Summation using several roots

Question

Don't these roots sum to $0$? Then how do I evaluate this sum? The question is: $U$ denotes a set of complex roots to $a^{1020} = 1$ and $P$ denotes the set of complex roots to $b^4 = 1$. $K = U - P$.

The equation is $$ \bigg\lvert \sum_{y \epsilon K}^{} \frac{1+y}{-y^3+y^2-y+1}\bigg\rvert^2 $$

Answer 1

Here's a somewhat more pedestrian answer than the very elegant one given by Bran Moehring. Although the underlying ideas are essentially the same, I've used more tedious (and unfortunately, therefore, more error-prone) calculation in place of some of the ingenuity needed to come up with his answer.

Since $\ K=\left\{ e^\frac{(255m+n)2\pi i}{1020}\big\vert\,m,n\in\mathbb{Z},0\le m\le3, 1\le n\le 254\right\}\ $, then if you put $\ r_{m,n}=e^\frac{(255m+n)2\pi i}{1020}\ $, and use Don Thousand's observation you can write your sum as $$ S=\bigg\lvert\sum_{m=0}^3\sum_{n=1}^{254}\frac{\left(1+r_{m,n}\right)^2}{1-r_{m,n}^4}\bigg\rvert^2\ , $$ which can be further simplified by using the facts that $\ r_{m,n}^4=e^\frac{2\pi in}{255}\ $, $\ r_{0,n}^2=r_{2,n}^2=-r_{1,n}^2=-r_{3,n}^2=e^\frac{2\pi in}{510}\ $, $\ r_{0,n}=-r_{2,n}=e^\frac{2\pi in}{1020}\ $, and $\ r_{1,n}=-r_{3,n}=e^\frac{2\pi i(255+n)}{1020}\ $, giving \begin{align} S&=\bigg\lvert\,4\sum_{n=1}^{254}\frac{1}{1-\rho_n}\bigg\rvert^2\ , \end{align} where $\ \rho_n=r_{0,n}^4=e^\frac{2\pi in}{255}\ $. Now observe that $\ \rho_{255-n}=e^{-\frac{2\pi in}{255}}=$$\frac{1}{\rho_n}\ $, so $\ \frac{1}{1-\rho_{255-n}}=\frac{1}{1-\rho_n^{-1}}=-\frac{\rho_n}{1-\rho_n}\ $, $\ \frac{1}{1-\rho_n}+\frac{1}{1-\rho_{255-n}}=1\ $, and the sum becomes $$ S=\bigg\lvert\,4\sum_{n=1}^{177}1\,\bigg\rvert^2=508^2\ . $$

Answer 2

You can apply a series of simplifications until the problem is trivial:

1. Note that $-1 \not\in K,$ and $$y\neq -1 \implies \frac{1+y}{-y^3+y^2-y+1} = \frac{(1+y)^2}{1-y^4}$$

2. Set $f(y) = \dfrac{(1+y)^2}{1-y^4}.$ Then note that $K$ is symmetric over the point $0,$ so $$\sum_{y\in K} f(y) = \sum_{y\in K} \frac{1}{2}(f(y) + f(-y)) = \sum_{y\in K}\frac{1+y^2}{1-y^4} = \sum_{y\in K}\frac{1}{1-y^2}$$

3. Set $g(y) = \dfrac{1}{1-y^2}.$ Then note that $K$ is symmetric over the real line, so $$\sum_{y\in K} g(y) = \sum_{y\in K} \frac{1}{2}(g(y) + g(\bar{y})) = \sum_{y\in K}\frac{1}{2} = \frac{1}{2}|K|$$

4. Since $|U| = 1020, |P| = 4,$ and $P \subset U,$ it follows that $$\frac{1}{2}|K| = \frac{1}{2}(1020 - 4) = 508$$

5. We conclude the answer is $508^2 = 258064$

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Point(s) of Clarification from the Comments: (before I delete them to tidy up)

• By "$K$ is symmetric over the point $0$" I mean $y \in K \iff -y\in K.$ Similarly, by "$K$ is symmetric over the real line" I mean $y \in K \iff \bar{y}\in K$ where $\bar{y}$ denotes the conjugate of $y.$ In both cases, the important part is that the symmetry gives a permutation of $K.$ This is because what follows is an instance of the following more general method: $$\rho : K \to K \text{ is a permutation } \\ \implies \sum_{y\in K} F(y) = \sum_{y\in K} F(\rho(y)) \\ \implies \sum_{y\in K} F(y) = \sum_{y\in K} \left(\frac{F(y) + F(\rho(y))}{2}\right)$$ When $\rho$ has order two (i.e. $\rho(\rho(y)) = y$), which is the case in steps 2 and 3 above, this averaging of $F(y)$ and $F(\rho(y))$ tends to simplify sums. For a more concrete example of this method, note that $\rho(k) = n+1-k$ is a permutation of $\{1,2,3,\ldots, n\},$ so $$\sum_{k=1}^n k = \sum_{k=1}^n \frac{k + (n+1-k)}{2} = \sum_{k=1}^n \frac{n+1}{2} = \frac{n(n+1)}{2}$$