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

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

Let ${\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right)$ and ${i^2} = -1$. Then $(x + y{\omega _3} + z{\omega _3}^2)(x + y{\omega _3}^2 + z{\omega _3})$ is equal to:

The value of $\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}$ is

If $Z \neq 0$ is a complex number such that $Z^2 + Z|Z| + |Z|^2 = 0$,then $Z$ is in the set (Here $\omega$ is a complex cube root of unity).

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2018}+z^{2017}+1=0$ satisfy the equation