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Given that $\zeta=e^{2\pi i/19}$, how to find the value of $$S=\sum_{m=1}^{19} \dfrac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$$?

All I could think of was to somehow factorize the denominator and apply some sort of a partial fraction method, but I didn't succeed in that too.

Answers and hints in the right direction appreciated.

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user1001001
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    Maybe it helps that your denominator is $\sum_{n=0}^3(\zeta^m)^n = \frac{(\zeta^m)^4-1}{\zeta^m-1} = \frac{(\zeta^m-1)(\zeta^m+1)(\zeta^m-i)(\zeta^m+i)}{\zeta^m-1} = (\zeta^m+1)(\zeta^m-i)(\zeta^m+i)$ – GDumphart Oct 13 '14 at 14:59
  • Just an idea: The polynomials $P(x)=x^3+x^2+x+1$ and $Q(x)=x^{19}-1$ have no commun roots in $\mathbb{C}$. Hence there exists polynomials $U,V$ such that $UP+VQ=1$. As $U(\zeta^m)=1/P(\zeta^m)$, if $U=u_0+\cdots$, $S=19u_0$. So you have to find $U$. – Kelenner Oct 13 '14 at 15:09

2 Answers2

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Let $X$ be the set $\{\; \zeta, \zeta^2, \zeta^3, \ldots, \zeta^{18}\; \}$. Since $19$ is a prime number, for any integer $m$ relative prime to $19$, the map

$$X \in x\quad \mapsto \quad x^m \in X$$ is a permutation of $X$. Together with the obvious identity $\sum\limits_{x\in X} x = -1$, we have

$$\begin{align} &\sum_{m=1}^{19} \frac{1}{\zeta^{3m} + \zeta^{2m}+\zeta^m + 1}\\ =& \frac14 + \sum_{x\in X} \frac{1}{x^3+x^2+x+1} = \frac14 + \sum_{x\in X} \frac{x-1}{x^4-1} = \frac14 + \sum_{x\in x} \frac{(x^4)^5 - 1}{x^4 - 1}\\ =& \frac14 + \sum_{y\in X} \frac{y^5-1}{y-1} = \frac14 + \sum_{y\in X} ( y^4 + y^3 + y^2 + y + 1 )\\ =& \frac14 + \sum_{z\in X} (z + z + z + z + 1)\\ =& \frac14 -4 + 18 = \frac{57}{4} \end{align} $$

achille hui
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  • Very nice! $\qquad$ – Start wearing purple Oct 13 '14 at 15:44
  • I don't quite understand the language in the answer – user1001001 Oct 13 '14 at 15:52
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    @Pkwssis $\zeta^k, k = 1,\ldots 19$ are all the roots of the polynomial $t^{19} - 1 = 0$. So the sum can be viewed as a sum of $\frac{1}{x^3+x^2+x+1}$ for $x$ over these 19 roots. If one exclude the root $x = 1$, it has the nice property that any sum of the form $\sum_{x\in X}f(x^m), \gcd(m,19) = 1$ over the remaining 18 roots in $X$ is the same as $\sum_{x\in X}f(x)$ (It is just a reordering of summation). In my derivation, I make the substitution $y = x^4$ and $z = y^k$, $k = 1,\ldots,4$ in various line and use this fact to simplify the expression. – achille hui Oct 13 '14 at 16:04
  • @achillehui Thanks! That was indeed helpful – user1001001 Oct 13 '14 at 16:09
  • line 3 is very convenient that $x \mapsto x^k$ is the same as rearranging the terms – cactus314 Oct 13 '14 at 17:19
5

We will use

  • the partial fraction decomposition $$\frac{1}{a^3+a^2+a+1}=-\frac{1}{2}\cdot\frac{1}{-1-a}+\frac{1}{2(1-i)}\cdot\frac{1}{i-a}+\frac{1}{2(1+i)}\cdot\frac{1}{-i-a} \tag{1}$$
  • the fact that for $\zeta=\exp\frac{2\pi i }{N}$ one has $$S_N(z):=\sum_{m=1}^N\frac{1}{z-\zeta^m}=\frac{N z^{N-1}}{z^N-1}.\tag{2}$$ This can be shown using that $$S_N(z)=\frac{d}{dz}\sum_{m=1}^N\ln(z-\zeta^m)=\frac{d}{dz}\ln\prod_{m=1}^N(z-\zeta^m)=\frac{d}{dz}\ln\left(z^N-1\right).$$

Combining (1) and (2), we obtain $$\sum_{m=1}^{19}\frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}=-\frac{1}{2}S_{19}(-1)+\frac{1}{2(1-i)} S_{19}(i)+\frac{1}{2(1+i)} S_{19}(-i)=\frac{57}{4}.$$

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