What is the special thing about |z|=2, will this point lie in the mandelbrot set?

Question

For the quadratic iteration $z \to z^2+4$, if you perform a few iterations letting $z_0 =0.5+1.936491673i$, the modulus of the points will be 2 ( or closer 2 two because of the inaccuracy of the decimals). Will the point lie in Mandelbrot set?

Answer 1

Your question doesn't isn't about the Mandelbrot set, because $z_0 \ne 0$ for your iteration. Instead, you are looking at the Julia set for $c = 4$, and the fixed points of the mapping $z \mapsto z^2 + c$, which are easily found via the quadratic formula: $$z_0 = \frac{1 \pm i\sqrt{15}}{2}.$$ Indeed, for any $c \in \mathbb C$, there are at most two such fixed points: $$z_0 = \frac{1 \pm \sqrt{1-4c}}{2}.$$ These are obviously contained in the Julia set $J_c$, but there is nothing special about their magnitudes in the sense that it does not need to equal $2$.

We note that $c \in \mathbb C$ is a member of the Mandelbrot set $M$ if and only if the orbit of the mapping $z \mapsto z^2 + c$ remains bounded for the initial value $z_0 = 0$. Thus, your computation does not say anything about whether $c = 4 \in M$.