Shape parameters of an ideal tetrahedron

Since $\mathrm{Im}z>0$, the imaginary parts of $z$, $z^{\mathrm{\prime }}=1\mathrm{/}(1-z)$ and $z^{\mathrm{\prime }\mathrm{\prime }}=1-1\mathrm{/}z$ are all positive: the second is $\mathrm{Im}z\mathrm{/}\mathrm{\mid }1-z{\mathrm{\mid }}^2$, and the third is $\mathrm{Im}z\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^2$. Thus their principal arguments lie in $(0,\pi )$.

Multiplication gives $z{z}^{\mathrm{\prime }}{z}^{\mathrm{\prime }\mathrm{\prime }}=z\cdot (1-z{)}^{-1}\cdot (z-1)z^{-1}=-1$. The sum of the three arguments is therefore an odd multiple of $\pi$; because it lies strictly between $0$ and $3\pi$, it must equal $\pi$. The arguments are the dihedral angles by the edge cross-ratio convention in [F1].

For $\omega =e^{i\pi \mathrm{/}3}$, $1-\omega ={\omega }^{-1}$ and $\omega -1={\omega }^2$. Hence $1\mathrm{/}(1-\omega )=\omega$ and $(\omega -1)\mathrm{/}\omega =\omega$; each parameter has argument $\pi \mathrm{/}3$. Each of the three parameter arguments occurs on a pair of opposite edges, proving the regular-angle assertion.