0$. Since $x=0$,the real part of $z$ is $0$,which means $z$ lies on the imaginary axis.

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

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