Mandelbrot set is the set of all sequences defined as follow :

$$c \in \mathbb{C} \qquad u_c(n+1) = u_c(n)^2 + c \qquad u_c(0) = 0$$

with a convergent module. This set can be represented as follow with, in black, values of \(c\) for which module converge, and in black, those for which module diverge :

This representation allows us to see that Mandelbrot set seems to be symmetric about the real axis. But how to prove it ? It is what I'm going to show you.

Proof

First, we have to prove that \(u_{\overline c}(n) = \overline{u_c(n)}\). To do it, we are going to use induction with the following statement :

$$\forall n \in \mathbb{N} \quad H_n : u_{\overline c}(n) = \overline{u_c(n)}$$

that leads us to the following reasoning :

Bases :
For \(n = 0\), \(u_{\overline c}(0) = 0 = \overline{u_c(0)}\). So, \(H_0\) holds.

Induction steps :
For \(n+1\) :

$$u_{\overline c}(n+1) = u_{\overline c}(n)^2 + \overline c$$And, as we assume \(H_m\) holds :

$$u_{\overline c}(n+1) = \overline{u_c(n)}^2 + \overline c = \overline{u_c(n)^2} + \overline c = \overline{u_c(n)^2 + c} = \overline{u_c(n+1)}$$

Consequently, \(H_n \Rightarrow H_{n+1}\).

Conclusion :
\(H_n\) holds \(\forall n \in \mathbb{N}\).

Now, we can easily prove what is interesting us. We just have to notice that \(|a| = |\overline a|\) and so \(|\overline{u_c(n)}| = |u_c(n)| = |u_{\overline c}(n)|\), that allows us to write :

$$\lim_{n \to +\infty} |u_c(n)| = \lim_{n \to +\infty} |u_{\overline c}(n)|$$

In other words, we have just proved that the convergence of the module of the sequence with \(c\) is the same than with \(\overline c\), where \(c\) and \(\overline c\) are 2 complex numbers symmetric about the real axis, what proves the symmetry.