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Let be $f:\mathbb C\rightarrow\overline {\mathbb C}$ a meromorphic function. The set of periods $\Omega_f$ is a discrete (additive) group and we have one of these possibilities:

i) $\Omega_f= \{0\}$

ii) $\Omega_f= \mathbb Z\omega$

iii) $\Omega_f= \mathbb Z\omega_1+\mathbb Z\omega_2$

In the case iii) we say that $f$ is an elliptic function and we now that the foundamental regions are compact subset of $\mathbb C$ (for example foundamental parallelograms). Elliptic fuctions, respect a fixed group of periodicities $\Omega_f=\Lambda$, form a field $E(\Lambda)$ and one can prove that $E(\Lambda)=\mathbb C(\wp,\wp')$.

Now I have not found in literature similar results for the case ii) of meromorphic simply periodic functions. In this case foudamental regions are not compact sets and, fixed the group $\Omega_f$, simply periodic functions respect $\Omega_f$ form a field. I ask if this field is in fact $\mathbb C\big(e^\frac{2\pi iz}{\omega}\big)$; moreover what is the relation beethween the meromorphic functions:

$$\varepsilon_k=\sum_{n\in\mathbb Z}(z-n)^{-k}$$

and the field of simply periodic funcion, where $\Omega_f=\mathbb Z$?

  • 2
    Assume for simplicity that $\Lambda_f=\mathbb{Z}$. Note that elliptic functions will automatically also be periodic (i.e. your case ii) above), but they will give rise to periodic functions with infinitely many poles in the strip $\mathbb{C}/\mathbb{Z}$. However I think it should be easy to prove that any function belonging to the field $\mathbb{C}(e^{2\pi i z})$ will only have finitely many poles in this strip, since $e^{2\pi i z}$ will be injective when restricted to this strip.2012-08-21
  • 0
    So $\mathbb C (e^{\frac {2\pi iz}{\omega}})$ is only contained in the field of simply periodic functions.....2012-08-21

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