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

The shifty conjugacy


Suppose that $f$ has a fixed point at $z_0$. Show that the function \(g\) obtained by conjugating $f$ with the function $\varphi(z)=z+z_0$ 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.


Maybe I'm misunderstanding the problem statement but should $g$ have a fixed point at $z_0$ and not 0?


@Levente Hmm... I suppose it depends a bit on the order in which. I think that you can work out the correct order in the process of solving the problem. If it were an exam question, I think I'd better be more precise and say something like
$$\text{this}\circ\text{that}\circ\text{this}^{-1} = \text{the other}.$$


If we consider $\phi(z) = z- z_0$ then $\phi^{-1} = z + z_0$. We conjugate with $f$ to get $g$.

$$g(z) = \phi(f(\phi^{-1}(z))) $$ Now we consider $g(0) = \phi(f(\phi^{-1}(0))) = \phi(f(z_0)) = \phi(z_0) = 0$, thus $g$ has a fixed point at $0$.

Now to show that this fixed point at $0$ can be classified identically to $z_0$ all we need do is take the derivative of $g'(0)$. By application of the chain rule we get
$g'(0) = \phi'(f(\phi^{-1}(0))) \cdot f'(\phi^{-1}(0)) \cdot \phi^{-1}(0)$

Since $\phi$ and $\phi^{-1}$ both have derivative equal to $1$ everywhere we get $$g'(0) = f'(z_0)$$ Thus the classification of the fixed point at $0$ for $g$ is identical to the classification of the fixed point $z_0$ under $f$.


@Levente This looks good. If we take a look at our definition of conjugacy as stated in our class notes, we see that $f$ is conjugate to $g$ if there is a $\varphi$ such that
$$f\circ \varphi = \varphi \circ g.$$
This notation is consistent with all the references I've seen.

The question I wrote, though, asks us to consider what happens if we conjugate $f$ with $\varphi$. I guess it's unclear if that means
$$\varphi^{-1}\circ f\circ \varphi = g \:\:\:\: \text{ or } \:\:\:\: \varphi\circ f\circ \varphi^{-1} =g.$$

I had been thinking the former and you chose the latter. I think either is fine, especially this distinction isn't particularly important when working with conjugacy (where $\varphi$ is a bijection), rather than just semi-conjgacy (where $\varphi$ need not be a bijection).