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

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 $$17$$

• 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 $$S = \sum_{r=1}^{25} \frac{z^{2r}+1}{z^{2r} + z^{r} + 1}$$ Apr 11 at 11:01
• Edit: 12 instead of 25 and + $(-i)$ Apr 11 at 11:11
• 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 $$S = -\frac{2 i}{\sqrt{3}}\sum_{r=1}^{25} \frac{1}{Z_r-j}-\frac{1}{Z_r-\bar j}$$ Apr 11 at 11:26
• I did tried that but even after breaking into partial fractions like you did, how will we proceed [email protected] Apr 11 at 11:34

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 The sum over $$1$$ is clear, $$25$$. We consider now for $$k=1,2,3,4,5,6,7,8$$ the sum $$\sum_{0 So the result is: $$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 $$17$$.

• This is a perfect answer Apr 11 at 11:42
• Thanks alot, this worked like a charm. Appreciate it a alot Apr 11 at 11:45