Let $\mathfrak{a}\subset \mathcal{O_K},K=\mathbb{Q}(\sqrt{D}),D<0$ with square free discriminant $D$. I want to calculate the dual lattice $\mathfrak{a}^*$ of $\mathfrak{a} $ relative to the bilinear form $(x,y)=Tr(x\overline y).$ Let $\mathfrak{a}$ $$ \mathfrak{a}=m\mathbb{Z}+(l+n\theta)\mathbb{Z} \ \ with \ \ \theta=\frac{1+\sqrt{D}}{2}, \ n|m,n|l,N(\mathfrak{a})=mn,m,l,n \in \mathbb{Z}$$ and $e_1=1,e_2=\theta,e_1^*=r+s\theta,e_2^*=p+q\theta$. After calculating $Tr(e_ie^*_j)=\delta_{ij}$, I get $$e_1^*=\frac{l+n\theta}{N(\mathfrak{a})\sqrt{D}},e_2^*=-\frac{m}{N(\mathfrak{a})\sqrt{D}}, $$ but I should get $e_1^*=\frac{l+n \overline\theta}{N(\mathfrak{a})\sqrt{D}} $ because if $\mathfrak{a}=(\alpha,\beta)$ then $\mathfrak{a}^*=(\frac{\overline\alpha}{N(\mathfrak{a})\sqrt{D}},\frac{\overline\beta}{N(\mathfrak{a})\sqrt{D}}).$
If $(x,y)=Tr(xy)$ I get the correct dual lattice.
We have the group isomorphism $$f : \Bbb{Z}^2\to O_K, \qquad f:\Bbb{Q}^2\to K $$ then $$Tr(f(a)f(b)) = a^\top M b$$ where $M_{ij}= Tr(f(e_i)f(e_j))$ is the trace matrix appearing in the definition of $$Disc(O_K)=\det(M)$$ It is immediate that the dual lattice is $$f(M^{-\top} \Bbb{Z}^2)$$
It works the same way when replacing $O_K$ by an ideal $f(A \Bbb{Z}^2)$ whose dual lattice is $f((MA)^{-\top} \Bbb{Z}^2)$.
We obtain that the dual ideal of $I$ is $\frac{I^{-1}}{Disc(O_K)}$.