If $n$ is a positive integer,then $(1+i \sqrt{3})^n+(1-i \sqrt{3})^n$ is equal to

Let $1, \omega$,and $\omega^2$ be the cube roots of unity. The least possible degree of a polynomial with real coefficients,having $2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2$,and $5-\omega-\omega^2$ as roots is

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

If $1, \omega, \omega^2$ are the cube roots of unity,$k$ is a positive integer and $(1-\omega+\omega^2)^{3k} + (1-\omega^2+\omega)^{3k} = (1-\omega+\omega^2)^{3k+1} + (1+\omega-\omega^2)^{3k+1}$,then $k=$

For $n \in N$,$\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^n=$

If $n$ is an integer and $Z = \cos \theta + i \sin \theta$,where $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\frac{1 + Z^{2n}}{1 - Z^{2n}} = $