For $K=\mathbb Q(\sqrt{-7})$,
$$
\zeta_K(s)=\zeta(s)L(s,\chi_3).
$$
The character is odd, $\chi_3(2)=1$, and
$$
\sum_{\substack{1\leq k<7/2\\(k,7)=1}}\chi_3(k)
=\chi_3(1)+\chi_3(2)+\chi_3(3)=1+1-1=1.
$$
The supplied odd-character formula gives
$$
L(1,\chi_3)=\frac{\pi}{\sqrt7}.
$$

The <analytic class number formula> for this imaginary quadratic field is
$$
\operatorname*{Res}_{s=1}\zeta_K(s)
=\frac{2\pi h_K}{w_K\sqrt{|d_K|}}
=\frac{\pi h_K}{\sqrt7},
$$
because $w_K=2$ and $d_K=-7$. Since $\operatorname*{Res}_{s=1}\zeta(s)=1$, comparison with $L(1,\chi_3)$ yields $\boxed{h_K=1}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 / 3 / f / Solution

{wiki}

For a number field of signature $(r_1,r_2)$, roots-of-unity count $w_K$, regulator $R_K$, class number $h_K$, and discriminant $d_K$,
$$
\operatorname*{Res}_{s=1}\zeta_K(s)
=\frac{2^{r_1}(2\pi)^{r_2}h_KR_K}{w_K\sqrt{|d_K|}}.
$$


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