Unfortunately, I can not write out the action for such a pendulum here, but I can show the solutions. To do this, take the quoted text "ParametricPlot[{Re[sn(t+it,5 -25 i)], Im[sn(t+it,5 -25 i)]},{t,-4,4}]" and place it in Wolframalpha. If the real part of the parameter of the elliptic sine is equal to half of one, for example here "ParametricPlot[{Re[sn(t+it, 0.5 + 5 i)], Im[sn(t+it, 0.5 + 5 i)]},{t,-4,4}]", then we get remarkable solutions in which, as the imaginary part of the parameter changes from minus infinity to plus infinity, the parametric plot of the pendulum changes so that the segment bends into an inverted eight (infinity sign) until the point of rupture closes. It is possible that the exclusive value (0,5) of the real part of the elliptic sine parameter for the oscillations of our pendulum has the same value (0,5) for the location of the zeros of the Riemann Zeta function.
One remarkable observation is that the parametric graphs of the function $\mathrm{sn} (t + \mathrm{i}t, \frac{1}{2} + \mathrm{i}\tau)$ at $\tau>>0$ and the functions $\mathrm{sn} (1 + \mathrm{i}, \frac{1}{2} + \mathrm{i}t)$ are remarkably similar (same eggs). I would not be surprised if in the limit $\tau\to\infty$ they match. By the way, the roots of the function $\ mathrm{sn} (1 + \mathrm{i}, \frac{1}{2} + \mathrm{i}t)$ lie in the region of negative $t$, but it would not hurt to act on it with an integral transformation and see where the roots of the image will go.
To make sure I'm not fooling you, please take a look at this chart: ParametricPlot[{Re[sn(1+i, 0.5 +it)], Im[sn(1+i, 0.5 + it)]}, {t,-100,500}]
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