I'm trying to prove that the $\mathbb{Z}-$action on $\mathbb{C}^{\times}:=(\mathbb{C}\setminus\{0\},\times)$ defined by $z\mapsto q^kz$ for $k\in \mathbb{Z}$ and $q\in \mathbb{C}^{\times}$ such that $0<|q|<1$, is properly discontinuous.
For all $z_0\in \mathbb{C}^{\times}$, let $U_{\epsilon}:=\{z\in\mathbb{C}^{\times}:|z-z_0|<\epsilon\}$ be an open nbhd of $z_0$. I want to find an $\epsilon$ such that $$|q^kz_0-z_0|=|z_0||q^k-1|\geq \epsilon>0$$ for all $k\neq 0$. Does it exist? is it just $\epsilon<|q|$?
Moreover, for $z_1,z_2\in \mathbb{C}^{\times}$ such that $\mathbb{Z}.z_1\neq \mathbb{Z}.z_2$, I want to prove that there exist nbhds $U_1,U_2$ such that $$\{q^kz:z\in U_1\cap U_2\}=\emptyset$$ for all $k\in \mathbb{Z}$. Any hint?
Consider the iterating sequence $S = (q^k z_0)_{k \in \Bbb{Z}_{\geq 0}}$. The first iterate is $z_0$, lying on the circle $\{z \in \Bbb{C} : |z| = |z_0|\}$, of radius $|z_0|$. The next iterate is $q z_0$, lying on the circle $\{z \in \Bbb{C} : |z| = |q||z_0|\}$, of radius $|q||z_0|$. Since $0<q<1$, this second circle is contained in the disk bounded by the first circle. Continuing, each iterate lies on a circle contained by the disk interior to the circles containing each prior iterate. This means that if you set $\varepsilon < |z_0| - |q||z_0|$, the difference between the radii of the first two circles, then $U_\varepsilon$ will contain no point of the second or subsequent circles, so $U_\varepsilon \cap S = \{z_0\}$, as desired.
In your "Moreover", replace the above "circles" by annuli. The first annulus has inner radius $\inf_{z \in U_1 \cap U_2} |z|$ and outer radius $\sup_{z \in U_1 \cap U_2} |z|$. Now construct the iterated sequence of annuli and choose the neighborhoods small enough that the annuli don't intersect.