# Finding Julia set of $f(z)=z^2-1$

I am trying to determine the Julia set of $$f(z)=z^2-1$$. The fixed points are at $$z=\frac{1\pm\sqrt{5}}{2}$$ and I can see that for $$|z|>\frac{1+\sqrt{5}}{2}$$, then $$|f(z)|>|z|$$, so that $$|f^n(z)|$$ is a strictly increasing sequence.

So either $$|f^n(z)|\rightarrow\infty\Rightarrow z\in A(\infty)\subset F_f$$ or $$|f^n(z)|\rightarrow \alpha\in (\frac{1+\sqrt{5}}{2},\infty)$$.

How do I rule out the second possibility of approaching some circle where it might be unstable?

PS: I can also see that both fixed points are repelling periodic points, so must be in the Julia set, and that $$z=0,z=-1$$ are attracting periodic points, so must be in the Fatou set, but how can this be used to determine whether other points are in Julia set?

Any help would be much appreciated.

Let $$\phi = \frac{1 + \sqrt{5}}{2}$$ and consider $$z = \phi + \epsilon; 0 < \epsilon \in \mathbb{R}$$. Now \begin{aligned}&f(z)\\=&z^2 - 1 \\=& (\phi + \epsilon)^2 - 1 \\=& \phi^2 + 2 \phi \epsilon + \epsilon^2 - 1 \\=& \phi + 1 + \epsilon + \sqrt{5} \epsilon + \epsilon^2 - 1 \\=& \phi + \epsilon(1 + \sqrt{5} + \epsilon) \\>& \phi + (1 + \sqrt{5})\epsilon\end{aligned} Therefore the sequence $$z_{n+1} = z_n^2 - 1$$ with $$z_0 = \phi + \epsilon$$ is bounded below by $$z_n > \phi + (1 + \sqrt{5})^n\epsilon$$ and thus $$z_n \to \infty$$.
A more well known "escape radius" argument shows that if $$|z_k| > R > \max\left\{2, |c|\right\}$$ then $$z_{n+1} = z_n^2 + c \to \infty \text{ as } n \to \infty$$.