Bizarre complex numbers

We have $z=e^{4 \pi i / 7}$ . By the geometric series we have $z^6+z^5+z^4+z^3+z^2+z+1=\dfrac{z^7-1}{z-1}=0$ . Let $S$ be the expression, we can solve this using two methods:

Brute force

Simply sum the fractions: $\begin{aligned} S &= \dfrac{z(1+z^4)(1+z^6) + z^2(1+z^2)(1+z^6) + z^3(1+z^2)(1+z^4)}{(1+z^2)(1+z^4)(1+z^6)} \\ &= \dfrac{z(1+z^4+z^6+z^3) + z^2(1+z^2+z^6+z) + z^3(1+z^2+z^4+z^6)}{1+z^2+z^4+z^6+z^6+z+z^3+z^5} \\ &= \dfrac{(z+z^5+1+z^4) + (z^2+z^4+z+z^3) + (z^3+z^5+1+z^2)}{(1+z+z^2+z^3+z^4+z^5+z^6)+z^6} \\ &= \dfrac{2(1+z+z^2+z^3+z^4+z^5)}{z^6} \\ &= \dfrac{2(-z^6)}{z^6} = -2\\ &\implies \lvert S \rvert = \boxed{2} \end{aligned}$

Constructing a polynomial

Notice that $S=\dfrac{1}{z+\dfrac{1}{z}} + \dfrac{1}{z^2+\dfrac{1}{z^2}} + \dfrac{1}{z^3+\dfrac{1}{z^3}} = \dfrac{1}{2\cos \dfrac{4\pi}{7}} + \dfrac{1}{2\cos \dfrac{8\pi}{7}} + \dfrac{1}{2\cos \dfrac{12\pi}{7}} = \dfrac{1}{2\cos \dfrac{2\pi}{7}} + \dfrac{1}{2\cos \dfrac{4\pi}{7}} + \dfrac{1}{2\cos \dfrac{6\pi}{7}}$ , so let's find a polynomial with roots $2\cos \dfrac{2\pi}{7}$ , $2\cos \dfrac{4\pi}{7}$ and $2\cos \dfrac{6\pi}{7}$ .

The equation $x^6+x^5+x^4+x^3+x^2+x+1=0$ has roots $x=e^{2 \pi k i / 7}$ for $1 \leq k \leq 6$ . Then, dividing by $x^3$ we have $x^3+x^2+x+1+\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}=0$ . Let's try to group the terms $c = x+\dfrac{1}{x}$ to get the cosine: $\begin{aligned} x^3+\dfrac{1}{x^3} + x^2+\dfrac{1}{x^2} + x+\dfrac{1}{x} + 1 &= 0 \\ \left(x+\dfrac{1}{x}\right)^3 - 3\left(x+\dfrac{1}{x}\right) + \left(x+\dfrac{1}{x}\right)^2 - 2 + \left(x+\dfrac{1}{x}\right) + 1 &= 0 \\ c^3 + c^2 - 2c - 1 &= 0 \end{aligned}$ But $c=2\cos \dfrac{2 \pi k}{7}$ for $1 \leq k \leq 3$ only, because for $4 \leq k \leq 6$ we get the same values, and we must have only three roots for a third degree equation. Finally, by Vieta's formulas, the sum of the inverses of its roots is: $\begin{aligned} S &= -\dfrac{\text{coefficient of }c}{\text{constant term}} \\ &= -\dfrac{-2}{-1} = -2\\ &\implies \lvert S \rvert = \boxed{2} \end{aligned}$

I'll show a solution applying some trigonmetry (But, there's a much gentler way to get the same result ;it will be shown later).

By De Moiver's formula . Note that and similarly So ......(1)

well the following steps are familiar and simple and I just do it: ......(2)

and and so ......(3)

Substitue (2)(3)into(1) to get the right answer 2.