If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on :

If $$z$$ and $$\omega $$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega $$ is equal to

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 :