is clear that there cannot be an absolute bound in a result like this. In fact, a detailed analysis of the proof indicates that the actual condition the authors use is $|{\kappa}|>1-C|\arg({\kappa})|$ with some absolute constant $C>0$. That is, “sufficiently close to 1” means “for any $\theta\in(0,\pi)\cup(\pi,2\pi)$ there exists $\delta$ such that $A_{\kappa}$ contains the closed unit disc for all ${\kappa}$ with $\arg({\kappa})=\theta$ and $|{\kappa}|>1-\delta$” [^3]. Theorem \[thm:interior\] overcomes this obstacle.

Given the two maps $T_m$ and $T_p$, there is a natural projection map from the set of all $\{m,p\}$ sequences to points on $A$. We define $\pi:\{m,p\}^{\mathbb{N}}\to A$ by $\pi(a_0 a_1 a_2 \dots) = \lim_{n\to\infty}T_{a_0} \circ T_{a_1} \circ\dots\circ T_{a_n}(0,0)$. Note that because both $T_m$ and $T_p$ are contraction maps, this yields a well defined point in $A$. We call $a_0a_1\dots$ an [*address*]{} for $(x,y)\in A$ if $\pi(a_0a_1\dots)=(x,y)$. We say that a point $(x,y) \in A$ is a [*point of uniqueness*]{} if it has a unique address.

The question on when this IFS has a large number of points of uniqueness depends somewhat on the nature of the eigenvalues. If $M$ has two complex eigenvalues, $\kappa$ and $\overline\kappa$ where $\mathrm{arg}(\kappa)/\pi \in {\mathbb{Q}}$ then it is possible for the IFS to 