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

$${}_2F_1\left(\frac{c}{2},\frac{c+1}{2};c;\frac{4z}{(1+z)^2}\right)=(1-z)^{-1}(1+z)^c{}_2F_1\left(0,1;c;\frac{z}{z-1}\right)$$ $$\quad\quad\quad\quad \quad=(1-z)^{-1}(1+z)^c $$

converting to a polynomialwould be great. – Mariano Suárez-Alvarez♦ Mar 9 '12 at 0:00