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It is well-known that when $a$ is any non-zero real number, the most general solution of $f(x+a)=f(x)$ should be $f(x)=\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $|a|$ .

Now when $a$ is a non-real complex number, what is the most general solution of $f(x+a)=f(x)$ ? I don't believe that $f(x)=c$ only, where $c$ is any complex number.

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It is $f(x)=\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $a$. This says nothing, like the statement in the question about the real case. A nonconstant example is $\sin\left(\dfrac{2\pi x}{a}\right)$. –  Jonas Meyer Jul 14 '12 at 23:34
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A nice example is the complex exponential function $e^z$. Here, $a=2\pi i$ is the non-real period. –  GEdgar Jul 14 '12 at 23:46

1 Answer 1

Let $$ S=\{x+\lambda a:x\in\mathbb{R},\lambda\in[0,1)\} $$ and define $f$ on $S$ any way you want. Then $f$ can be uniquely extended to $\mathbb{C}$.

On the other hand given $f$ satisfying $f(x+a)=f(x)$ you easily see that values of $f$ uniquely determined by values of $f$ on set $S$.

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