If $1, \omega, \omega^2$ are the cube roots of unity,then $(2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2=$

Let $\alpha, \beta$ be the roots of the equation $x^2-\sqrt{6}x+3=0$ such that $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Let $a, b$ be integers not divisible by $3$ and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+ib)$,where $i=\sqrt{-1}$. Then $n+a+b$ is equal to:

$f(x) = (\cos x + i \sin x) \cdot (\cos 3x + i \sin 3x) \cdots [\cos(2n-1)x + i \sin(2n-1)x]$,$n \in N$. Then $f''(x) = ?$ (Where $i = \sqrt{-1}$)

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8$ is equal to

One of the cube roots of unity is

If $|Z|=2$, $Z_1=\frac{Z}{2} e^{i \alpha}$ and $\theta$ is the $\operatorname{amp}(Z)$, then $\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}}=$