1

We know if $R(z)$ is a rational function, then it is in the form $R(z) = f(z)/g(z)$, where $f$ and $g$ are complex polynomials. If we want to find its fixed points, we can take $R(z) = z$, which gives the equation $f(z) - z\cdot g(z) = 0$. By the Fundamental Theorem of Algebra, this equation has finitely many roots; does that mean that those roots are fixed points of $R$?

Integrand
  • 8,457

1 Answers1

1

You are correct: the fixed points (in $\mathbb C$) of the rational fraction $R(X) = \frac{P(X)}{Q(X)}$ are exactly the roots of the polynomial $P(X) - XQ(X)$. In particular, unless $R(X) = X$, the number of fixed points is finite and bounded by $\max \{\deg P, \deg Q + 1\}$.

DodoDuQuercy
  • 1,760