I have been having fun (and frustration) in finding complex associative operators over the complex numbers. So far, I have found the 3 listed in the title (r is a constant), and also know about $\sqrt{x^2 + y^2}$, but have been having difficulty finding other interesting ones. An example of a failed attempt was to find out if $\frac{ax+by+c}{dx+ey+f}$ could be associative. After doing some horrifying algebra, I arrived at the frustratingly trivial answer of c = 1, a = d/f, b = e/f, which meant that the only such operator would be constant. Are there any other interesting complex associative operators, and are there any rules of thumb for finding them? (Note: I don't find the operator x*y := x interesting.)
Thanks very much!
-William
So, after thinking about this for a while, I arrived at the following answer:
Any operator of the form, $a*b = f(f^{-1}(a)\circ f^{-1}(b))$, where $\circ$ is an associative operator, is an associative operator.
Proof: $(a*b)*c = f\left (f^{-1}\left(f\left(f^{-1}(a)\circ f^{-1}(b)\right)\right)\circ f^{-1}(c)\right ) = f\left (\left(f^{-1}(a)\circ f^{-1}(b)\right)\circ f^{-1}(c)\right ) = f\left (f^{-1}(a)\circ \left(f^{-1}(b)\circ f^{-1}(c)\right)\right ) = f\left (f^{-1}(a)\circ f\left(f^{-1}\left(f^{-1}(b)\circ f^{-1}(c)\right)\right)\right ) = a*(b*c)$
In this way, you have an infinite family of operators that you can recursively build up from known associative operators. However, you will notice that all such operators are commutative, unless we know of basic non-commutative associative operators. I wonder if you could use a similar technique to construct non-commutative operators...