Consider a fractional linear transformation $$f(z)=\frac{az+b}{cz+d}$$ on the Riemann sphere $\mathbb C\cup \{\infty\}$. Here $a,b,c,d$ are complex numbers with $ad-bc=1$.
Given a point $w$ in $\mathbb C\cup \{\infty\}$, we say the orbit of $w$ is the set $$\mathrm{orb}(w)=\{f^k(w)| k\in\mathbb Z\}.$$
Assume that $f$ has a unique attracting fixed point $z_+$ and a unique repulsive fixed point $z_-$. That is, $\mathrm{orb}(z_+)=\{z_+\}$, $\mathrm{orb}(z_-)=\{z_-\}$, $|f'(z_+)|<1$, $|f'(z_-)|>1$, and $z_+$ and $z_-$ are the only points whose orbits contain only a single element.
Is it true that for any $z\in (\mathbb C\cup \{\infty\})\setminus \{z_-\}$, we have $z_+\in \overline{\mathrm{orb}(z)}$? If it is true, how do we prove this?
Things become easier if the fixed points are zero and infinity. I.e. instead of $f$ we consider the “conjugate” transformation $$ g = T \circ f \circ T^{-1} $$ where $T(z) = \frac{z - z_+}{z - z_-}$. Show that $g$ is necessarily of the form $$ g(w) = \lambda w $$ for a complex constant $\lambda$, and $$ |\lambda| = |g'(0)| = |f'(z_+)| < 1 \, . $$ It follows that for all $z \ne z_-$, $$ f^k(z) = T^{-1}(\lambda^k T(z)) \to z_+ $$ for $k \to \infty$.