1
$\begingroup$

Let $\Lambda := \{m + in : m,n ∈ \mathbb{Z}\} \subset \mathbb{C}$ and $x \in \mathbb{R}$. I am trying to find an elliptic function $f : \mathbb{C} \setminus \Lambda \rightarrow \mathbb{C}$ such that $\forall \lambda \in \Lambda $,

$$\lim_{z \rightarrow \lambda} \frac{f'(z)}{f(z)} = x $$

I figured that maybe the Weierstrass elliptic function

$$℘(z,\Lambda) := \frac{1}{z^2} + \sum_{\lambda\in\Lambda\setminus\{0\}}\left(\frac 1 {(z-\lambda)^2} - \frac 1 {\lambda^2}\right)$$

could be used but I've tried to write it down and I can't see how I could conclude. Could you please help me?

$\endgroup$
3
  • 1
    $\begingroup$ Is $x$ the real part of $z$? Or is it a constant? $\endgroup$
    – user52817
    Jan 16 at 17:35
  • 1
    $\begingroup$ @user52817: “Let ... and $x \in \Bbb R$” seems pretty clear to me, $x$ is a given real constant. $\endgroup$
    – Martin R
    Jan 16 at 17:38
  • $\begingroup$ @Martin R, yes it is clear, just seems like a typo. $\endgroup$
    – user52817
    Jan 16 at 17:44

2 Answers 2

Reset to default
2
$\begingroup$

A (non-constant) elliptic function $f : \mathbb{C} \setminus \Lambda \to \mathbb{C}$ has a pole at each $\lambda \in \Lambda$, which implies that $f'/f$ has a simple pole at all these points. Therefore $$ \lim_{z \to \lambda} \frac{f'(z)}{f(z)} = x \in \Bbb R $$ is not possible if $\lambda \in \Lambda$.

$\endgroup$
0
1
$\begingroup$

If you meant $f$ is a meromorphic $\Lambda$-periodic function,

Then the equation $x^2=\frac{(\wp_i'(s)^2}{\wp_i(s)^2} = 4\wp_i(s) - g_2(i)/\wp_i(s)$

gives that $f(z)=\wp_i(z+s_k)$ is solution to your problem with $s_k$ one of the two solutions of $\wp_i(s)=\frac{x^2\pm \sqrt{x^4+16 g_2(i)}}{8}$.

$\wp_i^{-1}(u)$ is given by an elliptic integral.

$\endgroup$
3
  • $\begingroup$ It is given that $f$ is holomorphic in $\mathbb{C} \setminus \Lambda$, which means that the poles are necessarily in $\Lambda$, so that $f'/f$ cannot have a finite limit at those points. Am I misunderstanding something? $\endgroup$
    – Martin R
    Jan 16 at 17:46
  • $\begingroup$ Ah maybe, I interpreted it the opposite way, that is $f$ is a meromorphic function on $\Bbb{C}/\Lambda$. Your interpretation makes the question a nonsense whereas mine makes it interesting. $\endgroup$
    – reuns
    Jan 16 at 17:47
  • $\begingroup$ That's how I interpret “$f : \mathbb{C} \setminus \Lambda \rightarrow \mathbb{C}$” ... $\endgroup$
    – Martin R
    Jan 16 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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