we will assume that $0 < \nu < 1$. In this case we take $u = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, which is, again, clearly a cyclic vector. We have $$\pi(a_0 a_1 a_2 \dots) = \left(\sum_{i=0}^\infty i a_i \nu^{i-1},\sum_{i=0}^\infty a_i\nu^i\right)$$ (see Lemma \[lem:jordan-exp\] below). There is a symmetry to the $\nu < 0$ case such that $A_{\nu}$ and $A_{-\nu}$ share all of the desired properties. To see this, write $(x,y) = (\sum i a_i \nu^{i-1} \sum a_i \nu^i) \in A_{-\nu}$. Taking $a'_i = (-1)^i a_i \in \{m, p\}$, we see that $(-x,y) = (\sum a'_i i (-\nu)^i, \sum a'_i (-\nu)) \in A_{-\nu}$. Hence $A_\nu$ and $A_{-\nu}$ are reflections of each other across the $y$-axis.

For the complex case, we let ${\kappa}= a+bi$ and consider $v = \begin{pmatrix} x\\ y \end{pmatrix}$ as $z = x+ yi$. We see that the maps in with $u = \begin{pmatrix}1 \\ 0 \end{pmatrix}$, are equivalent to the maps in ${\mathbb{C}}$, namely, $$T_m (z) = {\kappa}z - 1 \ \ \mathrm{or}\ \ T_p(z) = {\kappa}z + 1.$$ In the complex case we have $\pi(a_0 a_1 a_2 \dots) = \sum_{j=0}^\infty a_j \kappa^j \in {\mathbb{C}}$, i.e., the attractor $A_{\kappa}$ is the set of expansions in complex base ${\kappa}$ 