An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# Fixed point conjugacy

notneds
1. Suppose that f has a fixed point at z0.
Show that the function g obtained by conjugating f with the function φ(z)=z+z0 has a fixed point at zero. In addition, show that the conjugation preserves the nature of the fixed point as attractive, super-attractive, repulsive, or neutral.

Can anyone beat Mark to answer this question?

WillHeDoTheAssignmen

So what we have to work with is that we know $f(z_0) = z_0$, and that our conjugacy function is $\phi (z) = z+z_0$. We want to show that the conjugate $g(0)=0$.

The way I started this problem was setting up our conjugation, namely $f(\phi(z))=\phi(g(z))$ and substituting in what I know.

$\phi(g(z)) = g(z) + z_0$,
$f(\phi(z)) = f(z+z_0)$

We want to show that when $f(z_0) = z_0, g(0)=0$.

Substituting $z=0$ into the two equations, we get $g(0) + z_0 = f(0+z_0)$.
This turns into $g(0)+z_0 = f(z_0)$, and we know $f(z_0) = z_0.$
Thus, $g(0) +z_0 = z_0$, and after subtracting a $z_0$ from both sides we get $g(0) = 0$.

To show that the nature of the fixed point is preserved, I think we need to show that $f'(z_0)$ is related to $g'(z_0)$, but here's where I'm stuck.

HELP

RedCrayon

Suppose $z_0$ is a fixed point of $f$ so that $f(z_0)=z_0$.

Note that $\varphi^{-1}(z)=z-z_0$ since $$(\varphi\circ\varphi^{-1})(z)=(\varphi^{-1}\circ\varphi)(z)=z$$

Then, $$g(z)=\varphi^{-1}(f(\varphi(z)))=\varphi^{-1}(f(z+z_0))=f(z+z_0)-z_0$$

This preserves the result found by @WillHeDoTheAssignmen since $g(0)=0$.

Note that the behavior of $z_0$ is classified by the magnitude of the multiplier $|f^\prime(z_0)|$ and the behavior of $0$ is classified by the magnitude of the multiplier $|g^\prime(0)|$.

Therefore, to show that the conjugacy $\varphi$ preserves the behavior of the fixed point $0$, it is sufficient to show that $$|g^\prime(0)|=|f^\prime(z_0)|$$

By the chain rule, $$g^\prime(z)=f^\prime(z+z_0)$$ so $$|g^\prime(0)|=|f^\prime(z_0)|$$

Thus, the conjugacy $\varphi$ must preserve the behavior of the fixed point $0$.

$g′(z)=f′(z+z_0)$ by the chain rule, to $|g′(0)|=|f′(z_0)|$?
@Yousername Just plug in $z=0$. Then take the magnitude of both sides.