Hypergeometric Function on the Unit Circle

Question

The Gauss's hypergeometric function is given by the series :$$_2F_1\left(a,b;c;z\right)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!} \;\;\;\;\;\;\left | z \right |<1 $$ But the function admits an analytic continuation on and beyond the unit circle. My question is : how to express the function for : $$z=e^{it}\;\;\;\; t\in \mathbb{R}$$

Answer 1

For example, there is an identity $$ _2F_1(a,b;c;z)=(1-z)^{-a}{}_2F_1\left(a,c-b;c;\frac{z}{z-1}\right).$$ The hypergeometric function on the right is given by a power series in $\frac{z}{z-1}$ which converges in the half-plane $\Re z<\frac12$. The latter contains a large part of the unit circle, namely, $\frac{\pi}{3}<\arg z<\frac{5\pi}{3}$. The complementary part can be analogously obtained by transforming the argument into $1-z$.

If however your hope is that on the unit circle hypergeometric function simplifies in some way - alas, it doesn't.