$A$ value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is

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

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

The smallest positive integral value of $n$ such that $\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{n} = 1$ is:

If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1$ where $m, n \in N$,then the greatest common divisor of the least values of $m$ and $n$ is:

${\left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right)^{1000}} = $