Which theta series do I have to consider in order to show that $$\sum_{\mu\in a}e^{-2\pi t(\mu+u)(\mu'+u')}=\frac{1}{|\sqrt{DNorm(a^2)}|t}\sum_{\nu\in a^{-1}(\frac{1}{\sqrt{D}})}e^{\frac{-2\pi}{t}\nu \nu'+2\pi i(\nu u+\nu'u')} \ \ (1)$$ ,where $u,u' \in \mathbb{C}$ , $a $ is an ideal in $K=Q(\sqrt{D}),D<0 .$ The prime denotes the conjugate number . I considered the theta series $$\theta_a(\tau,\alpha,\beta)=\sum_{\lambda \in a}e^{2\pi i\tau(\lambda+\beta,\lambda+\beta)/2}e^{-2\pi i(\lambda+\beta/2,\alpha)} $$ for $\beta=0$ , where $(\lambda+\beta,\lambda+\beta)=trace((\overline\lambda+\beta)(\lambda+\beta))=2Norm(\lambda+\beta)$ .
I think $\theta_a(-1/\tau,0,\alpha)$ should give the left hand side of (1) for $\tau=i/t$ .
I think the theta transformation formula for even lattices $M$ says
$$\theta_M(-1/\tau,-\beta,\alpha)=\frac{\tau}{i\sqrt{|M'/M|}}\sum_{\lambda \in M'}e^{2\pi i \tau(\lambda+\beta,\lambda+\beta)/2}e^{-2\pi i(\lambda+\beta/2,\alpha)}$$ , where $M'$ is the dual lattice of $M$ .
Thanks for the help .