I was experimenting a bit with tetration escape visualization and tried to visualize the following equation on the complex plane:
$$z_{n+1} = z_{0}^{\sin(z_{n})}$$
This sequence visualized looks like this:
I noticed between $7.8-0.11i$ and $8.03+0.11i$ there was a Mandelbrot set hidden between all the other mess:
Also between $-4.81-0.15i$ and $4.61+0.15i$ there is a Mandelbrot-like structure:
How comes the Mandelbrot set randomly appears in there?
Consider iteration of $z \mapsto c^{\sin(z)}$ near $c$:
$$z \mapsto c^{\sin(c + z)} - c$$
The Taylor series near $z = 0$ is (via Wolfram Alpha):
$$ z \mapsto \left(c^{\sin(c)} - c\right) \\ + \left(\log(c) c^{\sin(c)} \cos(c)\right) z \\ + \left(\frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right)\right) z^2 \\ + O(z^3) $$
Locally ($z$ small), the non-linearity is dominated by the $z^2$ term (as higher order terms will be much smaller). This means miniature quadratic Mandelbrot set copies may appear. Completing the square to the form $z \mapsto a(z + b)^2 + d$ gives:
$$ ab^2 + d = c^{\sin(c)} - c \\ 2ab = \log(c) c^{\sin(c)} \cos(c) \\ a = \frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right) $$
Affine conjugation via $z \mapsto z - b$ and $z \mapsto \frac{z}{a}$ gives the form $z \mapsto z^2 + C$ where $C = a (d + b)$ which expands to:
$$ C = \frac{1}{2} \log(c) \left(c^{\sin(c)} - c\right) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right) \\ - \left(\frac{1}{2} \log(c) c^{\sin(c)} \cos(c)\right)^2 + \frac{1}{2} \log(c) c^{\sin(c)} \cos(c) $$
Solving numerically for $C = 0$ gives (via Wolfram Alpha) a number of solutions including: $$c \approx 7.85398163397448\ldots$$
Solving numerically for $C = 0$ near $c = -4.7$ (via Wolfram Alpha) the solution: $$c \approx -4.71238898038469\ldots$$
These both correspond to mini-Mandelbrot islands with period $1$, there should also be mini-Mandelbrot islands corresponding to higher periods, for example I think there should be a period $3$ near $18.155$ (with $C \approx -1.754\ldots$).