Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then :

If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then :

If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by :

z and w are two nonzero complex numbers such that $$\,\left| z \right| = \left| w \right|$$ and Arg z + Arg w =$$\pi $$ then z equals