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Consider the Jacobi elliptic function $sn(\cdot,k)$ restricted to the interval $(0,2K)$, where $K=K(k)$ is complete elliptic integral of the first kind. If $0<k<1$, then it is well known the this function first increases on $(0,K)$, from 0 to 1, then it achieves its global maximum at $u=K$, and then decreases again from 1 to 0. The graph of $sn(\cdot,k)$ on $(0,2K)$ is similar to the graph of sine on $(0,\pi)$.

There is a strong numerical evidence that this properties holds true also for complex $k$ taking the absolute value of $sn(\cdot,k)$. More precisely, my question is as follows: Let $|k|\leq 1$, $k\neq\pm1$, is it true that the function $$u\mapsto |sn(u,k)|,$$ restricted on $(0,2K)$, achieves its global maximum at $u=K$?

Remarks:

  1. I can prove it for purely imaginary $k$ using Jacobi imaginary transformation for $sn(u,ik)$.

  2. Since it holds that $sn(2K-u,k)=sn(u,k)$, it would be sufficient to show the function in question in increasing on $(0,K)$.

  3. Note that $K$ is complex in general, so the meaning of $(0,2K)$ is just the line connecting points $0$ and $2K$ in $\mathbb{C}$.

  4. On the picture, you can see the plot of $|sn(Kx,k)|$, with $k=0.7+0.6i$ and $x\in(0,2)$. enter image description here

At the first glance, it seems that it should not be hard to answer this (very concrete) question. However, I was not able to do that trying to do so for some time. Thank you for your time!

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It is not true.

Here is part of the plot (in Maple) of $|sn(tK,k)|$ in the case $k = 1.25$. A similar picture will occur for complex $k$ near $1.25$.

enter image description here

Note: Maple and Mathematica use different conventions for the second parameter of the elliptic functions. Mathematica will give the same picture as this using ${1.25}^2$ instead of $1.25$.

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  • $\begingroup$ Yes, the question has to be slightly rephrased if $|k|>1$. <b>Please restrict yourself to $|k|\leq1$, $k\neq\pm1$!</b> (as indicated in Remark 1.). Sorry, for this. With the modulus within the unit circle, numerical plotting did not yield to a contradiction in my experiments. $\endgroup$
    – Twi
    Commented Oct 8, 2015 at 8:28

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