The transformations on the nome and Landen's transformation Could someone please explain how to transform the nome $q = e^{-\pi K'/K}$ from $q^2$ to $q$ and then to $-q$? In other words, how does changing $q^2$ to $q$ and then $q$ to $-q$ affect $k$ and $K$. Here, $k$ is the elliptic modulus, $K = K(k)$ is the complete elliptic integral of the first kind, and $K' = K(k')$, where $k' = \sqrt{1 - k^2}$ is the complementary modulus. I know that the transformation of $q$ is somehow connected to Landen's transformation, but I do not understand how these two concepts are related. Thank you in advance.
 A: I tend to look at this using Jacobi thetanull functions,
as these are usually expressed in terms of the nome $q$:
$$\begin{align}
 \vartheta_2(q) &= q^{1/4}\sum_{n\in\mathbb{Z}} q^{n(n+1)}
& \vartheta_3(q) &= \sum_{n\in\mathbb{Z}} q^{n^2}
& \vartheta_4(q) &= \sum_{n\in\mathbb{Z}} (-q)^{n^2}
\end{align}$$
As to the connection of your question with Landen's transformation,
let me just remark that Landen's transformation gives rise to the
arithmetic-geometric mean iteration step, which in turn resembles
the transformation of $\vartheta_3^2$ and $\vartheta_4^2$
under $q\mapsto q^2$. To answer your question,
it suffices to do that step backwards.
I will focus on the Jacobi thetanulls here, and thus get along without
considering integrals and Landen's transformation as such.
Expect some ambiguity from inverse powers.
The formulae we need are
$$\begin{align}
 \vartheta_2(-q) &= \pm\sqrt{\pm\mathrm{i}}\,\vartheta_2(q)
& \vartheta_2^2(q) &= 2\,\vartheta_3(q^2)\,\vartheta_2(q^2)
\\ \vartheta_3(-q) &= \vartheta_4(q)
& \vartheta_3^2(q) &= \vartheta_3^2(q^2) + \vartheta_2^2(q^2)
\\ \vartheta_4(-q) &= \vartheta_3(q)
& \vartheta_4^2(q) &= \vartheta_3^2(q^2) - \vartheta_2^2(q^2)
\end{align}$$
The formulae for $q\mapsto-q$ are obvious,
and the last one of those for $q^2\mapsto q$ has been proven
elsewhere on this site.
The remaining ones follow in a similar manner.
For the relations of the Jacobi thetanulls
to the complete elliptic integral of the first kind,
one only needs to keep in mind that
$$\begin{align}
 k(q) &= \frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}
& k'(q) &= \frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}
& K\left(k(q)\right) &= \frac{\pi}{2}\vartheta_3^2(q)
\end{align}$$
From those identities we can deduce
$$\begin{align}
 k(-q) &= \pm\mathrm{i}\frac{k(q)}{k'(q)}
& k'(-q) &= \frac{1}{k'(q)}
\\ k(q) &= \pm\frac{2\sqrt{k(q^2)}}{1+k(q^2)}
& k'(q) &= \frac{1-k(q^2)}{1+k(q^2)}
\\\therefore\quad
 k(-q) &= \pm\mathrm{i}\frac{2\sqrt{k(q^2)}}{1-k(q^2)}
& k'(-q) &= \frac{1+k(q^2)}{1-k(q^2)}
\end{align}$$
and for the complete elliptic integral:
$$\begin{align}
 K\left(k(-q)\right) &=
 k'(q)\,K\left(k(q)\right)
\\ K\left(k(q)\right) &=
 \left(1+k(q^2)\right) K\left(k(q^2)\right)
\\\therefore\quad
 K\left(k(-q)\right) &=
 \left(1-k(q^2)\right) K\left(k(q^2)\right)
\end{align}$$
which is what you have asked for.
