- Show that the Julia set of $f(z)=z^3+1$ is totally disconnected.
- Show that the Julia set of $f(z)=z^3+i$ is connected.

# Connected vs totally disconnected cubic Julia sets

To figure out if the Julia set of a function is connected or totally disconnected we simply find the critical orbit. In this case both functions have a critical point at $z = 0$. For $f(z) = z^3 + 1$ we have $f^n (0) = f^{n-1}(0)^3 + 1$ which diverges to infinity. Thus the Julia of $z^3 + 1$ is totally disconnected. For $f(z) = z^3 + i$ iterating the critical orbit of $z = 0$ gives $f^n (0) = 0 $ when $n$ is even and $i$ when $n$ is odd. Thus this orbit remains bounded and therefore the Julia set of $f(z) = z^3 + i$ is connected.

@Levente Yes - this is good. I think it might be worth mentioning, though, that you need to check *all* critical orbits and it just so happens that this function has just one critical point. I just created this question to illustrate the point.