Let $\Gamma=u\mathbb{Z} \oplus v\mathbb{Z} \subset \mathbb{C}$ be a lattice with generators $u,v \in \mathbb{C}$ such that $\operatorname{Im}(v/u)>0$.
Let $A= \frac{(\text{Area of } \mathbb{C} / \Gamma)}{\pi} = \frac{1}{\pi} \operatorname{Im}(v/u)=\frac{1}{2\pi i}(v\overline{u} - u \overline{v}) \: \: \: \: (>0). $
We define the pairing for complex numbers $w,z$ by $$ \langle \, z , w \, \rangle_\Gamma:=\exp \left( \frac{\overline{w}z - w\overline{z}}{A} \right). $$
Now we identify the complex torus $\mathbb{T} :=\mathbb{C}/\Gamma$ with its dual through the isomophism \begin{align*} \mathbb{T} &\longrightarrow \operatorname{Hom} \left( \mathbb{T}, \mathbb{C}^\times \right) \\ z &\longmapsto \left( w \mapsto \langle \, w, z \, \rangle_\Gamma \right) \end{align*} So the Fourier transform of a (nice enough) function $f(z)$ would be given by $$ \widehat{f}(w)= \int_{\mathbb{T}} f(z) \overline{\langle \, w, z \, \rangle} \, dz = \int_{\mathbb{T}} f(z) \langle \, z, w \, \rangle \, dz$$ I am looking for a function that would satisfy $f= \widehat{f}$ in this setting. My question is : Does $f(z)= e^{-\frac{|z|^2}{A}}$ work ? I think there is something very obvious that I am having troubles seeing...
Any help would be appreciated!