The <holomorphic fixed-point index>, or residue index, is
$$
\iota(f,z_0)=\operatorname*{Res}_{z=z_0}\frac1{z-f(z)}.
$$
For a simple fixed point with multiplier $\lambda\ne1$ it equals $1/(1-\lambda)$. Three distinct fixed points of a quadratic rational map are simple, and the rational fixed-point formula gives
$$
\frac1{1-\lambda_0}+\frac1{1-\lambda_1}+\frac1{1-\lambda_2}=1.
$$
If, say, $\lambda_0\lambda_1=1$, then
$$
\frac1{1-\lambda_0}+\frac1{1-\lambda_0^{-1}}=1.
$$
The remaining term would have to be zero, which is impossible. Thus the <multiplier relation for three distinct fixed points of a quadratic rational map> is
$$
\boxed{\lambda_i\lambda_j\ne1\quad(i\ne j)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 157 / 4 / a / Solution

If a quadratic rational map has three distinct fixed points with multipliers $\lambda_0,\lambda_1,\lambda_2$, then $\lambda_i\lambda_j\ne1$ for $i\ne j$. Otherwise the two corresponding indices already sum to one, contradicting the nonzero third index.


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