Julia Sets of $z^k$.

Question

I'm trying to figure out a way to show that the Julia sets of $z^k$ with $k\geq 2$ are the unit circle. I had a question posted before, but I wanted to add more details and how I'm thinking about it differently.

Intuitively, I get why it must be the unit circle based on the discussion below. Let $R(z)=z^k$. Then $R^n(z)=z^{k^n}$. As $n \to \infty$ when $|z|>1$ we have that $R^n(z) \to \infty$. However, if $|z|<1$, we have that $R^n(z)\to 0$ as $n \to \infty$. So the place I have to consider are the points $|z|=1$. In my mind, it's clear that any open set contain a point on the unit circle will contain points that will both go to $0$ or $\infty$.

Theorem: Let $F$ be any family of maps, each mapping $(X,d)$ into $(X_1,d_1)$. Then there is a maximal open subset of $X$ on which $F$ is equicontinuous. In particular, if $f$ maps a metric space $(X,d)$ into itself, then there is a maximal open subset of $X$ on which the family of iterates $\{f^n\}$ is equicontinuous.

I'm not sure if this is the correct way of applying the above theorem, but $R^n(z)$ maps points outside of the unit circle to points still outside the unit circle and similarly to those within the unit circle. But I'm still not sure how to make the connection to the points on the unit circle.

Answer 1

Your question is essentially answered in comments but let me spell it out anyway. Let's start with the definition. Set $$ \hat{{\mathbb C}}= {\mathbb C}\cup \infty, $$ the Riemann sphere.

Definition. If $R\in {\mathbb C}(z)$ is a rational function, then the Fatou set $F(R)$ consists of all points $z_0\in \hat{{\mathbb C}}$ such that $z_0$ has a neighborhood $U$ in $\hat{{\mathbb C}}$ so that the sequence $(R^n(z))$ forms a normal family on $U$, i.e. every subsequence contains a further subsequence whose restriction to $U$ which converges uniformly to a holomorphic function $U\to \hat{{\mathbb C}}$. The complement $$ J(R)=\hat{\mathbb C} - F(R) $$ is called the Julia set of $R$.

You already know that for the function $R(z)=z^k$ the Fatou set $F(R)$ contains $$ \{z: |z|< 1\} \sqcup \{z: |z|>1\} \cup \{\infty\}. $$ Proving this is based on a standard Calculus exercise: If $a$ is a complex number such that $|a|<1$ then $$ \lim_{n\to\infty} a^n=0, $$ while if $|a|>1$ then $$ \lim_{n\to\infty} a^n=\infty, $$ and convergence is uniform on compacts in $D=\{z\in {\mathbb C}: |z|<1\}$ and in $D^*=\{z\in \hat{\mathbb C}: |z|>1\}$ respectively.

Thus, it remains to show that $F(R)$ is disjoint from the unit circle $S=\{z: |z|=1\}$. This would imply that $J(R)=S$.

Here is a proof of disjointness.

Consider any open disk $U=B(a, r)$ of radius $r>0$ centered at $a\in S$, $|a|=1$. Then for every $$ z\in U_0=B(a,r)\cap D\ne \emptyset$$ we get $$ \lim_{n\to\infty} R^n(z)=0, $$ while for every $$ z\in U_1=B(a,r)\cap D^*\ne\emptyset$$ we get $$ \lim_{n\to\infty} R^n(z)=\infty. $$

The same holds for any subsequence $R^{n_j}(z)$. Hence, such a subsequence cannot converge on $U$ (even pointwise) to a continuous function $f: U\to \hat{\mathbb C}$: $$ f(a)= 0= \lim_{j\to\infty} f(z_j), z_j\in U_0, z_j\to a, $$ $$ f(a)= \infty= \lim_{j\to\infty} f(z_j), z_j\in U_1, z_j\to a. $$ Thus, $S\cap F(R)=\emptyset$, hence, $J(R)=S$.

Answer 2

The theorem you state does not really help in finding the Fatou set. It just states that it exist, being the maximal open subset $F$ for which every point $z_0\in F$ has a neighborhood $D$ so that $\{f^n,n\geq 1\}$ is an equi-continuous family on $D$ (with respect to the spherical metric on the Riemann sphere, $\widehat{\Bbb C}$).

