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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

karthikeya kurella
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1 Answers1

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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 first two rows are given.

Each new row is computed by computing the rightmost element as the quotient of the leftmost elements in the previous two rows. The other elements are computed by subtracting the quotient just computed times the previous row from the row prior to that.

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.

robjohn
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  • Nice - how did you find a multiple of $z^4 + z^3 + 1$ in powers of $z^{11}$? – Paul Sinclair Apr 17 '20 at 16:12
  • Look at $\left{1,z^{11},z^{22},z^{33}\right}\left(\text{mod }z^4+z^3+1\right)$. These form a basis for polynomials mod $z^4+z^3+1$. Write $z^{44}\left(\text{mod }z^4+z^3+1\right)$ as a linear combination of these basis elements (using a $4\times4$ matrix inversion). – robjohn Apr 17 '20 at 16:20
  • Thanks - always good to learn a new trick. – Paul Sinclair Apr 17 '20 at 16:21
  • @PaulSinclair: I added the same solution without changing the denominator. This requires inverting an $8\times8$ matrix. – robjohn Apr 17 '20 at 23:30
  • Ahh, what wonderful days we live in, where inverting an $8\times 8$ matrix is something one is willing to do on a whim. – Paul Sinclair Apr 18 '20 at 00:04
  • These are math times we live in. Math! – robjohn Apr 18 '20 at 07:16
  • @PaulSinclair: I have added another approach using the Extended Euclidean Algorithm that is quite a bit simpler than inverting an $8\times8$ (and perhaps even a $4\times4$) matrix. – robjohn Apr 18 '20 at 13:02