Some -complex- sum

$\begin{aligned} z^{2m} + z^{2m-1} + z^{2m-2} + ... + z + 1 & = 0 \\ \Rightarrow z^{2m+1} + z^{2m} + z^{2m-1} + ... + z^2 + z & = 0 \\ z^{2m+1} + \color{#3D99F6}{z^{2m} + z^{2m-1} + ... + z^2 + z + 1} & = \color{#3D99F6}{1} \\ z^{2m+1} + \color{#3D99F6}{0} & = \color{#3D99F6}{1} \end{aligned}$

$\Rightarrow z_r = e^{\frac{2r\pi}{2m+1}i}$ are the $(2m+1)^{th}$ complex roots of unity.

$\begin{aligned} \Rightarrow \sum_{r=1}^{2m} \frac{1}{z_r -1} & = \sum_{r=1}^{2m} \frac{1}{e^{\frac{2r\pi}{2m+1}i} -1} \\ & = \sum_{r=1}^{2m} \frac{e^{-\frac{r\pi}{2m+1}i}}{e^{\frac{r\pi}{2m+1}i} -e^{-\frac{r\pi}{2m+1}i}} \\ & = \sum_{r=1}^{2m} \frac{\cos{\frac{r\pi}{2m+1}} - \sin{\frac{r\pi}{2m+1}}i}{\cos{\frac{r\pi}{2m+1}} + \sin{\frac{r\pi}{2m+1}}i - \left(\cos{\frac{r\pi}{2m+1}} - \sin{\frac{r\pi}{2m+1}}i\right)} \\ & = \sum_{r=1}^{2m} \frac{\cos{\frac{r\pi}{2m+1}} - \sin{\frac{r\pi}{2m+1}}i}{2 \sin{\frac{r\pi}{2m+1}}i} \\ & = - \frac{1}{2} \sum_{r=1}^{2m} \left[ \cot{\left(\frac{r\pi}{2m+1}\right)}i+1 \right] \\ & =- \frac{1}{2} \left[ \color{#3D99F6}{0} \sum_{r=1}^{2m} 1 \right] \quad \quad \small \color{#3D99F6}{\left[ \cot{\left(\frac{r\pi}{2m+1}\right)} = - \cot{\left(\frac{2m - r\pi}{2m+1}\right)}\right]} \\ & = \large \boxed{-m} \end{aligned}$

Since the given polynomial is monic, let us consider

$f(z) = z^{2m} + z^{2m-1} + \ldots + z + 1 = (z - z_1)(z-z_2)\ldots(z-z_{2m})$

$\Rightarrow f(z) = \displaystyle \prod_{r=1}^{2m} (z-z_r)$

$\Rightarrow f’(z) = (z-z_2)(z-z_3)\ldots(z-z_{2m}) + (z-z_1)(z-z_3)\ldots(z-z_{2m}) + \ldots + (z-z_1)(z-z_2)\ldots(z-z_{2m-1})$

$\Rightarrow f’(z) = (2m)z^{2m-1} + (2m-1)z^{2m-2} + \ldots + 2z + 1$

Let the expression to be found be $E$

$E = \displaystyle \sum_{r=1}^{2m} \dfrac{1}{z_r -1} = \dfrac{1}{z_1 - 1} + \dfrac{1}{z_2 - 1} + \ldots + \dfrac{1}{z_{2m} - 1}$

$E = \dfrac{(z_2-1)(z_3-1)\ldots(z_{2m}-1) + (z_1-1)(z_3-1)\ldots(z_{2m}-1) + \ldots + (z_1-1)(z_2-1)\ldots(z_{2m-1}-1)}{\displaystyle \prod_{r=1}^{2m} (z_r-1)}$

$E = -\dfrac{(1-z_2)(1-z_3)\ldots(1-z_{2m}) + (1-z_1)(1-z_3)\ldots(1-z_{2m}) + \ldots + (1-z_1)(1-z_2)\ldots(1-z_{2m-1})}{\displaystyle \prod_{r=1}^{2m} (1-z_r)}$

$E = -\dfrac{f’(1)}{f(1)} = -\dfrac{\frac{(2m)(2m+1)}{2}}{2m+1} = \boxed{-m}$

We are going to find an equation with roots $\dfrac{1}{z_r-1}$ , so let $y=\dfrac{1}{z-1}$ . Solving for $z$ we get $z=\dfrac{y+1}{y}$ .

Substituting that, and multipliying everything by $y^{2m}$ we obtain:

$(y+1)^{2m}+y(y+1)^{2m-1}+y^2(y+1)^{2m-2}+\cdots+y^{2m-1}(y+1)+y^{2m}=0$

Now, by Vieta's formulas, the answer that we want is the sum of all roots of the equation above: $-\dfrac{\text{coefficient of } y^{2m-1}}{\text{coefficient of } y^{2m}}$

The coefficient of $y^{2m}$ is $1+1+\cdots+1 \text{ (2m+1 times)}=2m+1$ , and the coefficient of $y^{2m-1}$ is $2m+2m-1+2m-2+\cdots+2+1=\dfrac{2m(2m+1)}{2}=m(2m+1)$

So, the answer is: $-\dfrac{m(2m+1)}{2m+1}=\boxed{-m}$

Note $z^{2m}+\cdots+z+1=\frac{z^{2m+1}-1}{z-1}$ , so the $z_r$ are the nontrivial $(2m+1)$ -th roots of unity. Let $x$ be a primitive $(2m+1)$ -th root of unity, so

$\displaystyle\sum_{r=1}^{2m}\frac{1}{z_r-1}=\sum_{k=1}^{2m}\left(\frac{1}{x^k-1}\right)=\sum_{k=1}^m\left(\frac{1}{x^k-1}+\frac{1}{x^{2m+1-k}-1}\right)$

$\displaystyle=\sum_{k=1}^{m}\left(\frac{1}{x^k-1}+\frac{1}{x^{-k}-1}\right)=\sum_{k=1}^m \frac{x^{-k}-2+x^k}{-x^{-k}+2-x^k}$

$\displaystyle=\sum_{k=1}^m(-1)=\boxed{-m}$ .

This is a stupid solution but I used intuitive 'feeling' as I call it: since one possible root of unity is -1, I took it as the average value for each Zr, so we have sum from 1 to 2m of 1/(-1-1) = 2m*(-0.5) = -m