Mobius transformation without fixed points Let $T$ be a Mobius transformation of the unit disk $D=\{ |z|<1 \}$ onto itself, with no fixed points. Show that there exists $s \in \partial D$ such that $T^n(z) \rightarrow s$ as $n\rightarrow \infty$, for all $z\in D$. 
 A: We take two cases according to whether $T$ has one fixed point or two fixed points. (Since any non-identity Mobius transformation has one or two fixed points, these cases are exhaustive.)
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Case 1:
Suppose that $T$ has one fixed point.  Let $f(z) = z+1$.
Since all Mobius transformations with the same number of fixed points are conjugate, there exists a Mobius transformation $g$ so that 
$$T = g \circ f \circ g^{-1}.$$
The iterates of $T$ are given by $T^n = g \circ f^n \circ g^{-1}$.  Fix $z \in D$.  We claim that $T^n(z) \to g(\infty)$.  To prove so, let $w_0 = g^{-1}(z)$, and define a sequence $\{w_n\}$ recursively by $w_{n+1} = f(w_n)$.  We observe that for any $n \in \mathbb{N}$, $T^n(z) = g(w_n)$. By definition of $f$, $f^n(z) = z + n$.  Hence, $f^n(z) \to \infty$ as $n \to \infty$.  By continuity of $g$, $g(w_n) \to g(\infty)$.  That is, $T^n(z) \to g(\infty)$.
It is immediate that $g(\infty) \in \partial D$; if it were in $\mathbb{C^*}-\overline{D}$, this would contradict continuity of $T$, whereas if it were in $D$, then this would contradict the assumption that the restriction of $T$ to $D$ has no fixed points.
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Case 2:
Define $f$ by $f(z) = az$, where $a \in \mathbb{C} - \{0\}$ and $|a| < 1$.  Since $f$ has two fixed points (namely $0$ and $\infty$), there exists a Mobius transformation $g$ so that 
$$T = g \circ f \circ g^{-1}.$$
The iterates of $T$ are given by $T^n = g \circ f^n \circ g^{-1}$. Since $|a| < 1$, $f^n(z) \to 0$ for any $z \in \mathbb{C}$.  By a similar argument to that above, for any $z \in D$, we have $T^n(z) \to g(0)$.
