What are the solutions $u$ to the equation $\sqrt{1-m} \operatorname{sn} (u \mid \frac{1-m}{1+m}) = 1$ for a given $m$? Because $\mbox{sn}$ is an elliptic function of order 2, we know it attains any value two times in the fundamental parallellogram, so that there are two solutions. Let's first assume $0 < m < 1$, in particular that $m$ is real, and denote $M = \frac{1-m}{1+m}$. Numerics suggest that indeed there are 2 solutions in the fundamental parallellogram which are related by symmetry, of the form $u^{\pm} = K(M) + \left[ 1 \pm \gamma(m) \right] i K'(M)$ for some $\gamma \in (0,1)$. A plot of the numerically obtained $\gamma$ may be found here (note: near $m=1$ the resolution becomes bad, but I believe $\gamma \rightarrow 0$ as $m \rightarrow 1$. also, it seems that $\gamma \rightarrow 1/2$ as $m \rightarrow 0$). We may identify the potential solutions because $\mbox{sn}$ is symmetric about the point $K(M) + i K'(M)$. Does an algebraic expression for $u$ exist?
Side notes: further numerical investigation of $\gamma$ leads me to conjecture that it has the form $\gamma(m) = \frac{1}{2} - \frac{1}{2 \pi} \left( m + \frac{11}{48}m^3 + \frac{599}{5120}m^5 + \mathcal{O}(m^7) \right)$ as $m \rightarrow 0$. If this is true and such an analytic expansion of $\gamma$ exists it also gives us the answer $\forall m \in \mathbb{C} \setminus \{0,1\}$ by analytic continuation. Because $\sqrt{1-m} \rightarrow 0$ when $m \rightarrow 1$, I believe $\gamma \rightarrow 0$ in this limit because $\operatorname{sn} (u \mid M)$ must become singular to satisfy the equation. So this places the constraint $1 + \frac{11}{48} + \frac{599}{5120} + \dots = \pi$ on the coefficients of the (hypothetical) expansion of the (hypothetical) $\gamma$. The burden of proof remains...
