TITLE: Generalized addition function
QUESTION [5 upvotes]: I would like to have an example (or a proof that there does not exist) of a function on the complex numbers, which for lack of a better term I'll call generalized addition, such that 
$$x\oplus y=y\oplus x$$
$$(x\oplus y)\oplus z=x\oplus(y\oplus z)$$
$$x\oplus (\pm i\cdot x)=0$$
$$(ax)\oplus(bx)=(|a|+|b|)x\oplus 0$$
Where $a$ and $b$ are real. Actually, it would be even nicer is the last identity could be substituted with 
$$(ax)\oplus(bx)=f(|a|+|b|)x\oplus 0$$
where $f(|a|+|b|)\ge (|a|+|b|)^\alpha$ and $\alpha\ge 1$. An example of a function that satisfies the first three identities (but not the forth one) is $x\oplus y=\sqrt(x^2+y^2)$. Is there a way to incorporate the last condition?
EDIT: Continuity would be nice, bar a(n almost necessary) branch cut.
EDIT: Changed the fourth condition since it seems to lead to a trivial contradiction.

REPLY [4 votes]: Suppose $x \oplus y$ is of the form $h\big(g(x) + g(y)\big)$ with $g\big(h(x)\big) = x$, which immediately guarantees commutativity and associativity:
$$x \oplus y = h\big(g(x) + g(y)\big) = h\big(g(y) + g(x)\big) = y \oplus x,$$
$$(x\oplus y)\oplus z = h\big(g(h\big(g(x) + g(y)\big))+g(z)\big) = h\big(g(x) + g(y) + g(z)\big) = h\big(g(x) + g(h\big(g(y) + g(z)\big))\big) = x\oplus(y\oplus z).$$
Now let $g(x)$ be the complex number with the same modulus as $x$ and twice its argument. That is, $g(0) = 0$, and $g(x) = \lvert x\rvert e^{2i\arg x}$ when $x \ne 0$. Then we can set $h(0) = 0$ and $h(x) = \lvert x\rvert e^{i(\arg x)/2}$ otherwise. Assuming $\arg$ has range $(-\pi,\pi]$, this has a discontinuity along the negative real line, and we have $g\big(h(x)\big) = x$, but $h\big(g(x)\big) = x \operatorname{sgn} \operatorname{Re} x$. Now observe that $$g(\pm ix) = e^{\pm i\pi}\lvert x\rvert e^{2i\arg x} = -g(x),$$
and for real $a$, $\arg ax$ is either $\arg x$ or $\arg x \pm \pi$ depending on the sign of $a$, but in either case
$$g(ax) = \lvert a\rvert\lvert x\rvert e^{2i\arg x}e^{\pm2i\pi} = \lvert a\rvert g(x).$$
Now we can verify the third and (original) fourth conditions:
$$x\oplus(\pm ix) = h\big(g(x) + g(\pm ix)\big) = h\big(g(x) - g(x)\big) = h(0) = 0.$$
$$(ax)\oplus(bx) = h\big(g(ax)+g(bx)\big) = h\big(\lvert a\rvert g(x) + \lvert b\rvert g(x)\big) = h\big((\lvert a\rvert + \lvert b\rvert)g(x)\big) = h\big(g((\lvert a\rvert+\lvert b\rvert)x)\big) = h\big(g((\lvert a\rvert+\lvert b\rvert)x) + g(0)\big) = (\lvert a\rvert+\lvert b\rvert)x \oplus 0.$$
(I don't understand why the original fourth condition is troublesome or implies that $x \oplus 0 = 0$. Perhaps there is an error in this answer.)