Möbius transformation that carries the real axis to the unit circle

Select any $3$ distinct points $z_1,z_2,z_3$ on the unit circle. We note that the transformation $$ S(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \frac{(z_2 - z_3)z + z_1(z_3 - z_2)}{(z_2 - z_1)z + z_3(z_2 - z_1)} $$ maps $z_1,z_2,z_3$ to $0,1,\infty$. It follows that the inverse transformation $$ T(z) = S^{-1}(z) = \frac{z_{3}(z_2 - z_1)\,z + z_1(z_2 - z_3)}{(z_1 - z_2)\,z + (z_2 - z_3)} $$ maps $0,1,\infty$ to the unit circle. It is clear that selecting distinct $z_1,z_2,z_3$ leads to distinct transformations. Moreover, because any bilinear transformation sends the points $0,1,\infty$ somewhere, we can conclude that every bilinear transformation that takes $\Bbb R$ to the unit circle has the above form for some choice of $z_1,z_2,z_3$ on the unit circle. In other words, the above defines a one-to-one correspondence between the maps we want and $\Bbb T^3$.

One particularly "nice" example of such a transformation is $$ f(z) = \frac{z + i}{z - i}, $$ which corresponds to selecting $z_1 = -1,z_2 = i, z_3 = 1$. Indeed, substituting these $z_j$ into the above yields $$ \frac{1(i-(-1))z + (-1)(i-1)}{(-1 - i)z + (i - 1)} = \frac{-(1 + i)z + (1-i)}{-(1 + i)z - (1 - i)}\\ = \frac{-(1 + i)z + (1-i)}{-(1 + i)z - (1 - i)} \cdot \frac {-(1-i)}{-(1-i)} \\ = \frac{2z + 2i}{2z - 2i} = \frac{z + i}{z - i}. $$