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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(z0)=z0, and that our conjugacy function is ϕ(z)=z+z0. We want to show that the conjugate g(0)=0.

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

ϕ(g(z))=g(z)+z0,
f(ϕ(z))=f(z+z0)

We want to show that when f(z0)=z0,g(0)=0.

Substituting z=0 into the two equations, we get g(0)+z0=f(0+z0).
This turns into g(0)+z0=f(z0), and we know f(z0)=z0.
Thus, g(0)+z0=z0, and after subtracting a z0 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(z0) is related to g(z0), but here's where I'm stuck.

HELP

RedCrayon

Suppose z0 is a fixed point of f so that f(z0)=z0.

Note that φ1(z)=zz0 since (φφ1)(z)=(φ1φ)(z)=z

Then, g(z)=φ1(f(φ(z)))=φ1(f(z+z0))=f(z+z0)z0

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

Note that the behavior of z0 is classified by the magnitude of the multiplier |f(z0)| and the behavior of 0 is classified by the magnitude of the multiplier |g(0)|.

Therefore, to show that the conjugacy φ preserves the behavior of the fixed point 0, it is sufficient to show that |g(0)|=|f(z0)|

By the chain rule, g(z)=f(z+z0) so |g(0)|=|f(z0)|

Thus, the conjugacy φ must preserve the behavior of the fixed point 0.

Yousername

@RedCrayon

Can you explain further how you got from
g(z)=f(z+z0) by the chain rule, to |g(0)|=|f(z0)|?
I'm a little confused about this part.


RedCrayon

@Yousername Just plug in z=0. Then take the magnitude of both sides.