Pfaff and Euler transformations

Remarks

The map $z \mapsto \frac{z}{z - 1}$ sends $\C \setminus [1, \infty)$ onto itself, so the right sides make sense. For $\operatorname{Re} c > \operatorname{Re} b > 0$, the first Pfaff formula follows from Euler's integral representation by the substitution $t \mapsto 1 - t$, and the general case by analytic continuation in the parameters. The second Pfaff formula is the first with $a$ and $b$ exchanged, and applying the two Pfaff formulas one after the other gives the Euler transformation.

Example: for $b = c$, the Euler transformation reads $(1 - z)^{-a} = (1 - z)^{-a} \, {}_2F_1(c - a, 0; c; z)$, which is true since ${}_2F_1(\cdot, 0; c; z) = 1$.

The Pfaff transformation turns the series at $z$ into a series at $\frac{z}{z - 1}$, which converges for $\operatorname{Re} z < \frac12$; this gives the continuation of ${}_2F_1$ to that half-plane.

Going deeper

For a student

Together with the identity, the Pfaff and Euler transformations form a group of four expressions for the same solution of the hypergeometric equation. Combined with the solutions at the other singular points, they give Kummer's 24 solutions. The Euler transformation also explains the factor $(1 - z)^{c - a - b}$ that governs the behaviour of ${}_2F_1$ near $z = 1$.