The Gauss Hypergeometric function is defined via a power series:
$$F(a,b,c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)}\frac{z^n}{n!}$$
In Abramowitz and Stegun they say the circle of convergence of this function is the unit circle $|z|=1$. I interpret that as a statement for all $a,b,c$, but the circle of convergence could be larger (not smaller) for specific choices of $a,b,c$. (correct me if that's wrong please!)
I'm working with the specific case of $F(1/2,1/2,1;x)$, where $x\in\mathbb{R}$. The following plot from Wolfram Alpha therefore confuses me:
So the real part of this seems to conform to my understanding - it appears not converge at x=1, but the radius of convergence is larger then $|z|=1$ because of the particular parameters of my problem. But the real/complex parts confuse me:
For $x<1$, $F(1/2,1/2,1,x)$ is real (the imaginary part is not just small, as in the plot, but it's zero). However, for $x>1$, $F(1/2,1/2,1,x)$ is complex. How is that possible? How can a power series of a real variable $x$ have both real and imaginary parts?
EDIT: I don't know much about the Gamma function, but I think for any real $x$, $\Gamma(x)$ is also real, as long as $x$ is not an integer (then it doesn't exist).
It can be found on Wikipedia that $$\Gamma (n) =(n-1)!$$ and $$\Gamma (n+0.5 ) =\frac{\sqrt{\pi} \Gamma (2n)}{2^{2n-1} \Gamma (n)}=\frac{\sqrt{\pi} (2n-1)!}{2^{2n-1}(n-1)!}$$ Therefore if we denote $$a_n =\frac{\Gamma (n+0.5)^2 }{\Gamma (n+1) n! }=\frac{\pi [(2n-1)!]^2}{2^{4n-2}[(n-1)!]^2 (n!)^2}$$ Hence $$\frac{a_{n+1}}{a_n }=\frac{\pi [(2n+1)!]^2}{2^{4n+2}[(n+1)!]^2 (n!)^2}\cdot \frac{2^{4n-2}[(n-1)!]^2 (n!)^2}{\pi [(2n-1)!]^2}=\frac{4n^2 (2n+1)^2}{2^4 n^2 (n+1)^2}\to 1$$ Thus in your case the radius of convergence is equal to $1.$