Suppose $d_K<-4$. Then $K$ has signature $(0,1)$, regulator $1$, and exactly $w_K=2$ roots of unity. The <analytic class number formula> becomes
$$
\operatorname*{res}_{s=1}\zeta_K(s)
=\frac{\pi h_K}{\sqrt{|d_K|}}.
$$
Because $\zeta_K(s)=\zeta(s)L(s,\chi_{d_K})$ and $\operatorname*{res}_{s=1}\zeta(s)=1$,
$$
h_K=\frac{\sqrt{|d_K|}}{\pi}|L(1,\chi_{d_K})|.
$$
Putting $m=|d_K|$ into the supplied odd-character identity yields the <quadratic class number formula>
$$
h_K
=\frac1{|2-\chi_{d_K}(2)|}
\left|
\sum_{\substack{a\in(\mathbb Z/m\mathbb Z)^\times\\a<m/2}}
\chi_{d_K}(a)
\right|.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 2 / 2 / 5 / Solution

For $K=\mathbb Q(\sqrt{-11})$, the fundamental discriminant is $-11$ and $\chi_{-11}(2)=-1$ because $-11\equiv5\pmod8$. The quadratic residues modulo $11$ are
$$
1,3,4,5,9.
$$
Therefore
$$
\sum_{1\leq a<11/2}\chi_{-11}(a)
=\chi(1)+\cdots+\chi(5)
=1-1+1+1+1=3.
$$
The <quadratic class number formula> gives
$$
h_K=\frac{|3|}{|2-(-1)|}=1.
$$

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 2 / 2 / 6 / Solution

For an imaginary quadratic field of fundamental discriminant $D<-4$,
$$
h(D)=\frac{\sqrt{|D|}}{\pi}L(1,\chi_D).
$$
Equivalent finite character-sum forms follow by evaluating the odd quadratic <Dirichlet L-function> at one.

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