= Solution
The <Möbius transformation> $w=z/(z-1)$ sends $0\mapsto0$, $\infty\mapsto1$ and $1\mapsto\infty$. The <hypergeometric function> $F(a,c-b;c;w)$ has local exponent pairs $(0,1-c)$ at $w=0$, $(0,b-a)$ at $w=1$, and $(a,c-b)$ at $w=\infty$. Pulling back to $z$ and multiplying by $(z-1)^{-a}$ gives the pairs
$$
z=0:(0,1-c),\qquad z=\infty:(a,b),\qquad z=1:(0,c-a-b).
$$
At infinity the prefactor adds $a$ to both exponents, while at $z=1$ it subtracts $a$ from both pulled-back exponents. These are exactly the requested <Riemann P-symbol> exponents. Equivalently, the <Pfaff transformation> identifies this branch, up to a nonzero phase depending on the chosen branches, with $F(a,b;c;z)$.
When $c\notin\mathbb Z$, an independent <Frobenius solution> is
$$
\boxed{z^{1-c}F(a-c+1,b-c+1;2-c;z).}
$$
Its exponent $1-c$ differs from the first branch's exponent zero, establishing <linear independence>. For resonant parameters, use a parameter limit giving the appropriate logarithmic branch. More directly, on a simply connected region avoiding the singularities and zeros of a nonzero first solution $y_1$, <reduction of order> gives
$$
\boxed{y_2(z)=y_1(z)\int^z\frac{t^{-c}(1-t)^{c-a-b-1}}{y_1(t)^2}\,dt.}
$$
The <Wronskian> is a nonzero multiple of $z^{-c}(1-z)^{c-a-b-1}$, so this also produces an independent local solution in the resonant cases for which the first branch is defined.
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