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.

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.

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.

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.

Let $f:C\to C$ and if such $$|f(x-y)|=|f(x)-f(y)|,\forall x,y\in C$$ Then is it true that $$f(x+y)=f(x)+f(y),\forall x,y\in C$$

I have prove when $f:R\to R$.But for complex,I can't solve it.It is not difficult to prove the real number, but I think it is not easy to solve the plural situation.and this problem is from a paper 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.