I am doing a small project in school, which is dedicated to exploring what shapes might Julia sets of rational functions take. However, as we've started investigating into the results we've got before presenting them, it turned out that what we actually did was studying filled Julia sets for rational functions, i.e. $$\mathcal{K}(R) = \{z \in \mathbb{C} \,|\, R^n(z) \not\to \infty \text{ as } n \to \infty\};$$ and obtaining Julia sets as $$\mathcal{J}(R) = \partial \mathcal{K}(R).$$ However, it seems to be incorrect as it works only in case of $R$ being polynomial. But is it? I got totally confused at this point because all the Internet has to offer about filled Julia sets applies only to polynomials.
So my questions is whether the following statement is true: For a rational function $\mathbf{R : \hat{\mathbb{C}} \mapsto \hat{\mathbb{C}}}$ with $\mathbf{\infty}$ being an attracted fixed point of $\mathbf{R}$, the following two definitions are the same: $$\mathcal{J}(R) = \partial \mathcal{K}(R) \text{ with } \mathcal{K}(R) = \{z \in \mathbb{C} \,|\, R^n(z) \not\to \infty \text{ as } n \to \infty\}$$ and (the canonical one) : $$\mathcal{J}(R) = \hat{\mathbb{C}} \setminus \mathcal{F}(R) \text{ with } \mathcal{F}(R) \text{ being the Fatou set of } R.$$
The answer seems to be yes, but actually proving (or disproving) that requires more knowledge in area of complex dynamics than I have now. I would be really happy if somebody had already determined that (and somebody surely did, because it is crucial for approximations and computations of Julia sets) and if the answer is yes, I would just give a reference to the proof because my school does not require all proofs to be written out explicitly.
I will appreciate any help answering the question above. Thank you in advance!