$(\sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1})^8 =$

If $1, a, a^2, \ldots, a^{n-1}$ are the $n$th roots of unity,then $\sum_{i=1}^{n-1} \frac{1}{2-a^i}$ is equal to

The value of $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} $ is

Express $\frac{(\cos 2\theta - i\sin 2\theta)^4 (\cos 4\theta + i\sin 4\theta)^{-5}}{(\cos 3\theta + i\sin 3\theta)^{-2} (\cos 3\theta - i\sin 3\theta)^{-9}}$ in the form $x + iy$.

If $x + \frac{1}{x} = 2\cos \alpha$,then $x^n + \frac{1}{x^n} = $

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