= Solution
Choose a branch of $t=1-z$ with $|\arg t|<\pi$ and write $d=c-a-b\notin\mathbb Z$. The <Gauss hypergeometric equation> is
$$
z(1-z)y''+[c-(a+b+1)z]y'-aby=0.
$$
Changing to $t$ turns this into the same equation with third parameter $a+b+1-c=1-d$. Its solution regular at $t=0$ is $y_1=F(a,b,1-d;t)$. Substitution of $y=t^dv$ gives the <Gauss hypergeometric equation> for $v$ with parameters $c-a,c-b,1+d$. Thus $y_2=t^dF(c-a,c-b,1+d;t)$ is also a solution. Their distinct nonintegrally separated <characteristic exponents at a regular singular point> $0,d$ imply <linear independence> near $z=1$, so
$$
F(a,b,c;z)=Ay_1(z)+By_2(z).
$$
This also follows from the exponent transformation in the <Papperitz symbol>.
Initially assume $\Re c>\Re b>0$, so the <Euler integral for the hypergeometric function> applies. If $\Re d>0$, taking $z\to1$ gives an integral evaluated by the <beta function> and determines
$$
A=\frac{\Gamma(c)\Gamma(d)}{\Gamma(c-a)\Gamma(c-b)}.
$$
Indeed $y_1\to1$ and $y_2\to0$ there. To find $B$, instead work in the nonempty parameter region $\Re d<0$, put $z=1-\epsilon$ and scale the integration variable by $1-s=\epsilon u$. The singular contribution is
$$
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\epsilon^d
\int_0^\infty u^{c-b-1}(1+u)^{-a}\,du.
$$
The defining integral of the <beta function> evaluates this <integral> to $\Gamma(c-b)\Gamma(-d)/\Gamma(a)$. Since $\epsilon^{-d}y_1\to0$ and $\epsilon^{-d}y_2\to1$, it determines $B$. <Analytic continuation> in the parameters then yields the <hypergeometric connection formula>
$$
\boxed{A=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},\qquad
B=\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}.}
$$
The usual power-series definition requires $c\notin\{0,-1,-2,\ldots\}$; exceptional parameters need a separately specified analytic <limit>. Reciprocal <gamma functions> make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in $z$.
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