Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g(f(x + ni)) = f(x + (n+1)i)$ ?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.