I am trying to find the leading order expression in an expansion for large $\Delta$ of ${}\_2F\_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right)$, where $z\in\mathbb{C}$.


The only helpful relation I could find in the literature *Higher transcendental functions* by H. Bateman and A. Erdelyi, p.77. There, I found:


If we define $\xi$ by $y\pm\sqrt{y^2-1}=e^{\pm\xi}$, where the upper sign is for $\text{Im}\,{y}>0$ and the lower sign for $\text{Im}\,y<0$, then for large $\lambda$ we have  

\begin{align\*}
 &\left(\frac{y}{2} - \frac{1}{2}\right)^{-a-\lambda}\cdot{}\_2 F\_1(a+\lambda,a-c+1+\lambda,a-b+1+2\lambda,2(1-y)^{-1})\\
 &\qquad = \frac{2^{a+b}\Gamma(a-b+1+2\lambda)\Gamma\left(\frac{1}{2}\right)\lambda^{-\frac{1}{2}}}{\Gamma(a-c+1+\lambda)\Gamma(c-b+\lambda)}\cdot e^{-(a+\lambda)\xi}(1-e^{-\xi})^{-c+\frac{1}{2}}(1+e^{-\xi})^{c-a-b-\frac{1}{2}}[1+\mathcal{O}(\lambda^{-1})].
\end{align\*}


For $\lambda=\frac{\Delta}{2}$, $a=0$, $b=1$, $c=\frac{1}{2}$ and $y=1-2z^2$ we get
\begin{equation\*}
 \left(-z^2\right)^{-\frac{\Delta}{2}}{}\_2F\_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right) = \frac{2\Gamma(\Delta)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\Delta+1}{2}\right)\Gamma\left(\frac{\Delta-1}{2}\right)\sqrt{\frac{\Delta}{2}}}\cdot e^{-\frac{\Delta}{2}\xi}\left(1+e^{-\xi}\right)^{-1}[1+\mathcal{O}(\Delta^{-1})].
\end{equation\*}


This can be further simplified by using relations for the Gamma function:
\begin{equation\*}
\frac{\Gamma(\Delta)}{\Gamma\left(\frac{\Delta+1}{2}\right)}=\frac{2^{\Delta-1}\Gamma\left(\frac{\Delta}{2}\right)}{\sqrt{\pi}}
\end{equation\*}
and
\begin{equation\*}
\frac{\Gamma\left(\frac{\Delta}{2}\right)}{\Gamma\left(\frac{\Delta-1}{2}\right)}=\frac{\sqrt{\Delta}}{\sqrt{2}}+\mathcal{O}\left(\Delta^{-\frac{1}{2}}\right)
\end{equation\*}
to get
\begin{equation\*}
{}\_2F\_1\left(\frac{\Delta}{2},\frac{\Delta+1}{2},\Delta,z^{-2}\right) = \left(-z^2\right)^{\frac{\Delta}{2}}2^\Delta e^{-\frac{\Delta}{2}\xi}\left(1+e^{-\xi}\right)^{-1}[1+\mathcal{O}(\Delta^{-1})].
\end{equation\*}
My question its still if there is no nicer form, that in particular avoids the two cases in the definition of $\xi$.


Thanks in advance for any help!



Asymptotic form of the Gauß Hypergeometric function 2F1 for three parameters approaching infinity
In fact your function is elementary and very simple, for its explicit form look at Brychkov,Marichev,Prudnikov, Integral and Series, vol.3 :
$$
F(a,a+1/2;2a;z)=\frac{1}{\sqrt{1-z}}\left(\frac{2}{1+\sqrt{1-z}}\right)^{2a-1}.
$$


Unfortunately I can not give a reference to the exact formula, this vol. 3 is the only one I have not an English version.


Also sorry for may be somehow akward post -- it was my first answer here, be patient. 


