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