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.
We can find that if the only fixed point on the Riemann sphere is $\infty$ then our transformation takes the form $T(z)=z+b$. Now, if we want to consider an arbitrary fixed point, call it $w$, then we can look at a new transformation $S$ which exchanges $w$ and $\infty$.
Now, we may write our new transformation which fixes $w$, $T_w$, as follows
$$T_w=S^{-1}TS$$
Even if $S$ fixes some points $T$ will unfix them, all but $w$ that is which will be fixed because of $T$. To find a transformation which exchanges $\infty$ and $w$ we consider
$$S=\frac{1}{z-w}$$
which is indeed a Mobius transformation making the above a composition of Mobius transformations and, thus, a Mobius transformation.