Let $Z$ be a non real complex number such that $Z^{11}=1$. If $$N = \sum_{k=1}^{10} \frac{1}{Z^{8k}+Z^{k}+1}]$$ Then find N.
$\DeclareMathOperator{cis}{cis}$ My trial
$Z = \cos(\theta) + i \sin(\theta) \Rightarrow Z=\cis(\theta) = \cis(\frac{2c\pi}{11})$ where $c=1,2....11$.
Now $$\begin{align}Z^{8k} + Z^{k} +1 &=\left(\cis\frac{2c\pi}{11}\right)^{8k}+\left(\cis \frac{2c\pi}{11}\right)^{k} +1\\&= \cis \frac{16kc\pi}{11}+\cis \frac{2kc\pi}{11}+1\\&= \left(\cos \frac{16kc\pi}{11}+\cos\frac{2kc\pi}{11}+1\right) +i\left(\sin\frac{16kc\pi}{11}+\sin\frac{2kc\pi}{11}\right)\end{align}$$
I am unable to simplify further after this step.
Trial 2
$Z^{8k} + Z^{k} +1 = Z^{-3k} +Z^{k} +1=\frac{Z^{3k}}{Z^{4k}+ Z^{3k} +1}$
...... Struck up after this. I even tried changing the denominator to the form $Z^{k} -1$ but was unable to do so. In all the ways I am stuck up after few steps. Any hints/help will be appreciated. Thanks in advance
Working with a Smaller Denominator
Note that $$ \begin{align} \sum_{z^{11}=1}\frac1{z^8+z+1} &=\sum_{z^{11}=1}\frac{z^3}{z^4+z^3+1}\\ \end{align} $$ Furthermore, $$ \begin{align} p(z) &=\frac{z^{44}+12z^{33}+44z^{22}+11z^{11}+1}{z^4+z^3+1}\\ &=z^{40}-z^{39}+z^{38}-z^{37}+z^{35}-2z^{34}+3z^{33}-3z^{32}+2z^{31}\\ &\phantom{=\ }+9z^{29}-6z^{28}+4z^{27}-4z^{26}-5z^{25}+11z^{24}-15z^{23}+19z^{22}-14z^{21}\\ &\phantom{=\ }+3z^{20}+12z^{19}+13z^{18}+z^{17}-4z^{16}-8z^{15}-5z^{14}+4z^{13}+8z^{11}\\ &\phantom{=\ }-3z^{10}-z^9+z^8+2z^7+z^6-z^4-z^3+1\\ &\equiv-20z^{10}+4z^9+13z^8+25z^7-5z^6+z^5-14z^4-11z^3+16z^2-17z+31\\ &=q(z)\quad\left(\text{mod }z^{11}-1\right) \end{align} $$ Thus, $$ \begin{align} \sum_{z^{11}=1}\frac{69\,z^3}{z^4+z^3+1} &=\sum_{z^{11}=1}\frac{\left(z^{44}+12z^{33}+44z^{22}+11z^{11}+1\right)z^3}{z^4+z^3+1}\\[3pt] &=\sum_{z^{11}=1}q(z)z^3\\ &=11\left[z^8\right]q(z)\\[9pt] &=143 \end{align} $$ which means $$ \sum_{z^{11}=1}\frac{z^3}{z^4+z^3+1}=\frac{143}{69} $$ Subtracting $\frac13$ for the $z=1$ term, we get an answer of $$ \sum_{\substack{z^{11}=1\\z\ne1}}\frac1{z^8+z+1}=\frac{40}{23} $$
Using the Extended Euclidean Algorithm
It finally dawned on me that what we are looking for is $$ \frac1{z^8+z+1}\quad\left(\text{mod }x^{11}-1\right) $$ and the easiest way to get that is with the Extended Euclidean Algorithm. Using Mathematica, the command
PolynomialExtendedGCD[z^11-1,z^8+z+1,z]returns
{1,{1/69(-56+17z-16z^2+11z^3+14z^4-z^5+5z^6-25z^7), 1/69(13+4z-20z^2+31z^3-17z^4+16z^5-11z^6-14z^7+z^8-5z^9+25z^10)}}which says that $$ \bbox[5px,border:2px solid #C0A000]{\textstyle\frac1{z^8+z+1}\equiv\frac{25z^{10}-5z^9+z^8-14z^7-11z^6+16z^5-17z^4+31z^3-20z^2+4z+13}{69}\quad\left(\text{mod }z^{11}-1\right)} $$ Using the algorithm outlined in this answer, and applying it to polynomials, we can see what Mathematica did: $$ \begin{array}{|c|c|c|c|c|} \hline{\begin{array}{c}\text{linear combination}\\[-6pt]\text{of $z^{11}-1$}\\[-6pt]\text{and $z^8+z+1$}\end{array}}&\text{coefficient of $z^{11}-1$}&\text{coefficient of $z^8+z+1$}&{\begin{array}{c}\text{quotient of}\\[-6pt]\text{the previous two}\\[-6pt]\text{linear combinations}\end{array}}\\\hline z^{11}-1&1&0\\ z^8+z+1&0&1\\ -z^4-z^3-1&1&-z^3&z^3\\ z^3-z^2+2z+1&z^4-z^3+z^2-z&-z^7+z^6-z^5+z^4+1&-z^4+z^3-z^2+z\\ 5z+1&z^5+z^4-z^3+z^2-2z+1&-z^8-z^7+z^6-z^5+2z^4-z^3+z+2&-z-2\\ \color{#090}{\frac{69}{125}}&\frac{-25z^7+5z^6-z^5+14z^4+11z^3-16z^2+17z-56}{125}&\color{#C00}{\frac{25z^{10}-5z^9+z^8-14z^7-11z^6+16z^5-17z^4+31z^3-20z^2+4z+13}{125}}&\frac{25z^2-30z+56}{125}\\ \textstyle 0&\frac{125}{69}\left(z^8+z+1\right)&-\frac{125}{69}\left(z^{11}-1\right)&\frac{125}{69}(5z+1)\\\hline \end{array} $$
The colored elements in the table above say that $$ \textstyle\frac{\color{#090}{\frac{69}{125}}}{z^8+z+1}\equiv\color{#C00}{\frac{25z^{10}-5z^9+z^8-14z^7-11z^6+16z^5-17z^4+31z^3-20z^2+4z+13}{125}}\quad\left(\text{mod }z^{11}-1\right) $$ Dividing by the green term, we get the boxed result we got from Mathematica.