What is the argument of the quarter period $iK^\prime(k)$ when $|k| = 1$, i.e. $k = e^{i\alpha}$? (Is it $\frac{\pi-\alpha}{2}$?) I have a research problem in mathematical physics describing a periodic condensate. I've determined a class of solutions wherein the density of the condensate is described by an elliptic function, with convenient expressions in terms of the Jacobi $\textrm{sn}(u,k)$ function. Getting the periodicity constraint correct is a condition involving (amongst other physical parameters of the problem) the quarter periods, particularly $iK^\prime$.
I've been teaching myself about elliptic functions from Whittaker and Watson, A Course of Modern Analysis. I've been manipulating the quarter periods essentially from the integral definition;
$$K(k) = \int_{0}^{1} \frac{\mathrm{d}z}{\sqrt{1-z^2}\sqrt{1-k^2z^2}} \quad \text{for} \: |k|^2 \in [0,1] \tag{1} \label{Kdef}$$
and $K^\prime(k) = K(k^\prime)$ where $k^2 + {k^\prime}^2 = 1$ and $k^\prime \rightarrow 1$ as $k \rightarrow 0$.
I'm comfortable with e.g. transforming this expressions to show that if $|k|^2 > 1$,
$$K(k) = \frac{1}{k} \left( K\left(\frac{1}{k}\right) - iK^\prime\left(\frac{1}{k}\right)\right),$$
and that for $k \in \mathbb{R}$,
$$K(ik) = \frac{1}{\sqrt{1+k^2}}K\left(\frac{k}{\sqrt{1+k^2}}\right).$$
(The latter is effected by the integral substitution
$$w = \frac{\sqrt{1+k^2}z}{\sqrt{1+k^2z^2}}. \tag{2} \label{wsub}$$
and considering terms of the form $1-w^2$ and $1-\frac{k^2}{1+k^2}w^2$.)
A subclass of my condensate solutions would have the complex elliptic modulus $k$ satisfy $|k|=1$, if the periodicity constraint can be satisfied in this case. So I've been trying to determine properties of $iK^\prime$ for such $k$. It would be quite helpful even to understand the complex argument of $iK^\prime$, because then I might be able to make periodicity arguments using a path in this direction on the torus the elliptic function $\mathrm{sn}$ describes.
Using Maple to plot $K^\prime(e^{i\alpha})$ seems to strongly suggest that, up to fixing your branch cuts,
$\arg\left(K^\prime(e^{i\alpha})\right) = -\frac{\alpha}{2}$. My first thought was that this might be straightforward to show from the integral definition of the quarter periods, with a substitution similar to the one mentioned above, reducing the desired quarter period to a real integrand times a constant complex phase. (Of course, this is not the only way to end up with the desired phase on the overall integral!) But the naive substitution  $k^2 = -\left(1-e^{2i\alpha}\right)$ into (\ref{wsub}) seems insufficient – it certainly doesn't give $\frac{k^2}{1+k^2}$ generically real, at any rate, but I haven't even been able to use it to correctly describe any sort of properties of the quarter period. When I perform that substitution, I derive an "identity" of the form
$$K^\prime\left(e^{i\alpha}\right) = e^{i(2\pi-\alpha)} K^\prime\left(e^{i(\pi-\alpha)}\right)$$
which is patently not the case by looking at e.g. $\alpha=\pi$ or $\alpha=\frac{\pi}{2}$. I presume that with my rusty complex analysis, I'm blundering over branch cuts during the derivation.
Obviously, an argument solely considering the integrand in (\ref{Kdef}) might not be sufficient. I'm wondering if anyone can point out a more effective method, or a known treatment of the quarter periods for elliptic moduli satisfying $|k|=1$?
Edit: I've determined that the answer to the parenthetic question in the title is essentially yes; I will write up an answer.
 A: Landen's transformation provides the affirmative answer. We can write it in the form
$$K\left( \frac{\sqrt{a^2-b^2}}{a} \right) = \frac{2a}{a+b}K\left(\frac{a-b}{a+b}\right).$$
Let $a=1$ and $b=k$. We obtain
$$K\left( \sqrt{1-k^2} \right) = K^\prime(k) = \frac{2}{1+k}K\left(\frac{1-k}{1+k}\right).$$
Now substitute $k = e^{i\alpha}$. Then $\frac{2}{1+e^{i\alpha}} = e^{-i\frac{\alpha}{2}}\sec(\frac{\alpha}{2})$ and $\frac{1-e^{i\alpha}}{1+e^{i\alpha}} = -i\tan(\frac{\alpha}{2})$.
Using the identity provided in the question for pure imaginary elliptic modulus, we have
$$K\left(-i\tan\left(\frac{\alpha}{2}\right)\right) = \frac{1}{\sqrt{1+\tan^2(\frac{\alpha}{2})}} K\left( \frac{\tan(\frac{\alpha}{2})}{\sqrt{1+\tan^2(\frac{\alpha}{2})}} \right) = \cos\left(\frac{\alpha}{2}\right) K \left(\sin\left(\frac{\alpha}{2}\right)\right),$$
although I've ignored the matter of the signs of the square roots. Nonetheless, putting this all together, we find a tentative answer of
$$K^\prime\left(e^{i\alpha}\right) = e^{-i\frac{\alpha}{2}} K\left(\sin\left(\frac{\alpha}{2}\right)\right).$$
Checking this against the numeric plots which I did in Maple that motivated my conjecture, we see that this is correct for the domain $\alpha \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right]$. This can be understood as follows: the integral definition of $K(k)$ as a function of $k$ assumes a branch cut on the positive real axis from $1$ to infinity. $K^\prime(k) = K(k^\prime)$, and $k=e^{i\alpha}$ gives ${k^\prime}^2 = 1-e^{2i\alpha}$, which described a circle of radius $1$ centred at $1$ in the complex plane. ${k^\prime}^2$ orbits this circle twice as $k$ orbits the unit circle once, and as a function of $\alpha$, the argument of ${k^\prime}^2$ is $\pm\left(\frac{\pi}{2}-\alpha\right)$. So ${k^\prime}^2$ crosses the branch cut when $\alpha = \frac{\pi}{2}$ (and again when $\alpha=\frac{3\pi}{2}$). The effect is that the expression derived above for $K^\prime(e^{i\alpha})$ is correct on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right]$ and repeats with period $\pi$.