Equicontinuity implies uniform convergence of a subsequence on compact subsets so suppose $z_0\in F$ and $|z_0|=1$. Then there is a closed (compact) ball $K=\overline{B}(z_0,r)\subset D$, $r>0$ and a subsequence $f^{n_k}$ converging uniformly on $K$ (with image in $\widehat{\Bbb C}$). The limit of such a subsequence $f_* = \lim_k f^{n_k}_{|K}$ is then continuous. This is not possible, however, since $|f_*(z_0)|=\lim_k|f^{n_k}(z_0)|=1$ but $f_*(z)=\lim_k f^{n_k}(z) = 0$ for all $z\in U=K\cap \{|z|<1\}$ and as there is a sequence $z_m\in U$ converging to $z_0$, such a limit $f_*$ can not be continuous at $z_0$.

It suffices here to look at what happens on the circle and in the disk to get the result. You don't have to consider what happens outside of the disk for the proof (but could of course have done so with the same conclusion).

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The map $z\mapsto z^2$ is perhaps deceptively simple. You might find it useful with a fairly simple criteria regarding the conformal derivative as described in the following:

Recall that the Riemann sphere $\hat{{\Bbb C}}={\Bbb C}\cup\{\infty\}$ is a complex manifold which may be described through two charts $\Omega_0={\Bbb C}$ and $\Omega_\infty = {\Bbb C}^*\cup\{\infty\}$. The mapping $z\mapsto w=\frac{1}{z}$ yields the coordinate transformation between the charts on their overlap $\Omega_1\cap \Omega_2 = {\Bbb C}^*$. There is a natural conformal Riemannian metric on $\hat{\Bbb C}$ associated with this description (which in particular is invariant under the chart map $z\mapsto w=1/z$): $$ds = \frac{|dz|}{1+|z|^2}=\frac{|dw|}{1+|w|^2}. $$ Then, given a holomorphic map $f$ from an open subset $D\subset \hat{\Bbb C}$ into $\hat{\Bbb C}$ we may associate a conformal (spherical) derivative to this map by setting: $$ Df(z) = |f'(z)| \frac {1+|z|^2}{1+|f(z)|^2}. $$

Example: The map $f(z)=z^2$ extends naturally to a map on $\hat{\Bbb C}$ since going to the chart $\Omega_\infty$ we have the self map (first defined for $w\neq 0$, then extended by continuity): $$ \hat{f}(w) = \frac{1}{f(1/w)} = \frac{1}{(1/w)^2}= w^2$$ The conformal derivative equals: $$ Df(z) = 2 |z| \frac{1+|z|^2}{1+|z|^4}.$$

Coming back to the criteria of being normal: Let $\{f_n\}$ be a family of holomorphic maps $f_n:D\rightarrow \hat{\Bbb C}$. The family is normal iff the sequence of conformal derivatives $Df_{n}$ stays uniformly bounded on compact subsets of $D$. [The underlying reason is that derivatives and equi-continuity are related through Cauchy-integrals]

In your case you find that $\lim_n |Df^{n}(z_0)|=\lim_n 2^{2^n}=+\infty$ for $z_0\in S^1$ so $S^1$ must be disjoint from the Fatou set. On the other hand $\lim_n |Df^n(z_0)|=0$ for every $|z|\neq 0$ (also outside the disk) so all such points belong to the Fatou set.

Answer 3

By the sound of it, the theorem you stated is theorem 3.1.2 from this book, page 50.

Then comes the definition, which states that $F$ and $J$ are complementary, or $J = \mathbb{C_{\infty}} \setminus F$. Then comes theorem 3.1.5, which is more powerful than "but $R^n(z)$ maps points outside of the unit circle to points still outside the unit circle and similarly to those within the unit circle". It basically says that $F(R)=\{z: |z|<1\} \bigcup \{z: |z|>1\}=F(R^n), \forall n \in \mathbb{N},n>0$. Then $J(R)$ is what is left of $\mathbb{C_{\infty}} \setminus F$, which is the unit circle.

Alternatively, this result

If $f$ is an entire function, then $J(f)$ is the boundary of the set of points which converge to infinity under iteration.

reconfirms the fact that $J(R)$ is the unit circle.