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There is an algorithm to find points on the boundary of the Mandelbrot set, given a particular hyperbolic component and the desired internal angle. It involves Newton's method in two complex variables to solve

\[ F^p(z,c) = z \\ \frac{\partial}{\partial z} F^p(z,c) = b \]

where \(F^0(z,c) = z\) and \(F^{q+1}(z,c) = F^q(F(z,c)^2+c)\), \(p\) is the period of the target component, and \(b=e^{2 \pi i \theta}\) with the \(\theta\) the desired internal angle. The resulting \(c\) is the coordinates of the point on the boundary. It can also be modified to find points in the interior, simply set \(b = r e^{2 \pi i \theta}\) with \(\left|r\right| \le 1\).

After seeing this fractal Mandelbrot in an IRC channel, I wondered if it was possible to invert the algorithm, namely find \(b\) given \(c\), and use \(b\) as polar texture coordinates to place copies of an image inside the Mandelbrot fractal. The answer is yes, and in some sense it's actually easier than the other direction.

The algorithm I developed works like this:

1. When \(c\) is outside the Mandelbrot set, give up now;
2. For each period \(p\), starting from \(1\) and increasing:
1. Find \(z_0\) such that \(F^p(z_0,c) = z_0\) using Newton's method in one complex variable;
2. Find \(b\) by evaluating \(\frac{\partial}{\partial z} F^p(z,c)\) at \(z_0\);
3. If \(\left|b\right| \le 1\) then return \(b\), otherwise continue with the next \(p\).

I wrote this implementation.