Prove that $N(a+b\xi)=\frac{a^5+b^5}{a+b}$

Question

Let $\xi=e^{\frac{2\pi i}{5}}$, and let $K=\mathbb{Q}(\alpha)$ be a number field. I want to prove that $N(a+b\xi)=\frac{a^5+b^5}{a+b}$, where $a+b\neq0$.

If I define the $4$ inmersion like this: $\sigma_j(\xi)=\xi^j$, where $j=1,\ldots,4$, then I think I could start by computing $\sigma_1(a+b\xi)\sigma_4(a+b\xi)$ and $\sigma_2(a+b\xi)\sigma_3(a+b\xi)$, and finally multiplying them together.

$$\sigma_1(a+b\xi)\sigma_4(a+b\xi)=a^2+b^2+ab(\xi+\xi^4)$$ $$\sigma_2(a+b\xi)\sigma_3(a+b\xi)=a^2+b^2+ab(\xi^2+\xi^3)$$ I know that $(\xi+\xi^4)=2Re(\xi)$, where $Re$ denotes the real part. However this doesn't help much. If I multiply them together I get a very long expression which doesn't ressemble the one I am aiming to. I've also tried some geometric arguments but I cannot get to the solution. Can someone help me?

Answer 1

\begin{align*} N(a+b\xi)&=\prod_{i=1}^4(a+b\xi^i)\\ &=(-b)^4\prod_{i=1}^4\left(-\frac ab-\xi^i\right)\\ &=b^4\Phi_5\left(-\frac ab\right). \end{align*}