Let $R(z)=\frac{a_0+a_1z+\cdots+a_n z^n}{b_0+b_1z+\cdots +b_m z^m}$ where $a_nb_m \neq 0$ and $n >m$. I understand that this fixes $\infty$ because the degree of the top is greater than the degree of the bottom. Where I'm stuck is the following comment:
In particular, $R(z)$ has exactly one fixed point at $\infty$ if one of the following are true: $n > m+1$, or if $n=m+1$ and $b_m \neq a_n$.
I'm just wondering if someone could provide some intuition to this.
Latest and Greatest Attempt
I started pouring through all of the details of my book. It got me thinking about the problem a bit more. I stumbled on an easy example: $z^3+z$ has three fixed points at the origin.
For this problem I'm considering the mobius transform $g(z)=1/z$. This way I'm sending $\infty$ to a finite point $0$.
So now I have $S(z)=g(R(g^{-1}(z)))$ and $S(\infty)=0.$ Unlike the example in my book, it seems like it would be rather difficult to find an explicit form for $S(z)$ that will enable me to compute the number at fixed points at $0$.
That's where I'm stuck. How would I count the number of fixed points at $0$ in this case.
Hopefully, your book has somewhere the definition of the multiplicity of a fixed point. If you don't have this definition it is completely pointless for you to think about this problem since it's talking about something about which you know nothing.
It should be something like this : If $f$ is defined on some open subset of $\Bbb C$ containing $0$ and $f(0)=0$, then the multiplicity of $0$ as a fixed point is the order of the zero of $f(z)-z$ at $0$, or more intuitively, to what order is $f$ similar to the identity at $0$.
Then you extend the definition to nonzero fixed points $a$ by conjugating $f$ with some automorphism $\tau$ that sends $0$ to $a$.
For this definition to make sense, this number has to be unchanged when you conjugate $f$ with some transformation $\tau$ invertible near $0$, satisfying $\tau(0)=0$. Proving this is a bit painful, but it works out in the end (I will skip it)
In your example, to apply this definition you need to conjugate $R(z)$ with the inversion $\tau(z) = z^{-1}$ and then study the multiplicity of $0$ as a fixed point of $R' = \tau^{-1} \circ R \circ \tau$.
Now, $R'(z) = \frac {b_0z^n + b_1z^{n-1} + \cdots + b_m z^{n-m}}{a_0z^n + a_1z^{n-1} + \cdots + a_n}$.
Then,
$(a_0z^n + a_1z^{n-1} + \cdots + a_n)(R'(z)-z) = \\ -a_0 z^{n+1} + ((b_0-a_1)z^n + \cdots + (b_m-a_{m+1})z^{n-m}) - (a_{m+2}z^{n-m-1} + \cdots + a_n z)$
$a_n$ is nonzero so the order of $0$ as a zero in the first factor is $0$, and so the multiplicity is the order of $0$ as a $0$ in the polynomial in the right hand side.
This multiplicity is exactly $1$ if and only if the coefficient of $z^1$ doesn't vanish (if it vanishes then the multiplicity is higher).
If $n=m+1$ then this coefficient is $b_m-a_{m+1} = b_m-a_n$, so in this case you need $b_m \neq a_n$.
If $n>m+1$ then this coefficient is $-a_n$, so in this case you need $a_n \neq 0$, which was already in the assumptions of the problem.
So the fixed point has multiplicity exactly $1$ if and only if $n>m+1$ or ($n=m+1$ and $b_m \neq a_n$)
One neat thing about this way of counting multiplicities is that a degree $n$ rational fraction that is not the identity should always have $n+1$ fixed points counted with multiplicity