Connection formula for the Gauss hypergeometric function from $\frac{1}{z}$ to $z$ I want to understand better the following connection formula:
${}_2F_1\left[\begin{matrix}a,b  \\ c
\end{matrix};\frac{1}{z}\right]=e^{-i \pi a}z^{a}\frac{\Gamma \left[b-a \right]\Gamma \left[c\right]}{\Gamma \left[b\right]\Gamma \left[c-a\right]}~{}_2F_1\left[\begin{matrix}a,a-c+1  \\ a-b+1
\end{matrix};z\right]+e^{-i \pi b}z^{b}\frac{\Gamma \left[a-b \right]\Gamma \left[c\right]}{\Gamma \left[a\right]\Gamma \left[c-b\right]}~{}_2F_1\left[\begin{matrix}b,b-c+1  \\ b-a+1
\end{matrix};z\right].$
The problem for me is that ${}_2F_1\left[\begin{matrix}a,b  \\ c
\end{matrix};\frac{1}{z}\right]$ is a convergent series outside of the unit circle, while ${}_2F_1\left[\begin{matrix}a,b  \\ c
\end{matrix};z\right]$ is a convergent series inside the unit circle. So they don't have a common domain of convergence.
So how to understand this formula?
 A: First, I think the formula should read
$${_2F_1}\left( a,b;c;\frac 1 z \right)=\\
 \left( -\frac 1 z \right) ^ {-a}
  \frac {\Gamma(b-a) \Gamma(c)} {\Gamma(b) \Gamma(c-a)}
  {_2F_1}(a,a-c+1;a-b+1;z) +\\
 \left( -\frac 1 z \right) ^ {-b}
  \frac {\Gamma(a-b) \Gamma(c)} {\Gamma(a) \Gamma(c-b)}
  {_2F_1}(b,b-c+1;b-a+1;z),\\
|z|<1.$$
If you replace $(-1/z)^{-a}$ and $(-1/z)^{-b}$ with $e^{-i \pi a} z^a$ and $e^{-i \pi b} z^b$, it won't be valid for all complex $z$ inside the unit circle.
Regarding the analytic continuation, consider the simple case
$$f(z) = \sum_{k=0}^\infty z^k.$$
The sum diverges for all $z$ with $|z|=1$, but it doesn't mean that the unit circle is the natural boundary for $f(z)$. The expansion of $f(z)$ around $z=-1/2$ will have the radius of convergence equal to $3/2$ (which will turn out to be the distance to the nearest singularity), extending $f(z)$ outside of the unit circle.
One can also say that the analytic continuation of $f(z)$ coincides with the analytic continuation of
$$g(z) = -\sum_{k=1}^\infty \left( {\frac 1 z} \right) ^ k,$$
even though the regions of convergence of the two sums do not overlap.
The usefulness of reflection formulas is in providing a simple way to represent the function outside of the original subdomain, once the domain has been extended by analytic continuation.
