If $a = \sqrt{2i}$,then which of the following is correct?

If $1, \omega, \omega^2$ are the cube roots of unity,$n \in \mathbb{N}$ and $n > 2$,then the least value of $n$ such that $1+\omega$ is a root of $x^n-x=0$ is

Let $\alpha$ and $\beta$ be the roots of $x^{2}+x+1=0$. If $n$ is a positive integer, then $\alpha^{n}+\beta^{n}$ is

$\sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ is equal to

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

The roots of the equation $x^3-3x^2+3x+7=0$ are $\alpha, \beta, \gamma$ and $\omega, \omega^2$ are complex cube roots of unity. If the terms containing $x^2$ and $x$ are missing in the transformed equation when each one of these roots is decreased by $h$,then $\frac{\alpha-h}{\beta-h}+\frac{\beta-h}{\gamma-h}+\frac{\gamma-h}{\alpha-h}=$