Complex summation?

This is actually not too bad: let $n = 2015,$ and find $g$ such that $3g \equiv 1 \pmod n.$ Now $\begin{aligned} \sum_{k=1}^{n-1} \frac1{1+\omega^k + \omega^{2k}} &= \sum_{k=1}^{n-1} \frac{1-\omega^k}{1-\omega^{3k}} \\ &= \sum_{k=1}^{n-1} \frac{1-\omega^{3gk}}{1-\omega^{3k}} \\ &= \sum_{k=1}^{n-1} \left( 1 + \omega^{3k} + \omega^{6k} + \cdots + \omega^{3(g-1)k} \right) \\ &= -g + \sum_{k=0}^{n-1} \left( 1 + \omega^{3k} + \omega^{6k} + \cdots + \omega^{3(g-1)k} \right) \\ &= n-g + \sum_{k=0}^{n-1} \left( \omega^{3k} + \omega^{6k} + \cdots + \omega^{3(g-1)k} \right) \\ &= n-g, \end{aligned}$ because for $\zeta$ any $n$ th root of unity, $\sum_{k=0}^{n-1} \zeta^k = \begin{cases} n &\text{if } \zeta=1 \\ 0 &\text{otherwise.}\end{cases}$ (This is a standard fact: multiplying the sum by $\zeta$ permutes the terms, so $\zeta S = S,$ so $S= 0$ unless $\zeta = 1.$ )

For $n=2015,$ we get $g = 672,$ and the answer is $n-g = \fbox{1343}.$

Edit: I thought this problem was familiar. I solved it five years ago , in pretty much the same way!

If $u,v$ are complex numbers of modulus $1$ such that $1 + u + u^2 = 1 + v + v^2$ , then $(u-v)(u+v+1) = 0$ , so either $u=v$ or $u+v+1=0$ , and this means that either $u=v$ or else $\{u,v\} = \{e^{\frac{2\pi i}{3}},e^{-\frac{2\pi i}{3}}\}$ . Thus if $N \ge 1$ is an integer such that $2N+1$ is not divisible by $3$ , and if we define $\omega = e^{\frac{2\pi i}{2N+1}}$ , then none of $\omega^k$ (for $0 \le k \le 2N$ ) are equal to $e^{\pm\frac{2\pi i}{3}}$ , and hence we deduce that $1 + \omega^k + \omega^{2k}$ , for $0 \le k \le 2N$ are distinct complex numbers. We shall find the polynomial (of degree $2N+1$ ) whose roots are $1 + \omega^k + \omega^{2k}$ for $0 \le k \le 2N$ .

If $X = 1 + \omega^k + \omega^{2k}$ , then $\big(\omega^k + \tfrac12\big)^2 = X - \tfrac34$ , and hence $\omega^k = -\tfrac12 \pm \sqrt{X - \tfrac34}$ , and hence $\begin{aligned} 0 & = \; 1 - \left(-\tfrac12 \pm \sqrt{X - \tfrac34}\right)^{2N+1} \; = \; 1 + \left(\tfrac12 \mp \sqrt{X - \tfrac34}\right)^{2N+1} \\ & = \; 1 + \sum_{j=0}^{2N+1}\binom{2N+1}{j}\left(\tfrac12\right)^{2N+1-j}\left(\mp \sqrt{X-\tfrac34}\right)^j \\ & = \; 1 + \sum_{j=0}^N \binom{2N+1}{2j} \left(\tfrac12\right)^{2(N-j)+1} \left(X - \tfrac34\right)^j \mp \sqrt{X-\tfrac34}\sum_{j=0}^N \binom{2N+1}{2j+1} \left(\tfrac{1}{2}\right)^{2(N-j)}\left(X - \tfrac34\right)^j \end{aligned}$ and hence $f(X) \; = \; \left(X - \tfrac34\right)\left[\sum_{j=0}^N \binom{2N+1}{2j+1} \left(\tfrac{1}{2}\right)^{2(N-j)}\left(X - \tfrac34\right)^j\right]^2 - \left[1 + \sum_{j=0}^N \binom{2N+1}{2j} \left(\tfrac12\right)^{2(N-j)+1} \left(X - \tfrac34\right)^j\right]^2 \; = \; 0$ Now $f(X)$ is a polynomial of degree $2N+1$ , and it has $1 + \omega^k + \omega^{2k}$ as a root for any $0 \le k \le 2N$ , and hence it follows that this is the complete list of its roots. Thus we deduce that $\sum_{j=0}^{2N}\frac{1}{1 + \omega^k + \omega^{2k}} \; = \; -\frac{f'(0)}{f(0)}$ and hence $\sum_{j=1}^{2N}\frac{1}{1 + \omega^k + \omega^{2k}} \; =\; -\frac{f'(0)}{f(0)} - \frac13$ Now standard combinatorial tricks tell us that $f(0) \; =\; -\tfrac34\left[\sum_{j=0}^N \binom{2N+1}{2j+1} \left(\tfrac{1}{2}\right)^{2(N-j)}\left(-\tfrac34\right)^j\right]^2 - \left[1 + \sum_{j=0}^N \binom{2N+1}{2j} \left(\tfrac12\right)^{2(N-j)+1} \left(-\tfrac34\right)^j\right]^2 \; = \; -2 - 2\cos\left(\tfrac{(2N+1)\pi}{3}\right)$ and, similarly, $f'(0) \; = \; \tfrac13(1 + 2N) \left(3 - 2\sqrt{3}\sin\left(\tfrac{2N\pi}{3}\right)\right)$ With $N = 1007$ , we deduce that the final answer is $-\frac{f'(0)}{f(0)} - \frac13 = \boxed{1343}$