Identity regarding roots of unity

Question

Let $\zeta$ be a primitive $n$-th root of unity and $m \in \{0,1,\dots,n-1\}$. I am interested in finding the value of the following expression:

$$\sum_{k=1}^{n-1}\frac{\zeta^{mk}}{1-\zeta^k}.$$

This has come up in a context where it should be a rational number (in fact it seems like it will be of the form $\frac{r}{2}$ where $r \in \mathbb Z$). For example for $m=0$ I can get the values $\frac{n-1}{2}$ by letting $f(x) = \frac{x^n-1}{x-1}$ and noticing that the desired sum equals $\frac{f'(1)}{f(1)}$. However I am not able to find such a "trick" when $m$ is nonzero.

Answer 1

An idea is to use a Bezout identity with the polynomials $1-x$ and $1+X+\cdots+X^{n-1}$ to express $1/(1-\zeta^k)$ as a polynomial of $\zeta^k$. $$n-(1+X+\cdots+X^{n-1}) = \sum_{k=1}^{n-1} (1-X^k) = (1-X)\sum_{k=1}^{n-1}\Big(\sum_{\ell=0}^{k-1}X^\ell\Big)$$ $$n-(1+X+\cdots+X^{n-1}) = (1-X)\sum_{\ell=0}^{n-2}(n-1-\ell)X^\ell.$$ Applying the equality to $\zeta^k$ for $1 \le k \le n$ yields $$n = (1-\zeta^k)\sum_{\ell=0}^{n-2}(n-1-\ell)\zeta^{k\ell}.$$ Hence $$\sum_{k=1}^{n-1} \frac{\zeta^{km}}{1-\zeta^k} = \frac{1}{n}\sum_{k=1}^{n-1}\zeta^{km}\sum_{\ell=0}^{n-2}(n-1-\ell)\zeta^{k\ell}.$$ $$\sum_{k=1}^{n-1} \frac{\zeta^{km}}{1-\zeta^k} = \frac{1}{n}\sum_{\ell=0}^{n-2}(n-1-\ell)\sum_{k=1}^{n-1}\zeta^{k(m+\ell)}.$$ But $$\sum_{k=1}^{n-1}\zeta^{k(m+\ell)} = \sum_{k=0}^{n-1}\zeta^{k(m+\ell)} - 1 = n 1_{[n \textrm{ divides } m+\ell]} - 1.$$ The only term for which $n$ divides $m+\ell$ is given by $\ell = n-m$. Hence $$\sum_{k=1}^{n-1} \frac{\zeta^{km}}{1-\zeta^k} = \frac{1}{n} \Big((m-1)n - \sum_{\ell=0}^{n-2}(n-1-\ell)\Big).$$ $$\sum_{k=1}^{n-1} \frac{\zeta^{km}}{1-\zeta^k} = \frac{1}{n} \Big((m-1)n - \frac{n(n-1)}{2}\Big) = (m-1) - \frac{n-1}{2}.$$

Answer 2

Note that with $\zeta$ being primitive the powers $\zeta^k$ with $0\le k\le n-1$ just permute the powers of $\zeta = \exp(2\pi i/n)$ so we may take that as our root. Next introduce

$$f(z) = \frac{z^m}{1-z} \frac{n/z}{z^n-1}.$$

We have for $1\le k\le n-1$

$$\mathrm{Res}(f(z); z = \zeta^k) = \frac{\zeta^{km}}{1-\zeta^k}.$$

Residues sum to zero and with $m\lt n$ the residue at infinity is zero. Hence the desired sum must be minus the residue at $z=1.$ We write (the minus from the residue cancels the minus from the $1/(1-z)$ term):

$$- \mathrm{Res}(f(z); z=1) = \mathrm{Res}\left(\frac{z^m}{z-1} \frac{n/z}{(z-1)(1+z+z^2+\cdots+z^{n-1})}; z=1\right) \\ = \left. \left( \frac{nz^{m-1}}{1+z+z^2+\cdots+z^{n-1}} \right)' \right|_{z=1} \\ = \left. \left( \frac{n(m-1)z^{m-2}}{1+z+z^2+\cdots+z^{n-1}} - \frac{nz^{m-1} (1+2z+3z^2+\cdots+(n-1)z^{n-2})} {(1+z+z^2+\cdots+z^{n-1})^2} \right)\right|_{z=1} \\ = \frac{n(m-1)}{n} - \frac{n \frac{1}{2} (n-1) n}{n^2} \\ = m-1 - \frac{1}{2} (n-1).$$