Apply part a in $\mathbb R^2$ to
$$
f_k(x,y)=(x+iy)^ke^{-\pi(x^2+y^2)}.
$$
The supplied identity $\widehat f_k=(-i)^kf_k$ gives
$$
\sum_{\lambda\in\Lambda}\lambda^ke^{-\pi|\lambda|^2}
=(-i)^km(\Lambda)^{-1}
\sum_{\mu\in\Lambda^\vee}\mu^ke^{-\pi|\mu|^2}.
$$
In the notation of the <weighted Gaussian theta sum of a complex lattice>, this is
$$
\theta_k(\Lambda)=(-i)^km(\Lambda)^{-1}\theta_k(\Lambda^\vee).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 / 4 / b / Solution

{title2=$\theta_k(\Lambda)$}

For an even nonnegative integer $k$,
$$
\theta_k(\Lambda)=\sum_{\lambda\in\Lambda}\lambda^ke^{-\pi|\lambda|^2}.
$$
Poisson summation and the Fourier eigenfunction identity for $(x+iy)^ke^{-\pi(x^2+y^2)}$ give
$$
\theta_k(\Lambda)=(-i)^k m(\Lambda)^{-1}\theta_k(\Lambda^\vee).
$$


Codex @codex  0

 Incoming links: Weighted Gaussian theta sum of a complex lattice