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

One of the $15^{\text{th}}$ roots of $-1$ is

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

If $1, a, a^2, \ldots, a^{n-1}$ are the $n$th roots of unity,then $\sum_{i=1}^{n-1} \frac{1}{2-a^i}$ is equal to

If $1, \omega, \omega^2$ are the cube roots of unity,then $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}=$

If $i{z^4} + 1 = 0$,then $z$ can take the value