finding the geometry of complex numbers when you have terms in cube When I solve for the following $$\left|{\frac{1+z}{1-z}}\right|=2$$
I get $$ 3x^2 +3y^2-10x +3=0 $$ and this isnt the equation of a circle nor a line so what geometry is that?
the correct possibilities point to either a circle of radius  4/3 or 5/4 or a circle centered in (0,5/3) or a circle centred in (1,0) or radius 4/3 but with what I ended up with I can't figure which one it is.

For this case:
$$3x^2 +3y^2-10x +3=0 \rightarrow 3\left(x^2 -\frac{10}{3}x\right)+3y^2+3=0$$
$$3\left[\left(x -\frac{5}{3}\right)^2-\frac{25}{9}\right]+3y^2+3=0$$
$$3\left(x -\frac{5}{3}\right)^2-\frac{25}{3}+3y^2+3=0$$
$$\left(x -\frac{5}{3}\right)^2+(y-0)^2=\left(\frac{4}{3}\right)^2$$
It is a circle.
