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,
$$ \sqrt{\zeta}=\begin{cases} e^{2\pi ik'/n} & \mbox{with}\; 1\le k'\le n,\; \mbox{if}\; n\;\mbox{is odd}\\ e^{\pi ik'/n} & \mbox{with}\; 1\le k'\le 2n,\; \mbox{if}\; n\;\mbox{is even} \end{cases} $$
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 $1 \le k' \le 2n$.