For a non-real root $z$ of the equation $z^4+z^2+1=0$,find the value of $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^3$.

If the complex numbers $z_1, z_2$ and the origin form an equilateral triangle,then $z_1^2 + z_2^2 = $

$\begin{aligned} & \text{If } z=e^{i \theta} \text{ and } \frac{3 \cos 3 \theta+2 \cos 2 \theta+5 \cos 5 \theta}{3 \sin 3 \theta+2 \sin 2 \theta+5 \sin 5 \theta} \\ & =\frac{i \sum_{r=0}^{10} a_r z^r}{\sum_{r=0}^{10} b_r z^r} \text{ then } \frac{\left(\sum_{r=0}^{10} a_r+\sum_{r=0}^{10} b_r\right)}{10}= \end{aligned}$

Let $n$ denote the number of solutions of the equation $z^{2}+3 \bar{z}=0$,where $z$ is a complex number. Then the value of $\sum_{k=0}^{\infty} \frac{1}{n^{k}}$ is equal to:

If $z=x+iy$ is a complex number such that $\bar{z}^{\frac{1}{3}}=a+ib$,then the value of $\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right)$ is equal to

If $z_1 = 2 - 3i$ and the roots of the equation $z^3 + bz^2 + cz + d = 0$ are $i$,$z_1$,and $\bar{z}_1$,then $b + c + d =$