If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$, where $i = \sqrt{-1}$, then $(z^{201} - i)^{8}$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity,then their product is

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity and $1, \alpha, \alpha^2, \alpha^3$ are the fourth roots of unity in usual notation,then $\alpha+\alpha \omega-\alpha^3 \omega^2=$

For $n > 1$ and $n \in N$, if $z_1, z_2, \ldots, z_n$ are the roots of the equation $(z+1)^n = z^n$, then $\sum_{i=1}^{n-1} \frac{\cot^{-1}(2|\operatorname{Im} z_i|) - 1}{2 \operatorname{Re} z_i} = $

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2014}+z^{2015}+1=0$ are