I've gotten stuck on a particular sum, to which I think I know the answer (thanks to Wolfram:Alpha), but not the method leading to it. I wonder if someone here can help me solve it.
Let $d$ be a positive, odd integer, and define $u$ to be the first nontrivial $d$th root of unity (that is, $u = \exp\left(\frac{2\pi i}{d}\right)$). Furthermore, let $k$ and $m$ be two integers s.t. $0\leq k,m \leq d-1$. I would like to evaluate the sum:
$$S =\sum_{j=0}^{d-1}\frac{u^{-kj}}{1+u^{mj}}$$ Now, after experimenting in W:A for a bit, I managed to find an interesting result. Letting $A = [0,\frac{d-1}{2}]$ and $B = [\frac{d+1}{2},d-1]$... $$S=\begin{cases}-\frac{1}{2}d \hspace{10pt}\text{ if } (k\in A\land m\in B)\lor (k\in B \land m\in A) \\ \frac{1}{2}d \hspace{18pt}\text{ if } (k\in A\land m\in A)\lor (k\in B \land m\in B) \text{ or } k=0\\ 0 \hspace{26pt}\text{ if } m=0 \end{cases}$$
Such an elegant solution indeed, but I have no earthly idea on how to solve this. I've tried to conjugate $S$, fiddling with the exponents a bit and such, which gave rise to a few nice identities, but none proving useful enough to actually solve this. Is this correct? How would I show this? Any suggestions?
The conjecture in the OP is not quite correct but the answer (for $m \ne 0$) is indeed $\pm d/2$ though the sign depends on the parity of the smallest nonnegative solution $n_0$ to $d|(mn-k)$; for $k=0$ obviously $n_0=0$ so indeed the sign is plus, but for example for $d=5, m=2, k=1$ we have $n_0=3$ so the sign is actually minus although $m,k \in A$ in the OP notation. Similarly, for $m=1$ we get $n_0=k$ recovering the answer in the comment above.
Proof: We assume $m \ne 0$ (as the result is clearly $0$ for $m=0$), let $r<1$ and consider $f(r)=\sum_{j=0}^{d-1}\frac{u^{-kj}}{1+ru^{mj}}$ so $f(r) \to S, r \to 1$. But (by absolute convergence interchanging sum with series): $$f(r)=\sum_{j=0}^{d-1}\sum_{n \ge 0}(-1)^nr^nu^{j(mn-k)}=\sum_{n \ge 0}(-1)^nr^n\sum_{j=0}^{d-1}u^{j(mn-k)}$$ so by orthogonality we have $$f(r)=d\sum_{n \ge 0, d|mn-k}(-1)^nr^n$$
If $n_0$ is the first $n \ge 0$ for which $d|(mn_0-k)$ then all such $n$ are of the form $n_0+lq, l =0,1..$ where $q=d/(d,m$) so since $d$ is odd, hence $q$ is odd and $(-1)^{lq}=(-1)^l$ we have $$f(r)=(-1)^{n_0}dr^{n_0}\sum_{l \ge 0}(-1)^lr^{lq}=(-1)^{n_0}dr^{n_0}\frac{1}{1+r^q}$$ Letting $r \to 1$ we get that $S=(-1)^{n_0}d/2$ as claimed and we are done!