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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.

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Any further constraints on $\oplus$? Should it, say, be continuous? – Johannes Kloos May 4 '12 at 21:14
Thanks, I added it in the statement of the problem. – Ivan May 4 '12 at 21:18
"Generalized addition" is not a very good name for it unless it's a generalization of addition. "A is a generalization of B" usually means every B is an A but not every A is a B. That would mean every instance of addition is an instance of "generalized addition". But that's inconsistent with the properties you specify. – Michael Hardy May 4 '12 at 21:42
From the fourth property alone: $x = x \oplus 0 = (-(-x)) \oplus 0 = -x$ for any complex number $x$. – D. Thomine May 4 '12 at 22:04
A trivial example: $x \oplus y = 0$ for all $x,y$. – Robert Israel May 5 '12 at 0:15
up vote 4 down vote accepted

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.)

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@anon, the question states that the fourth identity applies to real-valued $a$ and $b$. Commutativity and associativity follow immediately from the form of $x \oplus y = g(x) + g(y)$; I'll edit that into the answer. – Rahul May 5 '12 at 1:35
Sorry, you're right, I forgot that $a,b$ must be real! I see commutativity immediately, but how do you get associativity quickly? Wouldn't that involve e.g. $$g(g(x)+g(y))+g(z)=g(x)+g(g(x)+g(y))~?$$ That looks like a mess.. – anon May 5 '12 at 1:37
@anon: Oops. $\ $ – Rahul May 5 '12 at 1:42
@anon, I think I fixed it. Can you check? – Rahul May 6 '12 at 1:39
Your solution is correct, I made a mistake in my last edit, there is no problem with the last statement in its current form. – Ivan May 6 '12 at 4:30

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