$(\sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1})^8 =$

If $n$ is an integer which leaves remainder $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals

Let $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$,then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$, where $i = \sqrt{-1}$, then $(z^{201} - i)^{8}$ is equal to:

If $\omega$ is a cube root of unity but not equal to $1$,then the minimum value of $|a + b\omega + c\omega^2|$ (where $a, b, c$ are integers but not all equal) is

If $z = \frac{-1-i \sqrt{3}}{2}$,then $\sum_{k=1}^{2022} \left(z^k + \frac{1}{z^k}\right)^2 = $