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Question:

I am trying to solve the following summation problem involving complex numbers:

Given a complex number $z$ satisfying $z^{26} = 1$, and the summation:

$$ S = \sum_{r=1}^{25} \frac{1}{z^{18r} + z^{9r} + 1} $$

I've attempted to manipulate the terms inside the sum by expressing $z$ as $e^{i\frac{2\pi k}{26}}$, where $k$ is an integer, but nothing fruitful came out.

However, I'm unable to proceed further towards finding a solution. Could someone please provide guidance on how to approach this problem? Any insights or alternative methods for solving the summation would be greatly appreciated. Thank you!

Answer given is $17$

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    $\begingroup$ I also tried manipulating the expression, $$ S = \sum_{r=1}^{25} \frac{1}{z^{2r} + z^{r} + 1} $$ But couldn't solve any further, I had one more idea of clubbing $i$th term and $26-i$th term, but could reduce it to $$ S = \sum_{r=1}^{25} \frac{z^{2r}+1}{z^{2r} + z^{r} + 1} $$ $\endgroup$ Commented Apr 11 at 11:01
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    $\begingroup$ Edit: 12 instead of 25 and + $(-i)$ $\endgroup$ Commented Apr 11 at 11:11
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    $\begingroup$ Would it work if you write $Z_r^2 + Z_r+1 = (Z_r-j)(Z_r-\bar j)$ ? where $Z_r = z^{9r}$ then you write the sum $$S = -\frac{2 i}{\sqrt{3}}\sum_{r=1}^{25} \frac{1}{Z_r-j}-\frac{1}{Z_r-\bar j}$$ $\endgroup$
    – A. PI
    Commented Apr 11 at 11:26
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    $\begingroup$ I did tried that but even after breaking into partial fractions like you did, how will we proceed [email protected] $\endgroup$ Commented Apr 11 at 11:34

1 Answer 1

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The powers of $z$ can be considered modulo $26$, and $9$ is relatively prime to $20$, so when the power $r$ modulo $26$ runs in the set of numbers $1,2,\dots,25$, the replacement power $s=9r$ modulo $26$ also runs in the same set. We can thus instead compute $$ \begin{aligned} S &= \sum_{0<s<26}\frac 1{z^{2s}+z^s+1} = \sum_{0<s<26}\frac {z^s-1}{z^{3s}-1}\\ &= \sum_{0<s<26}\frac {z^{27s}-1}{z^{3s}-1}\qquad\text{ use now $t=3s$ modulo $26$}\\ &= \sum_{0<t<26}\frac {z^{9t}-1}{z^t-1} = \sum_{0<t<26}(1+z^t+z^{2t}+\dots+z^{8t})\ . \end{aligned} $$ The sum over $1$ is clear, $25$. We consider now for $k=1,2,3,4,5,6,7,8$ the sum $$ \sum_{0<t<26} z^{kt}= -1+\sum_{0\le t<26} z^{kt}=-1+\frac{z^{26k}-1}{z^k-1}=-1\ . $$ So the result is: $$S=25+\sum_{1\le k\le 8}-1=25-8=17\ .$$


Computer check: Using sage the result is confirmed as follows:

F.<z> = CyclotomicField(26)
sum([1/(z^(18*r) + z^(9*r) + 1) for r in [1..25]])

And we obtain the result $17$.

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    $\begingroup$ This is a perfect answer $\endgroup$
    – A. PI
    Commented Apr 11 at 11:42
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    $\begingroup$ Thanks alot, this worked like a charm. Appreciate it a alot $\endgroup$ Commented Apr 11 at 11:45

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