Here is the problem:
Show that if a Möbius transformation $f$ has exactly one fixed point, then it is conjugate to the translation $z \to z+1$.
My work so far:
Let $a$ be the fixed point of $f$ and let $f(z) = \frac{Az+B}{Cz+D}$.
It suffices to show that there is a Möbius transformation $h$ such that $h(f(z))=h(z)+1$. The author suggests we consider a Moebius transformation that maps the fixed point to $ \infty$ ( why is this?). So I try $h(z) = \frac{1}{z-a}$. We need to then check that:
$h(f(z))= \frac{1}{\frac{Az+B}{Cz+D}-a}= h(z)+1 =\frac{1}{z-a}+1.$ Noting that $f(a)=a \implies \frac{Aa+B}{Ca+D}=a \implies a(A-Ca)=Da-B$ the lhs becomes,
$ h(f(z))= \frac{C}{A-aC}(\frac{a+\frac{D}{C}}{z-a}+1)$, after some algebra.
But I do not see how this is equal to $h(z)+1$. Do I have to choose a different $h$ or I am doing something wrong here? Also why are we mapping the fixed point to infinity in the first place? Many thanks!
Best Answer
If $f$ has exactly one fixed point $a \in \Bbb C$ and $h(z) = 1/(z-a)$ then $$ T = h \circ f \circ h^{-1} $$ is a Möbiustransformation with $\infty$ as the only fixed point. Show that necessarily $$ T(z) = z + b $$ for some $b \in \Bbb C$, $b \ne 0$, and finally use $$ z + b = (\frac zb + 1) \cdot b $$
Your approach was not successful because you used $f(a) = a$, but not that $a$ is the only fixed point of $f$.