A circular julia set

Let $f(z) = 16/z^3$. Show that the Julia set of $f$ is exactly the circle of radius 2 centered at the origin.

I'll take a stab at this one.

Let $z_0 \in \mathbb{C}$ such that $|z_0|=2$. Then we note, $|z_0^3|=2^3=8 \Rightarrow |f(z_0)|=|\frac{16}{z_0^3}|=|\frac{16}{8}|=2$.

Now instead let $|z_0|>2$. Then $|z_0^3|>|2^3| \Rightarrow |z_0^3|>8$. This in turn implies that $|f(z_0)|=|\frac{16}{z_0^3}|<2$.

Now, let us assume that $|z_0|$ < 2. Then note $|z_0 ^3|<8 \Rightarrow |f(z_0)| > 2$.

So, if we choose a starting value of $z_0$ where $|z_0|$ =2, then our function $f(z)$ returns another complex number of absolute value 2, no matter how many times we iterate. On the other hand, if we choose some $z_0$ where $|z_0|\neq 2$, then with each iteration of our function our output goes from inside the circle to outside the circle, or vice versa. Thus, it is not possible to define a boundary for these values under iteration.

Accordingly the set of all of the elements $z_0 \in \mathbb{C}$ where $f(z_0)$ remains bounded under iteration have the trait $|z_0|=2$, and so the Julia Set of $f(z)$ is a circle of radius $2$, centered at the origin.

@SomeCallMeTim I think this looks close. I'd focus instead on
$$F(z) \equiv f(f(z)) = z^9/256.$$
Then, $F$ is just a polynomial with the same Julia set as $f$ and we have a simple description of the Julia set of $F$ - namely, it's the boundary of the basin of attraction of $\infty$.