The roots of the equation $x^4 - 1 = 0$ are

$A$ value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

Let ${z_1}$ and ${z_2}$ be $n^{th}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then $n$ must be of the form

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 $\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta =$

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is