If $\omega$ is a complex cube root of unity and $x = \omega^2 - \omega + 2$,then:

If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of unity,then $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega$ is equal to

The least positive integer $n$ for which $\left( \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} \right)^n = 1$ is?

If $\alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_n$ are real numbers,$\alpha_1 \neq 0$ and $z = \cos \theta + i \sin \theta$ is a root of the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$,then $\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta =$

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity and $\alpha = \omega + 2\omega^2 - 3$,then $\alpha^3 + 12\alpha^2 + 48\alpha + 3$ equals