./dev 
Original theme by orderedlist (CCBYSA)
Where applicable, all content is licensed under a CCBYSA.

Littlewood polynomials are polynomials whose coeffients are drawn from a finite set of ${1,1}$.
Plotting the roots of augmented Littlewood polynomials (allowing for 0 as a coefficient) for a restricted degree $n$ yields pretty pictures.
Claim:
$$ \begin{align} & c_k \in {1, 0, 1} \\ p(z) & = \sum c_k z^k \\ p(z_0) & = 0 & \\ \to z_{0} & \ge \frac{1}{2} \text{ or } z_{0} = 0 \\ \end{align} $$
Proof (by contradiction):
$$ \begin{align} & c_0 \ne 0 \\ \to p(z_0) & = c_0 + \sum_{k=1}^{n} c_k z_{0}^k \\ \to 1 & =  \sum_{k=1}^{n} c_k z_{0}^k  \\ \to 1 & < \sum_{k=1}^{n}  \frac{1}{2} ^k < 1 \\ \end{align} $$
This is sloppy and doesn't take into account degree, $n$, of the polynomial.
"In the limit", this is true, but this is violated for smaller degree polynomials.
In general, we can ask what are the symmetries of transformations on $z$ that will still admit Littlewood polynomials.
Some candidates for $p(z)$ are:
The complex conjugate gives the mirror symmetry about the xaxis. The negation and complex conjugation gives the symmetry about the yaxis ($p(\alpha + i \beta) = 0 \to q(\alpha + i \beta)=0$). The inversion gives a kind of projection onto the Riemann sphere and/or Mobius transformation like symmetry, giving a projection symmetry about the circle/sphere at radius 1.
$$ \begin{align} & c_k \in {1, 0, 1} \\ p(z) & = \sum c_k z^k \\ \end{align} $$
yaxis symmetry ($x \to x$):
$$ \begin{align} p_{}(z) & = \sum_{k=0}^n (1)^k c_{k} z^k \\ p(z) & = p_{}(z) \\ \to p(z_0) & = p_{}(z_0) = 0 \end{align} $$
xaxis symmetry ($x \to x^{ \dagger }$):
$$ \begin{align} p_{\dagger}(z) & = \sum_{k=0}^n c_{k} ( z^{ \dagger })^k \\ p(z) & = p_{ \dagger }( z^{ \dagger } ) \\ \to p(z_0) & = p_{ \dagger }( z_0^{ \dagger } ) = 0 \\ \end{align} $$
Since the coefficients are real, $p_{\dagger}(z)$ has the same roots.
Inversion symmetry ($x \to \frac{1}{x}$):
$$ \begin{align} p_{1}(z) & = \sum_{k=0}^n c_{nk} z^k \\ p(z) & = z^n p_{1}(z^{1}) \\ \to p(z_{\dagger}) & = z_{\dagger}^n p_{1}(z_{\dagger}^{1}) = 0 \\ \end{align} $$
Noting that $z^n p_{1}(\frac{1}{z})$ is also a Littlewood polynomial.
All of the above create a situation that for every $p(z)=0$, there's a corresponding $p_{}(z)=0$, $p_{\dagger}(z^{\dagger})=0$ and $p_{1}(z^{1})=0, (z \ne 0)$, establishing the gross level symmetry.
An explanation of how Dragon curves show up:
The idea is that the dragon curve can be created through in iterated function system (IFS) by choosing randomly between $f_{,z}(x)=1zx$ and $f_{+,z}(x)=1+zx$ for some complex z and initial $x=0$.
The iteration of ${f_{,z}, f_{+,z}}$ trace out all Littlewood polynomials for initial values of $f_{,z}(0)$ and $f_{+,z}(0)$.
Here's what I think the argument is in the category cafe thread:
$p(z_0)=0$ is the result of a finite cutoff IFS from above. If the IFS is a nice fractal, that means other polynomials from the IFS evaluated at the source point $z_0$ will trace out a fractal (dragonlike curve) around the image point 0. Since we're dealing with a small neighborhood, we can heuristically use linearity to shift the image points not landing on 0 to 0, moving their source points along with them. The source points will then create a dragonlike fractal curve in the source map, giving us the dragon curves that we see in zoomed in snapshots of the root map.
Also note that the 'twist' in the dragonlike curves might be explained with another heuristic argument that goes as follows:
Take $p(z_0)=0$ and a 'similar' polynomial Littlewood $q(z_0 + \delta) = 0$. $p'(z_0) \sim q'(z_0)$ (and perhaps even more so than $p \sim q$?), and $q(z_0 + \delta) \sim q(z_0) + \delta q'(z_0)$, we can try solving for $q(z_0) + \delta q'(z_0) = 0 \to \delta = \frac{q(z_0)}{q'(z_0)} \sim \frac{q(z_0)}{p'(z_0)}$.
So the 'correction' source change for $q(z)$ has the factor of $p'(z_0)$ which might induce a sort of 'twist' to the curve.
Attempts at explaining the holes around the unit circle:
The holes on the unit disc look to have bigger gaps in a Farey like sequence.
Take the upper quadrent of the roots ($z = a+ib, a>0, b>0$) to remove as much symmetry as possible and then take the complex logarithm. The points below the unit disc could also be discarded but the gaps are easier to see when they're left in, which introduces a mirror symmetry in the complex logarithm plot.
The lines correspond to the first few Farey sequence numbers with a factor of $\pi$.
If we have $p(z)=0$ and $z = u^m$, then we also have $p(u^m)=0$, creating another polynomial $q(\cdot)$ of $\deg(p) \cdot m$. That is $q(u)=p(u^m)=p(z)=q(z^{\frac{1}{m}})$.
This provides another form of symmetry, even if it's only "probabilistic" in nature.
I suspect this is the reason for the holes around the $z=1$ line but I'm having a hard time coming up with a good reason.