I'd like to know to what point is it possible to generalize
this method
for obtaining the **functional equation** for the Dedekind zeta function $\zeta_K(s)$ of a number field ?

Let $\mathfrak{C}$ be the ideal class group $$\zeta_K(s) =\sum_{I \subset \mathcal{O}_K} N(I)^{-s} = \sum_{C_j \in \mathfrak{C}} \ \ \sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s}$$ Assuming $\mathcal{O}_K^\times$ is finite, each ideal $C_j \in \mathfrak{C}$ being a rank $n=[K:\mathbb{Q}]$ free $\mathbb{Z}$-module with basis $b_{j,1}, \ldots, b_{j,n}$, letting $C_j C_j^{-1} = (d_j)$ : $$\zeta_{K,C_j}(s)=\sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s} = \frac{1}{|O_K^\times|} \sum_{a \in C_j^{-1}} N(\frac{a}{d_j} C_j)^{-s}$$ $$=\frac{1}{|O_K^\times|}\frac{N(C_j)^{-s}}{N(d_j)^{-s}}\sum_{m \in \mathbb{Z}^n \setminus (0)} N(\sum_{l} m_l b_{j,l})^{-s}$$ $$\Gamma(s) \zeta_{K,C_j}(s) = \frac{1}{|O_K^\times|} \frac{N(C_j)^{-s}}{N(d_j)^{-s}}\int_0^\infty x^{s-1} \sum_{m \in \mathbb{Z}^n \setminus (0)} e^{-x\ N(\sum_{l} m_l b_{j,l})}dx$$ Therefore it reduces to finding a Poisson summation formula for $\displaystyle\Theta_j(x) = \sum_{m \in \mathbb{Z}^n} e^{-x \, N(\sum_{l} m_l b_{j,l})}$.

I did it for $\mathbb{Q}(\sqrt{-5})$ whose ideal class group has two elements, obtaining that $$\Lambda(s)= 5^{s/2}\pi^{-s}\Gamma(s)4 \zeta_{\mathbb{Q}(\sqrt{-5})}(s)=\Lambda(1-s) $$ I guess this method works at least for any imaginary quadratic field ?

If $K$ is imaginary quadratic, $\mathcal{O}_K = \mathbb{Z}[w]$ and any ideal has the form $(k,l+ow)$ then due to the property of $2\times 2$ matrices (see how $\scriptstyle\begin{pmatrix} 2 & 1 \\ 0 & \sqrt{5}\end{pmatrix}$ is treated in the link), I'll get a functional equation $\Lambda_{K,C_j}(s) = \lambda_j^{s/2} \pi^{-s}\Gamma(s)\zeta_{K,C_k}(s)=\Lambda_{K,C_j}(1-s)$, but I don't know how to show $\lambda_j $ is the same for each ideal class $C_j$.

What about the other cases ($\mathcal{O}_K^\times$ infinite, cubic extension, non-monogenic field...) ?