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

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(C) Let $z=x+iy$. Given $|z|^2+1=|z^2-1|$.Substituting $z=x+iy$,we get $x^2+y^2+1 = |(x+iy)^2-1| = |x^2-y^2-1+2ixy|$.Squaring both sides: $(x^2+y^2+1)

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the imaginary axis

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(C) Let $z=x+iy$. Given $|z|^2+1=|z^2-1|$.Substituting $z=x+iy$,we get $x^2+y^2+1 = |(x+iy)^2-1| = |x^2-y^2-1+2ixy|$.Squaring both sides: $(x^2+y^2+1)^2 = (x^2-y^2-1)^2 + (2xy)^2$.$(x^2+y^2+1)^2 - (x^2-y^2-1)^2 = 4x^2y^2$.Using $a^2-b^2 = (a+b)(a-b)$:$[(x^2+y^2+1) + (x^2-y^2-1)] 	imes [(x^2+y^2+1) - (x^2-y^2-1)] = 4x^2y^2$.$(2x^2)(2y^2+2) = 4x^2y^2$.$4x^2y^2 + 4x^2 = 4x^2y^2$.$4x^2 = 0 \implies x=0$.The locus $x=0$ represents the imaginary axis.

If a complex number $z$ satisfies $|z|^2+1=|z^2-1|$,then the locus of $z$ is

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