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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?

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    $\begingroup$ Is $x$ the real part of $z$? Or is it a constant? $\endgroup$
    – user52817
    Jan 16 at 17:35
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    $\begingroup$ @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

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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 $\lambda \in \Lambda$.

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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.

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  • $\begingroup$ 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

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