How do unity roots relate to the summation?

Question

let $w_0,...,w_{2n}$ be the $(2n+1)$-th root of unity. Compute $$S=\sum\limits_{k=0}^{2n}\frac{1}{1+w_k}$$

I tried this many times but I just can't see how it's done.

My idea is simplify the function of $w_k$

$$S=\frac{1}{1+w_0}+…+\frac{1}{1+w_{2n}}$$

From $w^{(2n-1)}=1$

$$S=\frac{1}{2}+…+\frac{1}{1+w_{2n}}$$

At this point it seems tough to me. Can anyone explain or spare some hints on how it relates?

Edit: $(2n+1)$-th root

Answer 1

Since $w_k = e^{2\pi i k/m} $ where $m=2n+1$ (I am assuming the $2n-1$ is a typo),

$\begin{array}\\ w_{m-k} &=e^{2\pi i (m-k)/m}\\ &=e^{2\pi i (1-k/m)}\\ &=e^{-2\pi i k/m}\\ &=e^{-2\pi i k/m}\\ \end{array} $

so

$\begin{array}\\ w_k+w_{m-k} &=e^{2\pi i k/m}+e^{-2\pi i k/m}\\ &=\cos(2\pi k/m)+i\sin(2\pi k/m)+\cos(-2\pi k/m)+i\sin(-2\pi k/m)\\ &=\cos(2\pi k/m)+i\sin(2\pi k/m)+\cos(2\pi k/m)-i\sin(-2\pi k/m)\\ &=2\cos(2\pi k/m)\\ \text{and}\\ w_kw_{m-k} &=e^{2\pi i k/m}e^{-2\pi i k/m}\\ &=1\\ \text{so}\\ \dfrac1{1+w_k}+\dfrac1{1+w_{m-k}} &=\dfrac{1+w_k+1+w_{m-k}}{(1+w_k)(1+w_{m-k})}\\ &=\dfrac{2+2\cos(2\pi k/m)}{1+w_k+w_{m-k}+w_kw_{m-k}}\\ &=\dfrac{2+2\cos(2\pi k/m)}{2+2\cos(2\pi k/m)}\\ &=1\\ \end{array} $

so

$\begin{array}\\ S &=\sum\limits_{k=0}^{m-1}\dfrac{1}{1+w_k}\\ \text{so}\\ 2S &=\sum\limits_{k=0}^{m-1}\dfrac{1}{1+w_k}+\sum\limits_{k=0}^{m-1}\dfrac{1}{1+w_{k}}\\ &=\sum\limits_{k=0}^{m-1}\dfrac{1}{1+w_k}+\sum\limits_{k=1}^{m}\dfrac{1}{1+w_{m-k}}\\ &=\dfrac1{1+w_0}+\sum\limits_{k=1}^{m-1}\dfrac{1}{1+w_k}+\sum\limits_{k=1}^{m-1}\dfrac{1}{1+w_{m-k}}+\dfrac1{1+w_0}\\ &=\dfrac{2}{1+w_0}+\sum\limits_{k=1}^{m-1}\left(\dfrac{1}{1+w_k}+\dfrac{1}{1+w_{m-k}}\right)\\ &=1+\sum\limits_{k=1}^{m-1}\left(1\right)\\ &=m\\ \text{so}\\ S &=\dfrac{m}{2}\\ \end{array} $

Note that we need $m$ to be odd - if $m=2n$ then $1+w_n=0$ so the sum does not exist.

Answer 2

Expanding on my comment, the following proves the more general statement.

Let $P(w)$ be a complex polynomial of degree $n$ with roots $w_k \,\big|_{k = 0, 1, \dots,n - 1}$, and let $a, b \in \mathbb C\,, a \ne 0$ such that $aw_k + b \ne 0$ for all $k=0,1,\dots,n - 1$. Then: $$\sum_{k=0}^{n-1} \frac{1}{aw_k + b} = -\,\frac{P'\left(- \dfrac{b}{a}\right)}{a\,P\left(- \dfrac{b}{a}\right)}$$

Let $z= aw+b \iff w = \dfrac{z-b}{a}$ then $z_k = aw_k + b$ are the roots of $Q(z) = P\left(\dfrac{z-b}{a}\right)$ $=q_nz^n + q_{n-1}z^{n-1} + \dots + q_1 z + q_0$ where $q_0 = Q(0) = P\left(- \dfrac{b}{a}\right)\,$, $q_1 = Q'(0) = \dfrac{1}{a} P'\left(- \dfrac{b}{a}\right)$. It follows from Vieta's relations that $\sum_{k=0}^{n-1} \dfrac{1}{z_k} = - \dfrac{q_1}{q_0}$ which completes the proof.

OP's problem is the case $n \mapsto 2n+1$, $P(w) = w^{2n+1} - 1$, $P'(w) = (2n+1) w^{2n}$, $a=b=1$, so: $$\sum_{k=0}^{2n} \dfrac{1}{1 + w_k} = -\dfrac{P'(-1)}{P(-1)} = - \dfrac{2n+1}{-2} = \dfrac{2n+1}{2}$$