Question: Describe the locus of points z satisfying the given equation
$$|z-1|^2 = |z+1|^2 +6$$
Answer:
$$z=x+iy$$
$$z-1 = (x-1)+iy$$
$$z+1 = (x+1)+iy$$
$$|z-1|^2 = |z+1|^2+6 \implies (x-1)^2+y^2 = (x+1)^2 + y^2 +6$$
$$-4x=6\implies x =- \frac{3}{2}\\ z=- \frac{3}{2}+iy, y\in \mathbb{R}$$
The locus of points z: z is a verticle line with $x =- \frac{3}{2}$
