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Write the Papperitz symbol by listing pairs of exponents at $0,1,\infty$. For $F(a,c-b;c;w)$ they are $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. Under $w=z/(z-1)$ the points $0,1,\infty$ in the $z$-plane correspond respectively to $0,\infty ,1$ in the $w$-plane. Thus the pulled-back exponents are $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity.

Multiplication by $(1-z{)}^{-a}$ shifts the exponents at one by $-a$ and those at infinity by $+a$, yieldingThis is the original Gauss hypergeometric equation. Both functions are analytic and normalized to one at zero, so uniqueness of that normalized local solution proves the Pfaff transformationContinue the identity using the branch $|\arg (1-z)|<\pi$. As usual, the parameters must admit the normalized hypergeometric solution; excluded singular parameter values are treated only when their limiting solution exists.

First assume $a-b\notin \mathbb{Z}$ and nonsingular parameters, so the two infinity solutions $w_1$ and $w_2$, obtained by exchanging $a,b$, are independent. Write $F=A{w}_1+B{w}_2$. Since the power series at $1/z=0$ starts at one, $w_1\sim (-z{)}^{-a}$ and $w_2\sim (-z{)}^{-b}$.

In the parameter region $\Re (b-a)>0$, let $z\to -\infty$ in the Pfaff transformation. The $w_2$ term vanishes after multiplication by $(-z{)}^a$, while $z/(z-1)\to 1$ and $(-z)/(1-z)\to 1$. The supplied value at one of the Gauss hypergeometric function therefore givesAnalytic continuation in the parameters extends this coefficient to the nonresonant parameter domain. Exchange $a,b$ to determine $B$. Hence the hypergeometric connection formula at infinity isThe branch is $|\arg (-z)|<\pi$. The existence of $F(a,b;c;1)$ alone does not exclude integer $a-b$. At such resonant values the two separate displayed terms can be singular; the correct statement takes their combined parameter limit, which can contain logarithms. This qualification is essential, for example when $a=b$.

The Möbius transformation $w=z/(z-1)$ sends $0\mapsto 0$, $\infty \mapsto 1$ 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 pairsAt 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 isIts 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 givesThe 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.

The Papperitz symbol describes a homogeneous second-order Fuchsian differential equation on the Riemann sphere, with three regular singular points and the two indicial exponents at each. At a finite point $z_j$, an exponent $r$ corresponds locally to a factor $(z-z_j{)}^r$; at infinity it corresponds to $z^{-r}$. Repeated or resonant exponents can require logarithms. The six exponents obey the Fuchs relation that their sum is one. With three singularities there is no independent accessory parameter, so these data determine the normalized equation.

A Möbius transformation of a Papperitz symbol moves the singular points without changing their exponents. Multiplication of the dependent variable by a local power shifts the two exponents by that power, using the reciprocal local coordinate at infinity.

Set $w=z/(z-1)$ and $v(w)=F(a,c-b,c;w)$. At $w=0,1,\infty$, the exponent pairs for $v$ are respectively $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. The Möbius transformation sends $z=0,1,\infty$ to $w=0,\infty ,1$. Before rescaling, $v(w(z))$ therefore has exponent pairs $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(1-z{)}^{-a}$ subtracts $a$ at one and adds $a$ at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with $(1-z{)}^{-a}=1$ at zero. The resulting solution is analytic there and has value one. Uniqueness of the normalized analytic solution yieldsThis is the Pfaff transformation. The normalization presupposes the usual admissible parameters, in particular $c\notin \{0,-1,-2,\dots \}$. Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.

The Papperitz symbolspecifies a second-order Fuchsian differential equation. The first row lists its three distinct regular singular points $0,1,\infty$; $z$ is the independent variable. The two entries below each singular point are its characteristic exponents. Thus local solutions have leading behaviorsand, with the usual convention at infinity,When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the Fuchs relationFor a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the Gauss hypergeometric equation. Its distinguished solution $F(a,b;c;z)$ is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes $c\notin \{0,-1,-2,\dots \}$, with exceptional parameter values handled separately or by continuation.

Now putThe hypergeometric equation in $u$ has exponent pairsThe Möbius transformation of a Papperitz symbol sendsso $Y$ has symbol

On the other hand, $F(a,b;c;z)$ has exponent pairsat $0,1,\infty$. Multiplication by $(1-z{)}^a$ applies the dependent-variable rescaling of a Papperitz symbol: it adds $a$ to both exponents at $z=1$ and subtracts $a$ from both at infinity. Hencehas exactly the same three exponent pairs as $Y$.

Both functions are analytic near $z=0$ and equal one at $z=0$, so uniqueness of the normalized exponent-zero hypergeometric solution gives the Pfaff transformationThe identity first holds near zero with the branch of $(1-z{)}^a$ equal to one there, and then extends by analytic continuation on any domain where compatible branches are chosen.