Circle Inversion Fractals

Mandelbrot's Algorithm

No point of L(C₁, C₂, C₃, C₄, C₅) lies in the discs bounded by S₂, S₃, S₄, or S₅.

Consider the circle B, S₂ in the previous notation.

Inversion in S takes B to B', the horizontal line through the center of 5'.

Moreover, the disc bounded by B inverts to the half-plane below B'.

This is easy to see: the disc bounded by B is disjoint from 3 and 4, and 3 and 4 invert above B', so the disc bounded by B inverts to the half-plane below B'.

Inversion in 3' and 4' gives points above B'.

Inversion in 1' and 2' leaves the y-coordinate unchanged, so will not move points above B' to points below B'.

Finally, inversion in 5' will not take points above B' to points below B', because B' passes through the center of 5'.

Consequently, once a point inverts above B', it will never invert below B'.

Because any finite set of orbit points has empty limit set, we can ignore all finite collections of orbit points before inversion in 3' or 4' is applied.

The only remaining case is infinite orbits with no applications of 3' or 4'.

These generate limit points in L(1', 2', 5'), which is the circle B'.

Similar arguments work for the other circles S₃, S₄, and S₅, with appropriate circles replacing S.

Return to Mandelbrot's method.