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When all three points are finite, useThe distinctness assumptions ensure that it is a nonconstant Möbius transformation, and direct substitution gives the three required values. If one of the specified points is infinity, the limiting formulas areThese cover every case because the points are distinct.

For uniqueness, if $g$ and $\tilde{g}$ have the same three images, the Möbius map $\tilde{g}\circ g^{-1}$ fixes $0,1,\infty$. Fixing zero and infinity makes it $z\mapsto cz$ with $c\ne 0$; fixing one gives $c=1$. Therefore the normalized map is unique. This is the sharp three-transitivity of Möbius transformations.

For a holomorphic map, conformality means preservation of oriented angles locally; it holds at every point of $G$ exactly when $f^{\prime }(z)\ne 0$ there. A Möbius transformation is $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, extended to the Riemann sphere at its pole and at infinity.

For three distinct finite points definewhich sends $z_1,z_2,z_3$ to $0,1,\infty$. If one point is infinite, take the corresponding limit: for $z_3=\infty$ use $(z-z_1)/(z_2-z_1)$; for $z_1=\infty$ use $(z_2-z_3)/(z-z_3)$; for $z_2=\infty$ use $(z-z_1)/(z-z_3)$. Construct $T_w$ similarly. Then $M=T_w^{-1}\circ T_z$ has the required three images.

For uniqueness, normalize any two proposed maps by these same transformations. Composing one normalized map with the inverse of the other fixes $0,1,\infty$. A Möbius map fixing infinity has $c=0$, one fixing zero then has $b=0$, and fixing one forces $a=d$ up to the irrelevant common nonzero scalar. It is the identity. This proves sharp three-transitivity of Möbius transformations.