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