A simple 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.

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

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

This looks good. Here's an image of the Julia set with the weakly attractive and repulsive vectors shown:

This all makes sense to me, but can you explain exactly how you are plugging in the values of k into the exponential term and simplifying?

@Yousername

RedCrayon is using the fact that
$$e^{i\theta} = \cos(\theta)+i\sin(\theta).$$
Thus, for example,
$$e^{i\pi/3} = \cos(\pi/3)+i\sin(\pi/3) = \frac{1}{2}+i\frac{\sqrt{3}}{2}.$$

Keep in mind that, for an equation like $v^n=1$, we expect $n$ solutions equally spaced about the unit circle.