# A variant of Cauchy-type functional equation conjecture

Let $$f:\mathbb{C}\to \mathbb{C}$$ be a complex function such that $$|f(x-y)|=|f(x)-f(y)|,\qquad x,y\in\mathbb{C}.$$ Is it true that $$f(x+y)=f(x)+f(y),\qquad x,y\in\mathbb{C}?$$

The answer is affirmative when $$f:\mathbb{R}\to\mathbb{R}$$ is a real function, and the proof is not difficult. The above generalization seems harder: I inferred it from a conjecture that I saw in a paper.

• What about $f(x)=x+a$? In any case your question is not appropriate for this site, please use MathStackExchange. – abx Oct 13 '18 at 12:09
• @abx,if $f(x)=x+a$ not such condition,and this problem I think it's a research question. It's not simple. – inequality Oct 13 '18 at 12:32
• @CarloBeenakker: Actually, there are many (non-continuous) functions besides $f(x)=cx$ that satisfy the second equation (hence also the first equation). – GH from MO Oct 13 '18 at 12:52
• I voted to close, but after the edit it seems like a reasonable question (with the $\|$ corrected to $|$). – Nik Weaver Oct 13 '18 at 13:34
• I edited the question greatly (for clarity and language), and I voted to reopen it. At the same time, I ask the OP to make a bigger effort (e.g. next time) to formulate his/her question nicely. In particular, the post refers to some paper, hence a reference would be necessary. This site is for professionals, so participants should act like professionals. – GH from MO Oct 14 '18 at 13:36

counterexample $$f(x)=1-e^{\Re(x) i}$$