Given a function $f$, the Julia set of $f$, $J(f)$, is the boundary of $A(\infty)$, where $A(\infty)$ is the "basin of attraction to infinity", ie.

$$\begin{aligned} A(\infty) = \{ z \in \mathbb{C}: f^n(z) \rightarrow \infty \text{ as } n \rightarrow \infty \} \end{aligned}$$

The filled-in Julia set of $f$, $K(f)$, are those $z \in \mathbb{C}$ for which $f^n(z)$ stays bounded:

$$\begin{aligned} K(f) = \{ z \in \mathbb{C}: \{ f^n(z) \} \text{ is bounded}\} \end{aligned}$$

Given $f(z) = z^2 + c$, where $c \in \mathbb{C}$, how can we find $K(f)$ ? Turns out

$$\begin{aligned} R = \frac{1 + \sqrt{1 + 4\lvert c \rvert}}{2} \end{aligned}$$

may help !

Let $z_0 \in \mathbb{C}$. If for some $n > 0$ we have $\lvert f^n(z_0) \rvert$ > R, then $f^n(z_0) \rightarrow \infty$ as $n \rightarrow \infty$ !

That is, $z_0 \in A(\infty)$, so $z_0 \notin K(f)$.

You can find more info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 6:56 PM