It is given that the hypergeometric function $F(a, b ; c ; z)$ is the solution of the hypergeometric equation determined by the Papperitz symbol

$$P\left\{\begin{array}{ccc} 0 & \infty & 1 \\ 0 & a & 0 \\ 1-c & b & c-a-b \end{array}\right\}$$

that is analytic at $z=0$ and satisfies $F(a, b ; c ; 0)=1$, and that for $\operatorname{Re}(c-a-b)>0$

$$F(a, b ; c ; 1)=\frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)} .$$

[You may assume that $a, b, c$ are such that $F(a, b ; c ; 1)$ exists.]

(a) Show, by manipulating Papperitz symbols, that

$$F(a, b ; c ; z)=(1-z)^{-a} F\left(a, c-b ; c ; \frac{z}{z-1}\right) \quad(|\arg (1-z)|<\pi) .$$

(b) Let $w_{1}(z)=(-z)^{-a} F\left(a, 1+a-c ; 1+a-b ; \frac{1}{z}\right)$, where $|\arg (-z)|<\pi$. Show that $w_{1}(z)$ satisfies the hypergeometric equation determined by $(*)$.

(c) By considering the limit $z \rightarrow \infty$ in parts (a) and (b) above, deduce that, for $|\arg (-z)|<\pi$,

$$F(a, b ; c ; z)=\frac{\Gamma(c) \Gamma(b-a)}{\Gamma(b) \Gamma(c-a)} w_{1}(z)+(\text { a similar term with } a \text { and } b \text { interchanged })$$