A harder neutral fixed point problem

Let $f(z) = -z+z^4$. Use the the Leau-Fatou flower theorem to find (weakly) attracting and repulsive directions for the neutral fixed point at the origin.

Is there some way to conjugate this so it looks like $f(z)=z+az^{n+1}+O(z^{n+2})$?

@RedCrayon No - I recommend that you take a look at $f^2$.

Got it!

Note the Julia is invariant under composition of the function. $$f\circ f=z-4z^7+O(z^{10})$$

Note $n=6$ and $a=-4$. Suppose $$nav^n=-24v^6=1$$ Then $$v=(-1/24)^{1/6}=24^{-1/6}(e^{(2k+1)\pi i})^{1/6}=24^{-1/6}e^{(2k+1)\pi i/6}, \quad k=0,1,2,3,\cdots$$ but there exist unique values of $v$ only for $k=0,1,2,3,4,5$. To get the repulsive directions, substitute our values of $k$ into the exponential term and simplify. $f$ is repulsive in the directions $\sqrt{3}/2+i/2$, $i$, $-\sqrt{3}/2+i/2$, $-\sqrt{3}/2-i/2$, $-i$, and $\sqrt{3}/2-i/2$.

Suppose instead that $$nav^n=-24v^6=-1$$ Then $$v=(1/24)^{1/6}=24^{-1/6}(e^{2k\pi i})^{1/6}=24^{-1/6}e^{2k\pi i/6}, \quad k=0,1,2,3,\cdots$$ but there exist unique values of $v$ only for $k=0,1,2,3,4,5$. To get the attractive directions, substitute our values of $k$ into the exponential term and simplify. $f$ is attractive in the directions $1$, $1/2+i\sqrt{3}/2$, $-1/2+i\sqrt{3}/2$, $-1$, $-1/2-i\sqrt{3}/2$, and $1/2-i\sqrt{3}/2$.

Looks great again! Here's a picture of the Julia set with the weakly attractive and repulsive vectors:

Can you explain the first step where the Julia is invariant under the composition of this function $f∘f=z−4z^7+O(z^{10})$? I'm a little unclear about the Leau-Fatou flower theorem.

@Yousername The invariance of the Julia set under composition itself doesn't have anything to do with the flower theorem. It just helps us put our function in a form we can deal with. Remember, we need a function of the form $$f(z)=z+az^{n+1}+O(z^{n+2})$$ to apply the flower theorem.

I think, the invariance of the Julia set under composition of the function is a product of the Julia set being the closed set of repelling period points of the function. Conjugating just takes these periodic points and adjusts their period. If a periodic point $z_0$ used to have the orbit $$z_0\rightarrow z_1\rightarrow z_2\rightarrow z_3\rightarrow z_0,$$ after conjugating once it now goes $$z_0\rightarrow z_2\rightarrow z_0$$ It's tempting to claim that $z_1$ and $z_3$ have been ejected from the Julia set since they are no longer a part of the orbit containing $z_0$ and $z_2$, but actually they're just hanging out in a different orbit in the Julia set. So all the points in the Julia set might shuffle into different orbits when you conjugate, but all of the points are still there somewhere.