Răspuns:
Explicație pas cu pas:
[tex]\displaystyle z^4+z^3+z^2+z+1=0\\\dfrac{z^5-1}{z-1}=0, z\neq 1\\z^5=1\\z^5=\cos 0+i\sin 0\\\texttt{Radacinile ecuatiei sunt :}\\z_k=\cos\dfrac{2k\pi}{5}+i\sin\dfrac{2k\pi}{5},k=\overline{1,4}\\\texttt{Observatie: Cazul }k=0\texttt{ nu convine, deoarece z nu poate fi 1.}\\\texttt{Cum }z^5=1\texttt{ rezulta }|z|^5=1,\texttt{ deci }|z|=1.[/tex]
[tex]\texttt{Avem de calculat suma :}\displaystyle\\\sum_{k=1}^4\left(\left|z_k^n+\dfrac{1}{z_k^n}\right|\right)^2=\sum_{k=1}^4\dfrac{(|z_k^{2n}+1|)^2}{|z_k|^{2n}}=\sum_{k=1}^4(|z_k^{2n}+1|)^2\\\texttt{Conform fomulelor lui Moivre rezulta ca :}\\z_k^{2n}=\left(\cos\dfrac{2k\pi}{5}+i\sin\dfrac{2k\pi}{5}\right)^{2n}=\cos\dfrac{4kn\pi}{5}+i\sin\dfrac{4nk\pi}{5}\\\texttt{Revenind la exercitiu:}[/tex]
[tex]\displaystyle\sum_{k=1}^4\left|\cos\dfrac{4kn\pi}{5}+i\sin\dfrac{4kn\pi}{5}+1\right|^2=\sum_{k=1}^4{\left[\left(\cos\dfrac{4kn\pi}{5}+1\right)^2+\sin^2\left(\dfrac{4kn\pi}{5}\right)\right]}\\=\sum_{k=1}^4{\left[\cos^2\left(\dfrac{4kn\pi}{5}\right)+2\cos\left(\dfrac{4kn\pi}{5}\right)+1+\sin^2\left(\dfrac{4kn\pi}{5}\right)\right]}=\\=2\cdot \sum_{k=1}^4\cos\dfrac{4kn\pi}{5}+8\\\texttt{Sper ca te descurci mai departe.}[/tex]
