We can study the iteration of $$ f_c : z \mapsto z^2+c $$ Hence we can define $\displaystyle \left(z_n\right)_{n \in \mathbb{N}}$ $$ z_0=0,\ z_{n+1}=z_n^2+c=f_c\left(z_n\right) $$ The set of points $c$ containing the sequence $\displaystyle \left(z_n\right)_{n \in \mathbb{N}}$ that stays bounded, we get the Mandelbrot Set.
If i'm right, if we got a fixed point of $f_c$, let's call it $z^{\ast}$ ; The sequence $\displaystyle \left(z_n\right)_{n \in \mathbb{N}}$ converges if $\left|f_c'\left(z^{\ast}\right)\right|<1$, diverges if $\left|f_c'\left(z^{\ast}\right)\right|>1$ and has an unpredictable behavior if $\left|f_c'\left(z^{\ast}\right)\right|=1$.
My question is about understanding why the case $\left|f_c'\left(z^{\ast}\right)\right|=1$ corresponds to the boundary of the set in which the sequence $\displaystyle \left(z_n\right)_{n \in \mathbb{N}}$ converges. For me, it is a just a case in which we cannot conclude, why would it be the boundary ? Is there a way to prove it ?
For example, why is the main cardioid ( which is a boundary ) can be found using $\left|f_c'\left(z^{\ast}\right)\right|=1$
When $|f'_c(z^*)|\lt 1$ the fixed point is attractive. When you are close enough to the fixed point that the quadratic terms are small the distance to the fixed point is multiplied by $f'_c(z^*)$ at every iteration, so the iteration will converge to the fixed point. On the other hand, if $|f'_c(z^*)|\gt 1$ the distance will increase at every iteration, so it will not converge. The points where $|f'_c(z^*)=1$ are the boundary between these two regions so solving that equation gives you the boundary of the region where each point converges to the fixed point.
All the other regions (as distinct from the filaments) of the Mandelbrot set converge to a cycle of some length. The big circle to the left of the main cardioid is the set of points that converge to a cycle of length $2$. It is not hard to prove that the magnitude of the derivative is the same at all points of a cycle. You can find that boundary by solving $|(f_c^2)'(z^*)|=1$ using either of the points of the limit cycle as $z^*$. Each other region has a period as well. The period is the exponent of $f$ in the preceding.