After using FullSimplify in Mathematica, I got the left hand side of the following equation: $$(a-1)(z-1)\ _2F_1(1,1,1-a,\frac{1}{z})+a\ _2F_1(1,1,2-a,\frac{1}{z})-az+z=0$$
I series expanded it and it seems to be equal to zero.But I cannot find any transformation equations of the Hypergeometric functions to prove it. Could any one point out which equation of the hypergeometric functions I can use to prove this?
Your relation follows from Gauss' relations for contiguous functions. In particular, I am referring to 15.2.21 in Abramovitz and Stegun. The relation reads $$[ a-1 -(c-b-1)z ]{}_2F_1(a,b;c;z) +(c-a) {}_2F_1(a-1,b;c;z) -(c-1)(1-z){}_2F_1(a,b;c-1;z)=0 . $$
Using $a=1$, $b=1$, $c=2-\alpha$, $z=1/\zeta$ yields the relation $$ \alpha\, {}_2F_1(1,1;2-\alpha;1/\zeta) + (\alpha-1)(\zeta-1) {}_2F_1(1,1;1-\alpha;1/\zeta) = (1-\alpha) \zeta \,{}_2F_1(0,1;2-\alpha;1/\zeta) \tag{1}.$$
From the series definition of ${}_2F_1$, you obtain immediately that ${}_2F_1(0,b;c;z)=1$. Such that (1) yields your expression.