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

If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are

Let z and w be complex numbers such that $$\overline z + i\overline w = 0$$ and arg zw = $$\pi $$. Then arg z equals :

If $$z = x - iy$$ and $${z^{{1 \over 3}}} = p + iq$$, then

$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to :