0

The question in principle is:

Calculate $\displaystyle \sum_z \dfrac{1}{|1-z|^2}$, in which $z$ are the roots of $z^7 = 1$.

In my solution, I did: $z = cis(\theta) \Rightarrow |1 - z|^2 = |(1 - \cos\theta) - i\sin\theta|^2 = (1-\cos\theta)^2 + \sin^2\theta = 2(1 - \cos\theta)$. So the sum is simply $\displaystyle \dfrac{1}{2}\sum_\theta \dfrac{1}{1-\cos\theta}$.

And the angles $\theta$ for that polynomial are $\pm\dfrac{\pi}{7},\pm\dfrac{3\pi}{7},\dfrac{\pm5\pi}{7},\pi$. Since $\cos\pi = -1$ and $\cos(-x) = \cos x$, we get $S = \dfrac{1}{4} + \dfrac{1}{1-\cos(\pi/7)} + \dfrac{1}{1-\cos(3\pi/7)} +\dfrac{1}{1-\cos(5\pi/7)}$.

My problem resides in computing $\frac{1}{1 - \cos(\pi/7)} + \frac{1}{1 - \cos(3\pi/7)} + \frac{1}{1 - \cos(5\pi/7)}$ which happens to be 12. But how can I do it?

user26857
  • 52,094
Feripinho
  • 173

1 Answers1

2

By the double angle formula, $1-\cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)$, so the expression becomes $$\frac{1}{2}\left(\csc^2\left(\frac{\pi}{14}\right)+\csc^2\left(\frac{3\pi}{14}\right)+\csc^2\left(\frac{5\pi}{14}\right)\right).$$ Now by Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$, the roots of the equation $8x^3-4x^2-4x+1=0$ are $\alpha = \sin\left(\frac{\pi}{14}\right)$, $\beta = -\sin\left(\frac{3\pi}{14}\right)$, and $\gamma = \sin\left(\frac{5\pi}{14}\right)$, so the expression further reduces to $$\frac{1}{2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right) = \frac{1}{2}\left(\frac{\left(\alpha\beta\right)^2+\left(\beta\gamma\right)^2+\left(\gamma\alpha\right)^2}{\left(\alpha\beta\gamma\right)^2}\right) = \frac{1}{2}\left(\frac{\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)^2-2\alpha\beta\gamma\left(\alpha+\beta+\gamma\right)}{\left(\alpha\beta\gamma\right)^2}\right)$$ By Vieta's formulae, $\alpha+\beta+\gamma = -\frac{-4}{8} = \frac{1}{2}$, $\alpha\beta+\beta\gamma+\gamma\alpha = \frac{-4}{8} = -\frac{1}{2}$, and $\alpha\beta\gamma = -\frac{1}{8}$, and plugging in these values, we indeed get $12$.

Prasiortle
  • 2,347
  • 2
    Fixed now, thanks. – Prasiortle Jan 01 '24 at 13:45
  • 1
    Thank you very much. I just wanted to share this one https://math.stackexchange.com/questions/3339381/prove-that-solution-of-the-equation-8x3-4x2-4x1-0-has-roots-cos-frac-pi, which seems to be a little bit more organized thread (imo) – Feripinho Jan 02 '24 at 18:08
  • You're very welcome. The thread I linked to had many more different proofs (using a variety of trigonometric identities and complex-number-based approaches), but I can agree that the one you refer to is more streamlined if you just want to see a single proof. – Prasiortle Jan 02 '24 at 19:07