Fixed point of rational functions.

Question

Consider the following rational functions: $R_1(z)=z^2+z-1$ and $R_2(z)=z+\frac{1}{z}$. The first has fixed points at $z^2+z-1=z \Rightarrow z=\pm 1.$ The second has a single fixed point at $\infty.$

Show that neither of the two following propositions implies the other:

(i) $R$ has all of it's fixed points at $\xi$
(ii) $R^{-1}\{\xi\}=\xi$

My reasoning that $(i)$ and $(ii)$ are not related is because of the following:

Notice that $R_1^{-1}(-1)=\{0,-1\}$ since $z^2+z-1=-1 \Rightarrow z^2+z=0 \Rightarrow z=0,-1$ But $R_1(0)=-1$. So $0$ is not a fixed point.

But considering $R_2$, we have $R_2(0)=R_2(\infty)=\infty$. But $R^{-1}\{{\infty}\}=\{0, \infty\}$

Answer 1

Take for example the rational function

$R = \frac{z^2-2z+3}{z-1}$

$R = z \Leftrightarrow z^2-2z+3 = z(z-1) \Leftrightarrow 3z = 3 \Leftrightarrow z = 3$. This is the only fixed point, so in this case (i) is satisfied.

However, $R^{-1}(3)$ contains more than just $3$: $\frac{z^2-2z+3}{z-1} = 3 \Leftrightarrow z^2-2z+3 = 3z-3 \Leftrightarrow z^2-5z+6 = 0$ which has two real roots (3 and 2).

So (i) is satisfied but (ii) is not, so (i) does not imply (ii).

The logic to give this example was that I was looking for a function $R = \frac{P}{Q}$ that had only one fixed point: so $P/Q = z$ has only one root $\Leftrightarrow P - Qz = 0$ is a degree $1$ complex polynomial.

That (ii) does not imply (ii) take $R_2 = \frac{z^2+2z-1}{z^2+3z-2}$. Consider $R_2 = 1 \Leftrightarrow z = 1$. Thus $R_2^{-1}(1) = \{1\}$. But consider $R_2(z) = z \Leftrightarrow z^3+2z^2-4z+1 = 0$ so there are 3 fixed points (1 and two other).

For the examples that emka gave, I think $R_1$ is actually used to show that (ii) does not imply (i) and $R_2$ is used to show that (i) does not imply (ii). While $R_2$ is a correct example, $R_1$ is not, but I believe they just interpreted the problem of proving (ii) implies (i) to mean that if a rational function has a fixed point then it has one and only one.

Answer 2

For easy counterexamples, consider $R_1(z) = z$ and $R_2(z) = 0$, with $\xi = 0$.

Answer 3

You have already found that $R_2$ and $\xi = \infty$ is a situation where (i) is true and (ii) is false.

To complete the proof that they are independent (that neither implies the other) you need a situation where (i) is false and (ii) is true.

$R_1^{-1}(\{1\}) = \{1,-2\}$ and $R_1^{-1}(\{-1\}) = \{-1,0\}$ so in those cases, (ii) is false, so this doesn't work.
However, by looking at its third fixpoint, which is $\infty$, you have $R_1^{-1}(\{\infty\}) = \{\infty\}$, so there (ii) is true, and (i) is false because $R_1$ has two other fixpoints.