Let $\omega=\frac{-1+i\sqrt{3}}{2}$ be a complex cube root of unity. The Eisenstein integers $\mathbb{Z}[\omega]$ (a unique factorization domain) are of the forms $a+b\omega$ where $a$ and $b$ are ordinary integers. Let $f: \mathbb{Z}[\omega]\rightarrow \mathbb{R}$, I konw that the Fourier transform $$\hat{f}(k)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x+y \omega)e(k(x+y \omega)/\sqrt{-3})dxdy,$$ where $e(z)= \exp(2\pi i(z+\overline{z}))$.
Let $g: \mathbb{Z}[\omega]\rightarrow \mathbb{R}$. I would like to konw if the following is right. Please tell me something or some revelant references.
1) Mellin transform in $\mathbb{Z}[\omega]$
$$\int_{0}^\infty \int_{0}^\infty g(x+y \omega) N(x+y \omega)^{s-1} dxdy=\tilde{g}(s),$$where $N(z)$ means the norm of $z$, $\tilde{g}(s)$ is the ordinary Mellin transform of $g$ on $\mathbb{R}$ defined as $\tilde{g}(s)=\int_0^\infty g(x)x^{s-1}dx$ and we permit the convergence of the integrals.
2) Plancherel formula $$\int_{0}^\infty \int_{0}^\infty f(x+y \omega)g(x+y \omega)dxdy=\frac{1}{4\pi i} \int_{(c)} \tilde{f}(s)\tilde{g}(1-s)ds .$$ On $\mathbb{R}$ we know that $\int_0^\infty f(x)g(x)=\frac{1}{4\pi i} \int_{(c)} \tilde{f}(s)\tilde{g}(1-s)ds $ is a basic fact.