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Let $f:ℂ→ℂ$ be a function.

(a) How I can solve this functional equation:

$f(b-s)+f(s)=0,b∈ℝ$ with respect to $f$ and $b=1$.

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    No, I mean the case: $f(1-s)+f(s)=0$2012-12-30

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(a) For $s=\frac {1}{2}$, we have $f(\frac {1}{2}) +f(\frac {1}{2}) = 0$ , so we must have $f( \frac {1}{2}) =0$.

Consider $S = (\frac {1}{2}, \infty) \times (-\infty, \infty) \cup \frac {1}{2} \times (0, \infty)$.

Define $f(s): S \rightarrow \mathbb{C}$ however you wish. It does not even need to be continuous.

Consider $ T = \mathbb{C} \setminus S \setminus (\frac {1}{2},0)$

For $t \in T, 1-t \in S$, and so we define $ f(t) = -f(1-t)$. This then clearly satisfies your condition.

For (b), do the same with similar sets.

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    @CalvinLin: Hello, again, What about the case when I know that $f:C→C$ is defined in the whole complex plane. Does this affect the fact that its domain of definition maust be $S$ as you answred the question.2014-02-13