How to find the residue of $\frac{1}{z-f(z)}$?

Question

I am trying to find the residue of $\frac{1}{z-f(z)}$ at a fixed point w of f. f is a racional function and $|f'(w)|=1$. I am a little stuck, any help or hints will be greatly appreciated.

Answer 1

Let $w$ be your fixed point. If it turns out that $f'(w)\neq1$, then the answer is $\frac1{1-f'(w)}$. Otherwise, the answer is $\frac{2f'''(w)}{3(f''(w))^2}$. We can, without loss of generality, assume that $w=0$. Then $f(0)=0$ (because $0$ is a fixed point). Since and $f'(0)=1$, $f(z)=z+a_2z^2+a_3z^3+\cdots$, with $a_n=\frac{f^{(n)}(0)}{n!}$. Then\begin{align}\frac1{z-f(z)}&=\frac1{-a_2z^2-a_3z^3-\cdots}\\&=\frac1{z^2}\cdot\frac1{-a_2-a_3z-\cdots}\\&=\frac1{z^2}.(b_0+b_1z+b_2z^2+\cdots).\end{align}Clearly, $b_0=-\frac1{a_2}$. Besides, $-a_2b_1-a_3b_0=0$ and\begin{align}-a_2b_1-a_3b_0=0&\iff-a_2b_1+\frac{a_3}{a_2}=0\\&\iff b_1=\frac{a_3}{{a_2}^2}\\&\iff b_1=\frac{2f'''(0)}{3(f''(0))^2}.\end{align}This, of course, assumes that $f''(w)\neq0$. I don't know the answer if $f''(w)=0$.