 2\pi\tau_2 = %2\pi(\tau-\tau_1)= -2iM\log{z\over\Lambda_2},\qquad\qquad  \Lambda_2=z_{\infty}e^{2\pi i(c^{(0)} + \tau (b^{(0)}-1))/2M}, \label{p3bis}\gorilla\guinea-pig\rooster\ostritch \eagle\alligator \seal\alligator \seahorse\rabbit \tarantula\eel \seal