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?



Quadratic Transformation of the Hypergeometric Function 2F1
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. 


