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.

What is the product of all the values of $(1-i)^{\frac{2}{5}}$?

If the number of real roots of $x^9-x^5+x^4-1=0$ is $n$,the number of complex roots having argument on the imaginary axis is $m$,and the number of complex roots having argument in the $2^{nd}$ quadrant is $k$,then $m \cdot n \cdot k = $

For $n > 1$ and $n \in N$, if $z_1, z_2, \ldots, z_n$ are the roots of the equation $(z+1)^n = z^n$, then $\sum_{i=1}^{n-1} \frac{\cot^{-1}(2|\operatorname{Im} z_i|) - 1}{2 \operatorname{Re} z_i} = $

$(\sin \theta + i\cos \theta )^n$ is equal to

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to: