How do you find the locus defined by $abs(z-1)abs(z +1) =3$ ?

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Answer 1 · 2016-09-08T14:00:26

$(x^2+y^2+1)^2-4x^2=9$

$z = x + i y$

$abs(z-1) = sqrt((z-1) (z^*-1)) = sqrt((x+iy-1)(x-iy -1)) = sqrt(x^2+y^2-2x+1)$

also

$abs(z+1)=sqrt((z+1) (z^*+1)) = sqrt((x+iy+1)(x-iy +1)) =sqrt(x^2+y^2+2x+1)$

so the locus is

$sqrt(x^2+y^2-2x+1)sqrt(x^2+y^2+2x+1)=3$ squaring both sides

$(x^2+y^2+1)^2-4x^2=9$

Attached tle locus plot

enter image source here

How do you find the locus defined by abs(z-1)abs(z +1) =3 abs(z-1)abs(z +1) =3?