Minkowski's Bound Calculation I am studying some algebraic number theory and have reached Minkowski's Bound Theorem. I am not quite sure where some of the numbers come from.
Definitions. Let $F$ be of degree $n$ over $\mathbb{Q}$. Let $\sigma_{1},\ldots,\sigma_{n}$ be all $\mathbb{Q}$-homomorphisms of $F$ into $\mathbb{C}$. Let $\tau:\mathbb{C}\to\mathbb{C}$, $a+bi\mapsto a-bi$. Then $\tau\circ\sigma_{i}$ is a $\mathbb{Q}$-homomorphism of $F$ into $\mathbb{C}$, so it is equal to some $\sigma_{j}$. Let $r_{1}=\#\hom_{\mathbb{Q}}(F\to\mathbb{Q})$ be the number of $\mathbb{Q}$-homomorphisms of this type, say, after renumbering, $\sigma_{1},\ldots,\sigma_{r_{1}}$. For every $i>r_{1}$, $\tau\circ\sigma_{j}\neq\sigma_{j}$, so couples $(\sigma_{j},\tau\circ\sigma_{j})$ are formed. Let $r_{2}=\#\{(\sigma_{j},\tau\circ\sigma_{j})\}$ be the number of these pairs. Then $n-r_{1}=2r_{2}$ is an even number, and $r_{1}+2r_{2}=n$.
Theorem. Let $F$ be an algebraic number field of degree $n$ with parameters $r_{1}$ and $r_{2}$. Then every class of $C_{F}$ contains an ideal $I$ such that $$\text{N}(I)\leq\left(\frac{4}{\pi}\right)^{r_{2}}\frac{n!}{n^{n}}\sqrt{|d_{F}|}$$ where $d_{F}$ is the discriminant of $F$.
Example. Let $\mathbb{Q}(\sqrt{5})$, then $r_{1}=2$, $r_{2}=0$, $n=2$, and $|d_{F}|=5$. So $$\text{N}(I)\leq\left(\frac{4}{\pi}\right)^{r_{2}}\frac{n!}{n^{n}}\sqrt{|d_{F}|}=\frac{2}{4}\sqrt{5}\approx1.118<2.$$ Hence $\text{N}(I)=1$ and so $I=O_{F}$. Thus, every ideal of $O_{F}$ is principal and $C_{F}=\{1\}$.
My Question. I know it is a lot of reading for this simple question, but how to I calculate $n$, $r_{1}$, and $r_{2}$?

The number $n$ is the degree of the number field. In the usual examples, if $F=\mathbb Q(\alpha)$ where $\alpha$ satisfies an irreducible polynomial $f$ over $\mathbb Q$, then $n=\deg (f)$.
Keeping the same usual situation above, $r_1$ is the number of real roots of $f$ and $r_2$ is the number of pairs complex roots of $f$ The complex roots must come in pairs because $f$ has rational coefficients so $$f(z)=0 \Longleftrightarrow f(\overline z)=\overline{f(z)}=\overline 0=0.$$
Can you now see how this applies to your situation?
