Square root of a root of unity. Let $n$ be an integer, and let $\zeta=e^{\frac{2\pi ik}{n}}$ be a $n$-th root of unity, with $1\le k\le n$. Can someone explain to me why we have the following,
$$
\sqrt{\zeta}=\begin{cases}
  e^{2\pi ik'/n} & \mbox{with}\; 1\le k'\le n,\; \mbox{if}\; n\;\mbox{is odd}\\
  e^{\pi ik'/n} &  \mbox{with}\; 1\le k'\le 2n,\; \mbox{if}\; n\;\mbox{is even}
\end{cases}
$$
 A: Taking the square root means to take $1/2$ of the exponent  which leads to other roots of unity.
Observe $\zeta=e^{\frac{2\pi ik}{n}} =e^{\frac{2\pi i(k+n)}{n}}$.  So you have the choice of taking $1/2$ of the exponent either in the first or in the second formulation. Depending on wether $k$ and $n$ are even or odd, you want a consistent formulation where, in the exponent,  $k'$ is indeed a positive integer.
case 1: $n$ is odd. You can indeed satisfy $\sqrt\zeta=e^{\frac{\pi ik}{n}} =e^{\frac{\pi i(k+n)}{n}} = e^{\frac{2 \pi ik'}{n}} $, where $k'$ is  a positive integer, as folows: If $k$ is even, you have $k' = \frac{k}2$. This gives you $1 \le k' \le (n-1)/2$. If $k$ is odd, you have $k' = \frac{k+n}2$. This gives you $(1+n)/2 \le k' \le n$. So you cover $1 \le k' \le n$. 
case 2: $n$ is even. Comparing to case 1, you have the problem that for odd $k$, you cannot arrange that, in the exponent,  $k'$ is indeed a positive integer as laid out in case 1. Therefore you need the new formulation
$\sqrt\zeta=e^{\frac{\pi ik}{n}} =e^{\frac{\pi i(k+n)}{n}} = e^{\frac{\pi ik'}{n}} $. Now both choices of either  $k' = k$ or $k' = {k+n}$ solve the equation. So you cover $1 \le k' \le 2n$. 
