If a > 0 and z = $${{{{\left( {1 + i} \right)}^2}} \over {a - i}}$$, has magnitude $$\sqrt {{2 \over 5}} $$, then $$\overline z $$ is equal to :

Let z $$ \in $$ C be such that |z| < 1.

If $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z, then :

All the points in the set
$$S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )$$ lie on a :

If $$z = {{\sqrt 3 } \over 2} + {i \over 2}\left( {i = \sqrt { - 1} } \right)$$,

then (1 + iz + z⁵ + iz⁸)⁹ is equal to :