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I'm stuck with this problem: let $f(z)=z^2$ be a complex function, and $f^n(z)=z^{2^n}$ (iterations of $f$). $\forall z \in B:=B(0,1)$ (unit circumference) ,the set $ \cup_{n \in \mathbb{N}} f^{-n}(z) $ is dense in B, where $f^{-n}(z)$ is the inverse image of $f^n(z)$.

I have tried this: let $z=e^{i\theta}$, $f^{-n}(z)=\{e^{\frac{\theta + 2\pi k}{2^n}} : k=0,1,...,2^n-1\}$, and let $w \in B$ I have searched for $\forall r>0, \exists n,k$ so that, $|w-e^{\frac{\theta + 2\pi k}{2^n}}|<r$ (exists a inverse image of z in $B(w,r)$).

If we call $w=e^{i\alpha}$ the before inequality is $|e^{i\alpha}-e^{\frac{\theta + 2\pi k}{2^n}}|=|e^{i\alpha}(1-e^{\frac{\theta + 2\pi k}{\alpha2^n}})|=|1-e^{\frac{\theta + 2\pi k}{\alpha2^n}}|<r $ for such we search for $n,k$ so that $|\frac{\theta + 2\pi k}{\alpha2^n}|<r'$. If we fix $k$, $\exists N>0 $ so that $\forall n>N \rightarrow |\frac{\theta + 2\pi k}{\alpha2^n}|<r'$ why we have found $n$ and $k$.

However, if we get $w' \in B, w'\notin B(w,r)$ and let r'' so that $B(w,r)\cap B(w',r'')=\emptyset $. We fix $k$ sufficiently small, exists $N,N'$ so that $|w-e^{\frac{\theta + 2\pi k}{2^n}}|<r$ and $|w'-e^{\frac{\theta + 2\pi k}{2^{n'}}}|<r, \forall n>N$ and $n'>N'$. But this is a contradiction because $B(w,r)\cap B(w',r'')=\emptyset $.

thanks

P.S: this is a case of Julia set

Josegg
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2 Answers2

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For each $n$ the preimages $f^{-n}z$ are uniformly distributed on the circle, and when $n\to\infty$ the distance between consecutive preimages (which corresponds to the angle $\pi/2^{n-1}$) tends to zero. This immediately shows that the union of the preimages is dense.

John B
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This case is similar to show that $A=\{ \frac{m}{2^n} : m \in \mathbb{Z} , n \in \mathbb{N} \}$ is dense in $\mathbb{R}$. Show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N} \} $ is dense in $\mathbb {R}$ . I have already solved this problem, thanks.

Community
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Josegg
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