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

Sep 8, 2016

${\left({x}^{2} + {y}^{2} + 1\right)}^{2} - 4 {x}^{2} = 9$

#### Explanation:

$z = x + i y$

$\left\mid z - 1 \right\mid = \sqrt{\left(z - 1\right) \left({z}^{\cdot} - 1\right)} = \sqrt{\left(x + i y - 1\right) \left(x - i y - 1\right)} = \sqrt{{x}^{2} + {y}^{2} - 2 x + 1}$

also

$\left\mid z + 1 \right\mid = \sqrt{\left(z + 1\right) \left({z}^{\cdot} + 1\right)} = \sqrt{\left(x + i y + 1\right) \left(x - i y + 1\right)} = \sqrt{{x}^{2} + {y}^{2} + 2 x + 1}$

so the locus is

$\sqrt{{x}^{2} + {y}^{2} - 2 x + 1} \sqrt{{x}^{2} + {y}^{2} + 2 x + 1} = 3$ squaring both sides

${\left({x}^{2} + {y}^{2} + 1\right)}^{2} - 4 {x}^{2} = 9$

Attached tle locus plot