The area of the polygon, whose vertices are the non-real roots of the equation $$\overline z = i{z^2}$$ is :

Let $$A = \left\{ {z \in C:\left| {{{z + 1} \over {z - 1}}} \right| < 1} \right\}$$ and $$B = \left\{ {z \in C:\arg \left( {{{z - 1} \over {z + 1}}} \right) = {{2\pi } \over 3}} \right\}$$. Then A $$\cap$$ B is :

Let z_{1} and z_{2} be two complex numbers such that $${\overline z _1} = i{\overline z _2}$$ and $$\arg \left( {{{{z_1}} \over {{{\overline z }_2}}}} \right) = \pi $$. Then

Let a circle C in complex plane pass through the points $${z_1} = 3 + 4i$$, $${z_2} = 4 + 3i$$ and $${z_3} = 5i$$. If $$z( \ne {z_1})$$ is a point on C such that the line through z and z_{1} is perpendicular to the line through z_{2} and z_{3}, then $$arg(z)$$ is equal to: