I'm trying to evaluate
$$\zeta_5^3+\zeta_5^2+1$$
I got a mess of roots but I think it should be something simple involving $\sqrt{5}$. Any help?
I'm trying to evaluate
$$\zeta_5^3+\zeta_5^2+1$$
I got a mess of roots but I think it should be something simple involving $\sqrt{5}$. Any help?
On
You are absolutely correct about something simple involving $\sqrt{5}$, since $$ G(\chi_5)= \zeta_5-\zeta_5^2-\zeta_5^3+\zeta_5^4\tag{A} $$ is a Gauss sum, and $\left|G(\chi_5)\right|=\sqrt{5}$ simply follows by squaring. The real part of $G(\chi_5)$ is quite clearly positive, hence $G(\chi_5)=\sqrt{5}$. We also have $$ \zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4 = -1 \tag{B}$$ hence by considering the difference between $(B)$ and $(A)$ we have $$ 2(\zeta_5^2+\zeta_5^3) = -1-\sqrt{5},\qquad \zeta_5^2+\zeta_5^3+1=\color{red}{\frac{1-\sqrt{5}}{2}}.\tag{C} $$
$$\zeta^3+\zeta^2+1=-\zeta-\zeta^4=-\zeta-\zeta^{-1}=-\left(\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}+\cos\frac{2\pi}{5}-i\sin\frac{2\pi}{5}\right)=$$ $$=-2\cos72^{\circ}=-2\cdot\frac{\sqrt5-1}{4}=\frac{1-\sqrt5}{2}.$$