= Solution
For $F(a,1+a-c;1+a-b;w)$, the exponents at $w=0,1,\infty$ are $(0,b-a)$, $(0,c-a-b)$, and $(a,1+a-c)$. The substitution $w=1/z$ exchanges zero and infinity, giving $(a,1+a-c)$ at $z=0$, $(0,c-a-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(-z)^{-a}$ shifts the zero exponents by $-a$ and the infinity exponents by $+a$. The resulting <Papperitz symbol> is again $(0,1-c)$, $(0,c-a-b)$, $(a,b)$. Therefore \b[$w_1$ satisfies the same hypergeometric equation], with the stated branch $|\arg(-z)|<\pi$.
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