Exercise 6.19

Proof. Let $\gamma =\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕5)$, $\zeta =e^{2\mathit{i\pi }∕5}$ and $\alpha =\zeta +{\zeta }^4,\beta ={\zeta }^2+{\zeta }^3$.

Then $\gamma =\frac{\zeta +{\zeta }^{-1}}{2}=\frac{\zeta +{\zeta }^4}{2}=\frac{\alpha }{2}$.

$\alpha +\beta =\zeta +{\zeta }^4+{\zeta }^2+{\zeta }^3=-1$.

$\mathit{\alpha \beta }={\zeta }^3+{\zeta }^4+{\zeta }^6+{\zeta }^7={\zeta }^3+{\zeta }^4+\zeta +{\zeta }^2=-1$

So $\alpha ,\beta$ are the two roots of $x^2+x-1$ :

${\alpha }^2+\alpha -1$ = 0, so $4{(\alpha ∕2)}^2+2(\alpha ∕2)-1=0$ : $\gamma =\alpha ∕2$ is a root of

As $\mathrm{\Delta }=4\times 5$, the two roots of $f$ are irrational. $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$ and $f$ has no root in $\mathbb{Q}$, so $f(x)$ is irreducible in $\mathbb{Q}[x]$. Therefore the minimal polynomial of $\gamma =\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕5)$ is $f(x)=4x^2+2x-1.$ The other root of $f$ is $\beta ∕2=({\zeta }^2+{\zeta }^3)∕2=\cos <!--\; FUNCTION\; APPLICATION\; -->(4\pi ∕5)$.

Conclusion : the conjugates of $\gamma =\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕5)$ are $\gamma =\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕5)$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->(4\pi ∕5)$. □