The picture below was generated in Mathematica and shows the image of a rectilinear grid in $\mathbb C$ under the elliptic mapping $\operatorname{sn}(z,m)$. The question here concerns the highlighted circle.
Using the terminology of the Wikipedia article on Jacobi elliptic functions, let $m\in (0,1],K=\operatorname{K}(m), K'=\operatorname{K}(1-m),$ and $k=\sqrt m.$
If $L$ is the line $\operatorname{Im}(z)=\tfrac 1 2K'$ and $C$ is the circle $\left \|z^4\right \|=1/m$, it looks like $C$ is the circle in the diagram, and
$$\operatorname {sn(,m)}: L \rightarrow C$$
so that for $x\in \mathbb R$
$$\left \|\operatorname{sn}^4(x+\tfrac 1 2K'i,m)\right \|=1/m.$$
Can anybody point me to a proof or exposition of the above?
Example 8 (pg 529) of Watson and Whittaker: A Course of Modern Analysis states that
$$ \operatorname{sn}(u+iK'/2)=k^{-1/2}\dfrac{(1+k)\operatorname{sn}u+i\operatorname{cn}u\operatorname{dn}u}{1+k\operatorname{sn}^2u}$$
This can be derived from $$\operatorname{sn}(iK'/2)=ik^{-1/2}$$ (same book, pg 206, Example 2) and the addition formula for $\operatorname{sn}$.
Since $$ \begin{aligned} \operatorname{cn}(u) &= \cos(\operatorname{am}(u)) \\ \operatorname{dn}(u) &= \sqrt{1-k^2 \sin^2(\operatorname{am}(u))} \\ \operatorname{sn}(u) &= \sin(\operatorname{am}(u)) \end{aligned} $$
we can use a change of variable $v=\operatorname{am}$ to rewrite as
$$ \operatorname{sn}(u+iK'/2)=k^{-1/2}\dfrac{(1+k)\sin v+i\cos v\sqrt{1-k^2\sin^2 v}}{1+k\sin^2u}. $$ Setting $$\begin{aligned} a &= (1+k)\sin v\\ b &= \cos v\sqrt{1-k^2\sin^2 v}\\ c &= 1+k\sin^2v\\ d &= k^{-1/2} \end{aligned} $$ we can further rewrite as $$\begin{aligned} \operatorname{sn}(u+iK'/2) &= d\left(\dfrac{a+b i}{c}\right) \end{aligned} $$
For $u\in\mathbb{R}$, if we can show that $\left \|d\left(\dfrac{a+b i}{c}\right)\right\|$ is a constant for a given $k$ , then we will have shown that $\operatorname{sn}(u+iK'/2)$ is a circle.
But if $u\in\mathbb{R}$ then $a,b,c,d\in\mathbb{R}$. So
$$ \begin{aligned} \left \|\dfrac{a+b i}{c}\right\|^2 &= \dfrac{a^2+b^2}{c^2} \\ &= \dfrac{(1+k)^2\sin^2 v+\cos^2 v(1-k^2\sin^2 v)}{(1+k\sin^2v)^2} \\ &=\dfrac{(1+k)^2\sin^2 v+(1-\sin^2 v)(1-k^2\sin^2 v)}{(1+k\sin^2v)^2} \\ &= 1 \end{aligned} $$
and finally $$ \left \|\operatorname{sn}(u+iK'/2)\right\|=\left \|d\left(\dfrac{a+b i}{c}\right)\right\|=d= k^{-1/2}=m^{-1/4}\quad\blacksquare $$