To compute the sum using 2011 th roots of unity

Question

Question:

Let $x$ be a complex number such that $x^{2011}=1$ and $x\neq1$ then, compute the sum

$$S=\dfrac{x^2}{x-1}+\dfrac{x^4}{x^2-1}+\dfrac{x^6}{x^3-1}+\dots+\dfrac{x^{4020}}{x^{2010}-1}$$

My attempt:

$$x=(1)^{\frac{1}{2011}}\implies x=1,e^{\frac{2\pi i}{2011}},e^{\frac{4\pi i}{2011}},\dots,e^{\frac{2010\pi i}{2011}}$$

let, $\alpha= e^{\frac{2\pi i}{2011}},$ then roots of given equation are $x=\alpha,\alpha^2,\alpha^3,\dots,\alpha^{2010}$

also

$$\alpha^{2010}=\dfrac{1}{\alpha};\ \alpha^{2009}=\dfrac{1}{\alpha^2};\ \dots; \ \alpha^{1006}=\dfrac{1}{\alpha^{1005}}$$

now, we have to compute:

$$S= \dfrac{\alpha^2}{\alpha-1}+\dfrac{\alpha^4}{\alpha^2-1}+\dfrac{\alpha^6}{\alpha^3-1}+\dots+\dfrac{\alpha^{4020}}{\alpha^{2010}-1}$$

then, I combined first term and last term, similarly second term and second last term and so on in this way i ended up here

$$S=\dfrac{\alpha^{2}-\alpha^{-3}}{\alpha-1}+\dfrac{\alpha^{4}-\alpha^{-6}}{\alpha^2-1}+.........+\dfrac{\alpha^{2010}-\alpha^{-1006}}{\alpha^{1005}-1}$$

here i got stuck , somehow i just wanted to exploit the property that sum of all roots of unity (who itself are in G.P) is zero but I couldn't

please give me hint or provide right way to solve this question.

thank you!

edit:

I also tried taking logarithmic derivative of $f(x)=x^{2011}-1$ but no progress further

Answer 1

$$ \sum_{k=1}^{2010}\frac{x^{2k}}{x^k-1}=\sum_{k=1}^{2010}\left(1+x^k+\frac{1}{x^k-1}\right) \tag{1}$$ and the minimal polynomial of a primitive $2011$-th root of unity is $\frac{x^{2011}-1}{x-1}=1+x+x^2+\ldots+x^{2010}$.
By denoting the primitive $2011$-th roots of unity as $\zeta^1,\zeta^2,\ldots,\zeta^{2010}$, the RHS of $(1)$ equals $$ 2010-1+\sum_{k=1}^{2010}\frac{1}{\zeta^k-1}\tag{2}$$ with the last sum being the sum of the reciprocal of the roots of $\frac{(x+1)^{2011}-1}{x}$.
Vieta's formulas hence convert the RHS of $(2)$ into $$ 2009 - \frac{1}{2011}\binom{2011}{2} = 2009-1005=\color{red}{1004}.\tag{3}$$

Answer 2

Supposing that $\zeta$ is a primitive $n$th root of unity we consider

$$f(z) = \frac{z^2}{z-1} \frac{n z^{n-1}}{z^n-1} = \frac{1}{z-1} \frac{n z^{n+1}}{z^n-1} = \frac{n}{z-1} \left(z + \frac{z}{z^n-1}\right) \\ = n \left(1 + \frac{1}{z-1} \right) \left(1 + \frac{1}{z^n-1}\right).$$

We seek $$S_n = \sum_{k=1}^{n-1} \frac{\zeta^{2k}}{\zeta^k-1}.$$

We have by inspection that

$$S_n = \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$

With residues summing to zero this implies

$$S_n = - \mathrm{Res}_{z=1} f(z) - \mathrm{Res}_{z=\infty} f(z).$$

For the first residue we note that

$$\frac{1}{z^n-1} = \frac{1}{n} \frac{1}{z-1} + \cdots$$

and the constant term is

$$\left. \frac{1}{z^n-1} - \frac{1}{n} \frac{1}{z-1} \right|_{z=1} \\ = \left. \frac{1}{z-1} \left(\frac{1}{1+z+\cdots+z^{n-1}} - \frac{1}{n}\right)\right|_{z=1}$$

which is by L'Hopital

$$\left. - \frac{1+2z+\cdots+(n-1)z^{n-2}}{(1+z+\cdots+z^{n-1})^2} \right|_{z=1} = - \frac{1}{2} (n-1) n \frac{1}{n^2} = - \frac{1}{2n} (n-1)$$

and we may write

$$f(z) = n \left(1 + \frac{1}{z-1} \right) \left(1 + \frac{1}{n} \frac{1}{z-1} - \frac{1}{2n} (n-1) \cdots\right)$$

so that the residue is

$$n \left( 1 + \frac{1}{n} - \frac{1}{2n} (n-1) \right) = n + 1 - \frac{1}{2} (n-1) = \frac{1}{2} n + \frac{3}{2}.$$

For the residue at infinity we find with $n\ge 2$

$$- \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) = -n \mathrm{Res}_{z=0} \frac{1}{z^2} \left(1 + \frac{1}{1/z-1} \right) \left(1 + \frac{1}{1/z^n-1}\right) \\ = -n \mathrm{Res}_{z=0} \frac{1}{z^2} \left(1 + \frac{z}{1-z} \right) \left(1 + \frac{z^n}{1-z^n}\right) \\ = -n [z^1] \left(1 + \frac{z}{1-z} \right) \left(1 + \frac{z^n}{1-z^n}\right) = - n.$$

Collecting everything yields the closed form

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2} n - \frac{3}{2}.}$$

This will produce $1004$ for $n=2011$, confirming the answer by @JackDAurizio that appeared first.