Originally Posted by

**Plato** Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation.

$e^z=\exp(z)=|z|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$

This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$

For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$

Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$.

Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$.

For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$

Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$.