Is there any transformation to convert each of the following versions of ${}_2F_1$ to a polynomial?
The first one is $${}_2F_1\left(\frac{1-a}{2}, -\frac{a}{2}; b;\frac{4z}{(1+z)^2} \right), \quad a\in\mathbb{R},\quad b\in\mathbb{Z},\quad b\ge 0,\quad z\in\mathbb{R} $$
The second one is
$${}_2F_1\left(\frac{b-a-1}{2}, \frac{b-a}{2}; b+1;\frac{4z}{(1+z)^2} \right) $$
I checked the transformations reported in Mizan Rahman's paper (Quadratic Transformation Formulas for Basic Hypergeometric Series), but couldn't find a method.
Further Explanation: The type of polynomial I am looking for is not an orthogonal polynomial. Instead I am looking for transformations such as
$${}_2F_1\left(1-b,\frac{3}{2}-b;2-b;\frac{4z}{(1+z)^2}\right)=(1-z)^{2b-1}(1+z)^{2-2b}{}_2F_1\left(1,b;2-b;z\right)$$
and since ${}_2F_1\left(\frac{c}{2},\frac{c+1}{2};c;\frac{4z}{(1+z)^2}\right)=(1-z)^{-1}(1+z)^c{}_2F_1\left(0,1;c;\frac{z}{z-1}\right)$$ ${}_2F_1\left(1,b;2-b;z\right)=1$, then the equation resolves into the product of polynomials.$\quad\quad\quad\quad \quad=(1-z)^{-1}(1+z)^c $$
