Let $K=\mathbb{Q} (\sqrt{d})$ be a quadratic field , $d<0$ squarefree .For a lattice $L=\mathbb{a}\subset K$ I have to show that its dual lattice is $$\mathbb{a}^*=\frac{1}{\sqrt{d}}\mathbb{a}^{-1} .$$
For $L$ a lattice in $K$ its dual lattice is $$L^*=\lbrace \alpha\in K:Tr_{K/\mathbb{Q}}(\alpha L)\subset\mathbb{Z} \rbrace $$
Approach :
If $d\equiv2,3(4)$ then a basis for $K/\mathbb{Q}$ is given by $\lbrace 1,\sqrt{d}\rbrace$ and $\mathbb{a}=\mathbb{Z}+\mathbb{Z}\sqrt{d}$ .
Then $Tr_{K/\mathbb{Q}}((a+b\sqrt{d)}\cdot1)$=$2a$ and $Tr_{K/\mathbb{Q}}((a+b\sqrt{d}) \sqrt{d})=Tr_{K/\mathbb{Q}}(bd+a \sqrt{d})=2bd $ , so that the dual basis is $\lbrace \frac{1}{2},\frac{1}{2d}\rbrace$ and $\mathbb{a^*}=\frac{1}{2\sqrt{d}}(\mathbb{Z}+\mathbb{Z}\sqrt{d})=\frac{1}{2\sqrt{d}}\mathbb{a}$ .
My problem is that this is not equivalent to what I have to show .