In a comment to a previous answer it has been mentioned that the boundary of the Mandelbrot set contains the cardioid
$$
c = e^{it} \, \frac{2 – e^{it}}{4}
$$
but how can we prove this?
Complex Numbers – Why Mandelbrot Set Boundary Contains a Cardioid
complex numbersfractals
Related Solutions
I really like this question! I can't yet upvote, so I'll offer an answer instead. This is only a partial answer, as I don't fully understand this material myself.
Suppose that $f$ is a quadratic polynomial. Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a "quadratic-like map." Then we'll call $f$ renormalizable. (See Chapter 7 of McMullen's book "Complex dynamics and renormalization" for more precise definitions.) Now renormalization preserves the property of having a connected Julia set. Also the parameter space of "quadratic-like maps" is basically a copy of $\mathbb{C}$.
So, fix a quadratic polynomial $f$ and suppose that it renormalizes. Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$. This gives a map from a small region about $f$ to the space of quadratic-like maps. This gives a partial map from the small region to the Mandelbrot set and so explains the "local" self-similarity.
To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way. (I believe that there are issues as you approach the place where the baby is attached to the parent.) Thus renormalization explains why the baby Mandelbrot set appears.
Here's something I tried, thanks to Adam's comment for the basic idea for $p \ge 3$. This answer is missing some details, comments suggesting improvements are welcome, as would be other answers that fill in the gaps. It also relies on the fact (?) that:
$$\forall 0 \neq a \in \mathbb{C}, 0 \neq b \in \mathbb{C}, 1 < m \in \mathbb{N} . \exists x . |a e^{i x} + b e^{i m x}| \neq 1$$
Let $F(z, c) = z^2 + c$ with $F^{p+1}(z, c) = F^p(F(z, c), c)$. Now the boundary of a hyperbolic component can be parameterized by $\theta \in \mathbb{R}$ by the solution of the equation system: $$ F^p(z,c) = z \\ \frac{\partial}{\partial z}F^p(z,c) = e^{i \theta} $$ Now the question reduces to showing $c$ is of the form $c = c_0 + r_0 e^{i \phi}$ where $c_0 \in \mathbb{C}$ and $r_0 \in \mathbb{R}$ are constants and $\phi \in \mathbb{R}$.
$F^p(z,c) = z$ defines a polynomial of even degree $P(z) = 0$, whose constant coefficient is the product of its roots and is a polynomial in $c$ of degree $2^{p-1}$. The roots include those of $F^q(z, c) = z$ where $q | p$. Also, $\frac{\partial}{\partial z}F^p(z, c) = 2^p \Pi z_k$ where the $z_k$ are the $p$ roots in the periodic orbit of the desired solution $z$ (all $z_k$ are roots of $F^p(z, c) = z$, the remaining roots have lower period).
Case $p = 1$: $$ z^2 + c_0 + r_0 e^{i \phi} = z \\ \therefore z = \frac{1 \pm \sqrt{1 - 4(c_0 + r_0 e^{i \phi})}}{2} \\ \frac{\partial}{\partial z} = 2 z = e^{i \theta} $$
Now $|e^{i \theta}| = 1$ but $\exists x . |2 z| = |1 \pm \sqrt{x}| \neq 1$, so conclude that period $1$ component is not a perfect circle.
Case $p = 2$:
The equations reduce to $$ 4(1 + c_0 + r_0 e^{i \phi}) = e^{i \theta} $$ with obvious solution $c_0 = -1, r_0 = \frac{1}{4}, \phi = \theta$, so conclude that the period 2 component is a perfect circle.
Case $p = 3$:
The equations reduce to $$ 8 (c^3 + 2c^2 + c + 1) = e^{i \theta} \text{ where } c = c_0 + r_0 e^{i \phi}$$ For this to hold, the coefficients of $e^{i k \phi}$ must be zero for all $k > 1$. But setting $k = 3$ implies $r_0^3 = 0$ but we know that $r_0 > 0$ as hyperbolic components have non-empty interior. Contradiction, conclude that no period 3 component is a perfect circle.
Case $p > 3$:
Similarly to the $p = 3$ case, get a polynomial of degree $m > 1$ in $e^{i \phi}$ whose highest term has coefficient $r_0^m$. It remains to show that the polynomial really does have degree greater than $1$. The constant coefficient (product of roots) is a polynomial of degree $2^{p-1}$ in $c$, divided by the corresponding constant coefficient of all smaller divisors of the period gives:
$$m = 2^{p-1} - \sum_{q | p, q < p} 2^{q-1}$$
which solved numerically gives:
$$\begin{aligned} p & & & m \\ 1 & & & 1 \\ 2 & & & 1 \\ 3 & & & 3 \\ 4 & & & 5 \\ 5 & & & 15 \\ 6 & & & 25 \\ \vdots \end{aligned}$$
Finally, $m > 1$ for all $p \ge 3$ because $\exists q > 1 . q \nmid p$.
Best Answer
The cardioid is the boundary of the set of $c$s for which the iteration of $$ z\mapsto z^2+c $$ has an attractive fixpoint.
Given $c$ we can find the fixpoints easily by solving a quadratic equation, and we get $$ z_* = \frac{1\pm\sqrt{1-4c}}2 $$ From the general theory of iterated systems, we know that the fixpoint is attractive if $\left|\frac d{dz}(z^2+c) \right| < 1$ at $z_*$, which works out to $|z_*|<\frac12$. Thus we ought to find the boundary of the area in question by equating $z_*$ to $\frac12 e^{it}$. From $$ \frac12 e^{it} = \frac{1\pm\sqrt{1-4c}}{2} $$ a bit of simple algebra will give you $$ c = e^{it}\frac{\pm2-e^{it}}4 $$ as claimed. (And the solution with $\pm2=-2$ just corresponds to taking $e^{i(t+\pi)}$ instead).
Knowing that the boundary of "the area where the iteration has an attractive fixpoint" is included in the boundary of the entire Mandelbrot set is more involved, and I don't know how to prove that.