Questions mathématiques diverses

Question 22:
Soit $z$ un nombre complexe vérifiant : $z^2+1=\sqrt{3}z$.
Calculer la somme $S=\sum\limits_{k=1}^{24}{(z^k-z^{-k})^2}$

Réponse 22:
$z^2-\sqrt{3}z+1=0$, $\Delta=-1=i^2$ donc $z=\dfrac{\sqrt{3}\pm i}{2}=e^{\pm i\frac{\pi}{6}}$ $$\begin{align*} S&=\sum\limits_{k=1}^{24}{\left({z^k-z^{-k}}\right)^2}\\ &=\sum\limits_{k=1}^{24}{\left({\left({e^{\pm i\frac{\pi}{6}}}\right)^k-\left({e^{\pm i\frac{\pi}{6}}}\right)^{-k}}\right)^2}\\ &=\sum\limits_{k=1}^{24}\left({e^{\pm i\frac{k\pi}{6}}}-e^{\mp i\frac{k\pi}{6}}\right)^2\\ &=\sum\limits_{k=1}^{24}{\left({e^{i\frac{2k\pi}{6}}+e^{-i\frac{2k\pi}{6}}-2}\right)}\\ &=\sum\limits_{k=1}^{24}{\left({2\cos\dfrac{2k\pi}{6}-2}\right)}\\ &=2\sum\limits_{k=1}^{24}{\left({\cos\dfrac{2k\pi}{6}}\right)}-48\\ &=2\sum\limits_{k=0}^{23}{\left({\cos\dfrac{2k\pi}{6}}\right)}-2+2-48\\ &On\;pose \;k=6q+r \;où \;r\in\left\{{0,1,2,3,4,5}\right\}\\ S&=-48+2\sum\limits_{q=0}^{3}\sum\limits_{r=0}^{5}{\left({\cos\left({\dfrac{2(6q+r)\pi}{6}}\right)}\right)}\\ &=-48+2\sum\limits_{q=0}^{3}\sum\limits_{r=0}^{5}{\left({\cos\left({2q\pi+\dfrac{2r\pi}{6}}\right)}\right)}\\ &=-48+2\sum\limits_{q=0}^{3}\sum\limits_{r=0}^{5}{\cos\left({\dfrac{2r\pi}{6}}\right)}\\ &=-48+2\times 4\sum\limits_{r=0}^{5}{\cos\left({\dfrac{2r\pi}{6}}\right)} \end{align*}$$ Or $e^{i\frac{2r\pi}{6}}$ avec $r\in \left\{{0,1,2,3,4,5}\right\}$ sont les racines sixièmes de l'unité donc $\sum\limits_{r=0}^{5}{e^{i\frac{2r\pi}{6}}}=0$ $\Longrightarrow \sum\limits_{r=0}^{5}{\cos\left({\dfrac{2r\pi}{6}}\right)}=0$
Conclusion: $\boxed{S=-48}$