Relation between hypergeometric functions? Is there any relations between the following hypergeometric functions?
$$\ _2F_1(1,-a,1-a,\frac{1}{1-z})$$
$$\ _2F_1(1,-a,1-a,{1-z})$$
$$\ _2F_1(1,a,1+a,\frac{1}{1-z})$$
$$\ _2F_1(1,a,1+a,{1-z})$$
 A: First, you need to use Barnes integral representation 
$${}_2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac{1}{2\pi i}
\int_{-i\infty}^{+i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds. $$
The Gauss hypergeometric function ${}_2F_1(a,b;c;z)$ is usually defined by a power series that converges only for $|z|<1$, but you have to extend the definition on the whole complex plane such that the function can be evaluated at both $1-z$ and $1/(1-z)$.
Using the integral representation,
\begin{align*}
\ _2F_1(1,-a;1-a;{1-z}) &=
\frac{\Gamma(1-a)}{\Gamma(1)\Gamma(-a)}\frac{1}{2\pi i}
\int_{-i\infty}^{i\infty} \frac{\Gamma(1+s)\Gamma(-a+s)\Gamma(-s)}{\Gamma(1-a+s)}(z-1)^sds \cr
&=\frac{(-a)}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1+s)\Gamma(-s)}{s-a}(z-1)^sds,
\end{align*}
and
\begin{align*}
\ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right) &=
\frac{\Gamma(1+a)}{\Gamma(1)\Gamma(a)}\frac{1}{2\pi i}
\int_{-i\infty}^{i\infty} \frac{\Gamma(1+s)\Gamma(a+s)\Gamma(-s)}{\Gamma(1+a+s)}(z-1)^{-s}ds \cr
&=\frac{a}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1+s)\Gamma(-s)}{s+a}(z-1)^{-s}ds.
\end{align*}
In the second relation, change $s$ to $-s$, we get (keeping track of all the minus signs)
$$
\ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right) =
\frac{a}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1-s)\Gamma(s)}{a-s}(z-1)^{s}ds.
$$
Finally using the relation 
$$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s}
\quad \mbox{and}\quad \Gamma(-s)\Gamma(1+s)=-\frac{\pi}{\sin \pi(-s)} = -\frac{\pi}{\sin \pi s},$$
we get
$$\ _2F_1(1,-a;1-a;{1-z}) = \ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right)$$
and similar the equivalence for the rest two (or just change $a$ to $-a$).
