= Solution
In the hypergeometric symbol, select the exponent $1-c$ at zero and factor out its local power by writing
$$
y(z)=z^{1-c}v(z).
$$
The <Exponential shifting of a Papperitz symbol> makes the exponents of $v$ equal to $0,c-1$ at zero, $0,c-a-b$ at one, and $a-c+1,b-c+1$ at infinity. These are the hypergeometric parameters
$$
A=a-c+1,
\qquad
B=b-c+1,
\qquad
C=2-c,
$$
because $1-C=c-1$ and $C-A-B=c-a-b$. The solution for $v$ that is analytic at zero and equals one there is therefore the <Gauss hypergeometric function> $F(A,B,C;z)$. Thus the <second local hypergeometric solution> is
$$
\boxed{y(z)=z^{1-c}F(a-c+1,b-c+1,2-c;z)}.
$$
For noninteger $c$ this is linearly independent from $F(a,b,c;z)$; at resonant values, the usual limiting or logarithmic modification may be required.
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