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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

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

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Polynomial in what? ${}_2F_1\left(\frac{1-a}{2}, -\frac{a}{2}; b;\frac{4z}{(1+z)^2} \right)$ is not a polynomial in $z$ in most cases. –  Gerald Edgar Mar 8 '12 at 20:36
    
The ${}_2F_1$ can be converted to a polynomial form (sometimes multiplied with ratios of gamma functions), for example, if the function got transformed such that one of the first two arguements can be converted to 0. –  Remy Mar 8 '12 at 22:14
2  
In order for your question to be more attractive to non-experts (this is of course not necessary), providing an explanation and maybe an example of what you mean by converting to a polynomial would be great. –  Mariano Suárez-Alvarez Mar 9 '12 at 0:00
    
@Mariano Suárez-Alvarez and @Gerald Edgar : Thanks for your suggestions. I have just added more explanation. –  Remy Mar 9 '12 at 11:29
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