Julia set of $z^2+2z$ using conjugation of $z^2$

Question

I wanted to calculate the Julia set of $S: z \mapsto z^2+2z$. I found that for $ \varphi: z \mapsto z-1 $ and $\varphi^{-1}: z \mapsto z+1$ the map $R: z \mapsto z^2$ is conjugate: $$ S(z) = z^2 + 2z = z^2+2z+1-1=(z+1)^2-1 = (\varphi^{-1}(z))^2-1 = (\varphi \ \circ \ R \ \circ \ \varphi^{-1})(z) $$ So the Julia set should just be $$J(S) = \varphi(J(R)) = \varphi(\mathbb{S}^1) = \left\{z-1 \ \middle| \ \ |z| = 1 \right\}$$ the one unit to the left shifted unit circle. But for example for $z=1, \varphi(z) = 0$ but $0$ is for sure not in the Julia set as it is a superattracting fixed point. I think I am missing something very obvious here, so please can you help me spot my mistake?

Answer 1

As @Towers pointed out $0$ is in fact a repelling fixed point as it has derivative $R'(z) = 2z+2$ and therefore $R'(0) = 2 > 1$, so it must be part of the Julia set. On the other hand the center of the circle $J(S)$ is the superattracting fixed point for $z=-1$ with $R'(-1) = 0$ which $\varphi$ maps to $0$, so there is no reason to doubt the proof I gave. Sorry for the confusion.