If $z^2+z+1=0$,then find the value of $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$,where $z$ is a complex cube root of unity.

The product of all the values of $(\sqrt{3}-i)^{\frac{3}{7}}$ is

If $\alpha$ is a complex number satisfying the equation $\alpha^{2}+\alpha+1=0$,then $\alpha^{31}$ is equal to

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

Let $\alpha$ and $\beta$ be the roots of $x^2 + \omega x + \omega^2 = 0$,where $\omega$ is an imaginary cube root of unity. If $z = \alpha^9 + i\beta^9$,then the value of $|z|$ is:

If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of unity,then $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega$ is equal to