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$$\sum _{z^{11}=1}\frac{1}{z^{8k}+z^k+1}$$

I saw the answer to this question here which requires advanced computations using algebra systems. I was thinking if there was a more elementary way of solving this. I have tried to come up with a method, considering a quadratic equation where $\alpha_k=z^{8k}+z^k+1$. $x^2-(\sum 1/\alpha_k)x+\prod1/\alpha_k=0$. But to no avail. Besides considering a $10$ degree polynomial is not beneficial because it's roots are $1/\alpha_k$ identically, so we cannot put in a value of $x$ to get the sum. Also it is unclear as to what the roots of the quadratic will be. Or possibly can this expression be factorized.

I have also tried using symmetry, $k\mapsto 11-k$ which gives the following equality. But it is not obvious how to use this equality to arrive at the answer.

$$\sum_{k=1}^{10}\frac{1}{z^{8k}+z^k+1}=\sum_{k=1}^{10}\frac{z^{8k}}{z^{8k}+z^{7k}+1}$$

Any hints on how to proceed are appreciated. Thanks.

Paras Khosla
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  • Also asked here: https://math.stackexchange.com/questions/3610673/summation-of-rational-complex-function Why are so many people so interested in this particular sum? Is it on some exam or something? – Gerry Myerson Apr 19 '20 at 07:34
  • @GerryMyerson I am not able to find the dependence $S_{1,n}$ on $n$, considering the same notation as in the link you shared. Besides, this summand seems a bit more difficult to deal with because of the $8k$ power of $z$ in the denominator. – Paras Khosla Apr 19 '20 at 07:57
  • Sorry, you're right, it's a different sum. – Gerry Myerson Apr 19 '20 at 09:58
  • https://math.stackexchange.com/questions/2116732/let-omega-ei2-pi-2015-evaluate-sum-limits-k-12014-frac11-omega and https://math.stackexchange.com/questions/3630776/evaluate-sum-k-11007-frac2-cos-left-frac2k-pi2015-right12-cos-l – lab bhattacharjee Apr 19 '20 at 15:50
  • @labbhattacharjee This is a different question, the expression has $z^{8k}$ term, while the linked question has $z^{2k}$ term. – Paras Khosla Apr 19 '20 at 18:25
  • I don't think you can avoid calculations. The second approach in linked answer based on Euclidean algorithm seems to be the most reasonable. Essentially you need to find the equation whose roots are $w_k=1/(z_k^8+z_k+1)$ where $z_k$ are roots of $z^{10}+z^{9}+\dots +z+1=0$. – Paramanand Singh Apr 20 '20 at 11:41

0 Answers0