Question

The function ${}_2F_1\Big(\frac{a-b}{2},\frac{a+b-1}{2};c;y\Big)$ can be transformed (as reported by A. Erdélyi) by the following formula

${}_2F_1\Big(\frac{a-b}{2},\frac{a+b-1}{2};c;y\Big)= (1-z)^{1-b}{}_2F_1\Big(a,a-1;b;z\Big) $

for

$y =4z(1-z)$, and $\quad a,b,c \in\mathbb{R}$

The problem is that solving the equation $y =4z(1-z)$ for the value of $z$ yields two solutions

$z_1 = \frac{1}{2} \left(1-\sqrt{1-y}\right)$

and

$z_2 = \frac{1}{2} \left(1+\sqrt{1-y}\right)$

Therefore, should the resultant transformation include both values of $z_1$ and $z_2$? or just use one value of them and ignore the second one.

How to include both values in one answer?

Answer 1

Here is a general explanation (which I can make more precise once you fix the formulas): the identity will hold "everywhere" except on the branch cut for the hypergeometric function (i.e. on the real line, from 1 to $+\infty$). On the branch cut, the choice of sign will flip the orientation of the cut, so that you get continuity in a clockwise/counter-clockwise manner.