Let $d$ be an integer larger than 2 and let $N$ be the newton's method function for $f(z)=z^d-1$. Show that zero is in the Julia set of $N$.
Two facts you may assume:
$$N(z)\equiv z-\frac{f(z)}{f^\prime(z)}=\frac{z^2(-1/z+d/z+1/z^{1+d})}{d}$$ We must show $(1)$ $\infty$ is a fixed point of $N$, $(2)$ $\infty$ is a repelling fixed point of $N$, and $(3)$ zero maps to $\infty$. Since $d>2$, it's fairly obvious to see that $N(\infty)=\infty$. The derivative of $N(z)$ is $$N^\prime(z)=\frac{1/z^d-d/z^d-1+d}{d}$$ Since $0<|N^\prime(\infty)|=1-1/d<1$ for $d>2$, $\infty$ appears to be an attractive fixed point for $N$. Can somebody explain this?
@RedCrayon I would take a look at
$$1/N(1/z) = \frac{d z}{\left(\frac{1}{z}\right)^{-d}+d-1}.$$
I have a general idea of how to find the three things @RedCrayon listed but why do we need to find them?
@Person
First of all, I really should have conjugated so $\infty$ mapped to zero and zero mapped to some other finite point, like $1$. Here's a conjugacy that does that: $$\phi(z)=\frac{1}{z+1}$$ We want to show three things: that $\infty$ is a fixed point of $N$, that $\infty$ is a repulsive, and that zero maps to $\infty$.
Equivalently, we must show that zero is a fixed point of $\phi\circ N\circ\phi^{-1}$, that $zero$ is a repulsive, and that $1$ maps to zero.
Recall that the Julia set of $N$ is the closed set of repelling periodic points under iteration of $N$. So we want to show that $\infty$ has repulsive behavior of some kind. Showing that it is a repulsive fixed point of $N$ suffices.
If we then show that $N(0)=\infty$, we also show that zero is in the Julia set of $N$ since the Julia set of $N$ is closed under iteration of $N$. This related to why composing $N$ with itself any finite number of times gives the same Julia set.
I haven't tried sowing that zero is in the Julia set of $N$ using a direct method (like showing it is a periodic point), but I assume you'll get something messy like a $1/0$ or indeterminant form.