$Re[f(x + n i)] = 0$ and $f(z)$ is not periodic

Let $z$ be a complex number and $f(z)$ an entire function such that

For $x$ real and $n$ any integer.

$Re[f(x + n i)] = 0$ and $f(z)$ is not periodic.

What are typical examples of such $f(z)$ ? Is there a way to express the general solution ? Is it possible that there is an entire function $g(z)$ such that $g(f(x + ni)) = f(x + (n+1)i)$ ?

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An entire function $f$ for which $f(x + n i) \in i \mathbb{R}$ for all $x \in \mathbb{R}$ and $n \in \mathbb{Z}$ is necessarily periodic with period $2 i$. This follows from Schwarz' reflection principle.

Let $g_n(z) = i \, f(z + n i)$ then $g_n(\mathbb{R}) \subseteq \mathbb{R}$ and according to this principle $g_n(\overline{z}) = \overline{g_n(z)}$ for all $z \in \mathbb{C}$. For $n=0$ this shows that

$$f(\overline{z}) = -\overline{f(z)}$$

and for $n=1$ that

$$f(\overline{z - i}) = f(\overline{z} + i) = -\overline{f(z + i)}.$$

Combining these equalities we get

$$\overline{f(z + i)} = -f(\overline{z - i}) = \overline{f(z - i)}$$

and after substituting $z \leftarrow z + i$ and conjugation

$$f(z + 2 i) = f(z).$$.

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Thanks. That really explains why I could not find any :) – mick Nov 24 '12 at 23:00