m $v=cz+d$ with the integers $(c,d)$ co-prime.

Now given the positive lattice basis $v=cz+d$ and $v'=az+b$, form the integer matrix $\g=\begin{pmatrix} a&b\\ c&d\end{pmatrix}$ , which has $\det(\g)=+1$ since $\{v,v'\}$ form a positive basis of the lattice. Thus we get a matrix in the modular group $\G=SL_2(\Z)$. Then with $\g$ applied as a Möbius transformation to $z$, the length of $v$ can be computed via $${\operatorname{Im}}(\g z) = \frac{{\operatorname{Im}}(z)}{|cz+d|^2}=\frac{{\operatorname{area}}(L)}{|v|^2}$$ The signed ratio between the lengths of $v$ and $v'$ (when $v'$ is chosen of minimal length) is $$\rho{(v)} = \pm |\gamma z| \;.$$ where the sign is $+$ if ${\operatorname{Re}}(\g z)>0$ and $-$ otherwise. Moreover, we have $${\operatorname{sk}}(v,v') = {\operatorname{Re}}(\g z)$$ Indeed, $${\operatorname{Re}}(\g z) = \frac{ac(x^2+y^2) +(ad+bc)x +bd}{|cz+d|^2}$$ which is ${\operatorname{sk}}(v,v')$ in view of . Consequently, the uniform distribution modulo one of ${\operatorname{sk}}(v)$ as $|v|\to\infty$ is then exactly the uniform distribution modulo one of ${\operatorname{Re}}(\g z)$ as $\g$ varies over $\GinfmodG$ with ${\operatorname{Im}}(\g z )\to 0$, that is Theorem \[equidistribution\].

A sketch of a proof of Good’s theorem {#sec:spectral}
=====================================

To prove Theorem \[equidistribution\], we use Weyl’s criterion to reduce it to showing that the corresponding “Weyl sums” satisfy $$\label{character asymptotics} 
  \sum_{\g \in(\GinfmodG)_{\varepsilon,z}}e(m{\operatorname{Re}}{\g
z})=\delta_{m=0}\frac{t_\G}{\vol{(\GmodH)}}\frac 1\varepsilon 
+o(1/\varepsilon)$$ as $\varepsilon\to 0$. Here $t_\G$ equals $2$ if $-I\in \G$ and $1$ otherwise. In turn, will follow, by a more or less standard Tauberian theorem (see e.g. [@PetridisRisager:2004a p. 1035-1038]) from knowing the analytic properties of the series $$V_m(z,s):=\sum_{\g\in\GinfmodG}{\operatorname{Im}}(\g z)^se(m{\operatorname{Re}}(\g z)) \;.$$ studied also in [@Good:1981b; @Neunhoffer:1973a] Here $e(x)=\e