The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle.

If three complex numbers are in A.P. then they lie on a circle in the complex plane.

If the complex numbers, $${Z_1},{Z_2}$$ and $${Z_3}$$ represent the vertics of an equilateral triangle such that
$$\left| {{Z_1}} \right| = \left| {{Z_2}} \right| = \left| {{Z_3}} \right|$$ then $${Z_1} + {Z_2} + {Z_3} = 0.$$

For complex number $${z_1} = {x_1} + i{y_1}$$ and $${z_2} = {x_2} + i{y_2},$$ we write $${z_1} \cap {z_2},\,\,if\,\,{x_1} \le {x_2}\,\,and\,\,{y_1} \le {y_2}.$$
Then for all complex numbers $$z\,\,with\,\,1 \cap z,$$ we have $${{1 - z} \over {1 + z}} \cap 0.$$