Let $T$ be a mobius transformation with exactly one fixed point on $\mathbb{C} \cup \{\infty\}$. What form does $T$ take? Find a fomrula for $T^n(z)$. What happens to $T^n(z)$ as $n \to \infty$.
Attempt
(we will assume $w \neq \infty$ since this case I covered already). If we examine a generic Mobius transformation with exactly one fixed point then we will have
$$\frac{az+b}{cz+d}=z$$
which is solvable by the quadratic equation
$$\frac{a-d \pm \sqrt{(d-a)^2-4cb}}{2c}$$
Since there is only one fixed point this corresponds to the discrimanant being $0$ and so we have the following conditions
\begin{align*}
(d-a)^2-4cb&=0\\
\frac{a-d}{2c}&=z
\end{align*}
If $z=0$ then we have $b=0$ and our transformation is of the form
$$\frac{az}{cz+d}$$ . . .
confusion
Frome here the algebra seems to get a bit nasty and I'm not convinced I'm not on the right track going in that direction.
Best Answer
After conjugating by an appropriate transformation $S$, you can assume that the fixed point is $\infty$. Then what form does $T$ take?