\[z(1-z)\left(\ifrac{\mathrm{d}F\left(a,b;c;z\right)}{\mathrm{d}z}\right)=(c-a)F%
\left(a-1,b;c;z\right)+(a-c+bz)F\left(a,b;c;z\right)=(c-b)F\left(a,b-1;c;z%
\right)+(b-c+az)F\left(a,b;c;z\right),\]