Quadratic transformations of hypergeometric functions

Remarks

In each formula both sides solve the same second-order hypergeometric equation in $z$. Indeed, the substitution of a quadratic rational function of $z$ into a hypergeometric equation with special parameters gives back a hypergeometric equation. Both sides are holomorphic near $z = 0$ with value $1$ there and the exponent $0$ there, which singles out one solution; analytic continuation then extends the identity to the stated domains. For instance in part 1, $4z(1 - z) = 1 - (1 - 2z)^2$ maps the half-plane $\operatorname{Re} z < \frac12$ into $\C \setminus [1, \infty)$.

Such transformations exist only for special parameters. Consider the exponent differences $1 - c$, $c - a - b$ and $a - b$ of the equation. A quadratic transformation exists if two of them are equal up to sign or one of them is $\pm\frac12$. All three parts fit this pattern: in part 1, the right side has $c - a - b = \frac12$; in part 2, $c = 2b$ gives $c - a - b = -(a - b)$; in part 3, $c = 1 + a - b$ gives $1 - c = -(a - b)$.

Example: with $a = b = \frac12$ in part 3, $1 + a - b = 1$ and the formula expresses ${}_2F_1(\frac12, \frac12; 1; z)$, which is $\frac{2}{\pi}$ times the complete elliptic integral of the first kind with parameter $z$, through $(1 - z)^{-1/2} \, {}_2F_1(\frac14, \frac14; 1; \frac{-4z}{(1-z)^2})$.

Going deeper

For a student

Linear transformations such as Pfaff and Euler exist for all parameters and come from the Möbius symmetries of $\set{0, 1, \infty}$. Quadratic transformations come instead from rational maps of degree $2$ of the sphere that send singular points to singular points, which is possible only for special exponents. Kummer and Goursat studied them systematically; for more special parameters there are also transformations of higher degree.