Evaluate: $[\sqrt{2}(\cos 56^{\circ} 15^{\prime} + i \sin 56^{\circ} 15^{\prime})]^8$

If $\alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_n$ are real numbers,$\alpha_1 \neq 0$ and $z = \cos \theta + i \sin \theta$ is a root of the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$,then $\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta =$

If ${\left( {\frac{{1 + \cos \theta + i\sin \theta }}{{i + \sin \theta + i\cos \theta }}} \right)^4} = \cos n\theta + i\sin n\theta $,then $n$ is equal to

Let $r$ be a real number and $n \in N$ be such that the polynomial $2x^2+2x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be

Express the following expression in the form $A + iB$: $(\cos 2\theta + i\sin 2\theta )^{ - 5} (\cos 3\theta - i\sin 3\theta )^6 (\sin \theta - i\cos \theta )^3$.

If $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to