If $\omega$ is a complex cube root of unity,then $\cos \left[\left(\omega^{1234}+\omega^{2021}\right) \pi-\frac{\pi}{4}\right]$ is equal to

The least positive integral value of $n$ such that $\left[\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right]^n=1$ is

If $\omega_0, \omega_1, \ldots, \omega_{n-1}$ are the $n$-th roots of unity,then $(1+2 \omega_0)(1+2 \omega_1)(1+2 \omega_2) \ldots (1+2 \omega_{n-1})=$

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:

If $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ and $\alpha, \beta$ are the complex cube roots of unity,then $x^3+y^3+z^3=$

Let $\alpha = \frac{-1 + i \sqrt{3}}{2}$. If $a = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k}$ and $b = \sum_{k=0}^{100} \alpha^{3k}$,then $a$ and $b$ are the roots of the quadratic equation: