Show that the Julia set of $f(z) = -2 z^2 + z^4$ is connected.

An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# A connected quartic Julia sets

mark

Levente

We take the derivative of $f$ to be $f' = -4z + 4z^3 $, setting this to $0$ we obtain $f'(z) = 0$ at $z = 0, -1 , 1$. Iterating each of these critical points under $f$ we have $f^n (0) = 0$, $f(1) = 1$ and for $n>1$ we have $f^n(1) = -1$, similarly $f^n ( -1) = -1$. Thus all of these critical orbits remain bounded and therefore the Julia Set for $f$ is connected.

Question: Is it required for all the critical orbits to remain bounded? Can we use the polynomial escape criterion? ( or does that only tell us that if $|z_0| > R$ then iteration of $f$ under $z_0$ diverges?).

mark

@Levente For a polynomial, there are essentially three possibilities:

- All the (finite) critical points remain bounded, in which case the Julia set is connected,
- All the critical points diverge to $\infty$, in which case the Julia set is totally disconnected, or
- Some of the finite critical points remain bounded and some diverge to $\infty$, in which case the Julia set comes in infinitely many non-trivial components.

Of course, the third case cannot happen when the polynomial has degree two.