Julia set

Let $U$ be an open subset of the complex plane and let $f:U\to U$be analytic. Denote the $n$-th iterate of $f$ by $f^n$, i.e. $f^1=f$and $f^{n+1}=f\circ f^n$. Then the Julia set of $f$ is thesubset $J$ of $U$ characterized by the following property: if $z\in J$then the restriction of $\{f^n\mid n\in \mathbb{N}\}$ to any neighborhoodof $z$ is not a normal family.

It can also be shown that the Julia set of $f$ is the closure of the set ofrepelling periodic points of $f$. (Repelling periodic point means that, forsome $n$, we have $f^n(z)=z$ and $|f^{\prime }(z)|>1$.)

A simple example is afforded by the map $f(z)=z^2$; in this case, the Juliaset is the unit circle. In general, however, things are much more complicatedand the Julia set is a fractal.

From the definition, it follows that the Julia set is closed under $f$ andits inverse — $f(J)=J$ and $f^{-1}(J)=J$. Topologically, Julia setsare perfect and have empty interior.