What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$? 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.
 A: 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
$$\sum_{m=1}^{19}\frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}=-\frac{1}{2}S_{19}(-1)+\frac{1}{2(1-i)} S_{19}(i)+\frac{1}{2(1+i)} S_{19}(-i)=\frac{57}{4}.$$ 
A: 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
$$\begin{align}
&\sum_{m=1}^{19} \frac{1}{\zeta^{3m} + \zeta^{2m}+\zeta^m + 1}\\
=& \frac14 + \sum_{x\in X} \frac{1}{x^3+x^2+x+1} 
= \frac14 + \sum_{x\in X} \frac{x-1}{x^4-1}
= \frac14 + \sum_{x\in x} \frac{(x^4)^5 - 1}{x^4 - 1}\\
=& \frac14 + \sum_{y\in X} \frac{y^5-1}{y-1}
=  \frac14 + \sum_{y\in X} ( y^4 + y^3 + y^2 + y + 1 )\\
=& \frac14 + \sum_{z\in X} (z + z + z + z + 1)\\
=& \frac14 -4 + 18 = \frac{57}{4}
\end{align}
$$
