In tranforming the variable from $x$ to $1/x$ in the solution, $_2F_1(a,b;c;x)$ of hypergeometric equation of Gauss, $\mathit{i.e.,}$
\begin{equation} x(1-x)\frac{d^2y}{dx^2}+[c-(a+b+1)x]\frac{dy}{dx}-a\,b\,y=0 \\ \tag{1}\label{eq1} \end{equation}
I used the following method.
Firstly, set $x=1/z$. Then \eqref{eq1} becomes,
\begin{equation} (z-1)\,z^2\,\frac{d^2y}{dz^2}+[-(c\,z^2-(a+b+1)z)+2\,z\,(z-1)]\frac{dy}{dz}-a\,b\,y=0 \tag{2}\label{eq2} \end{equation}
Secondly, transformation from $y$ to $w$ is employed using the relation, $y=z^a w$. Then \eqref{eq2} is transformed into,
\begin{equation} (1-z)z^{a+2}\frac{d^2w}{dz^2}+z^{a+1}[1+a-b-(2+2\,a-c)\,z]\frac{dw}{dz}-a\,(1+a-c)\,z^{a+1}w=0 \tag{3}\label{eq3} \end{equation}
When divided by $z^{a+1}$, \eqref{eq3} becomes,
\begin{equation} (1-z)z\frac{d^2w}{dz^2}+[1+a-b-(2+2\,a-c)\,z]\frac{dw}{dz}-a\,(1+a-c)\,w=0 \tag{4}\label{eq4} \end{equation}
which should give the solution form, $z^a_2F_1(a,1+a-c;1+a-b,z)= x^{-a}\,_2F_1(a,1+a-c;1+a-b,1/x)$
According to several textbooks however, the transformed solution is referred to as $(-x)^{-a}\,_2F_1(a,1+a-c;1+a-b,1/x)$.
I would like to know whether the method I used was wrong, and if so, where the factor $(-1)^a$ comes from. Is it something to do with the pole, as $x=0$ becomes singularity. I am not familiar with this concept.
(*)In respone to the first part of Maxim's comment:
If, $x=1/z$, then,
\begin{equation} \begin{aligned} \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=-z^2\frac{dy}{dz} \end{aligned} \end{equation}
\begin{equation} \begin{aligned} \frac{d^2y}{dx^2}= \left[ \frac{d}{dz}(-z^2)\frac{dy}{dz}-z^2\frac{d^2y}{dz^2} \right]\frac{dz}{dx}=z^4\frac{d^2y}{dz^2}+2z^3\frac{dy}{dz} \end{aligned} \end{equation}
If you insert these together with $x(1-x)=(z-1)/z^2$, into \eqref{eq1}, it should give \eqref{eq2}.
Second part:I understand that in case considering $z$ as complex number, and assuming $z=\vert z \vert e^{i(\theta+2\,n\pi)}$, then $z^\alpha=\vert z \vert^\alpha e^{i(\theta\alpha+2\,n\alpha\pi)}$
So, does that mean that $(-)^\alpha$ factor comes from the part, $e^{i(2\,n\alpha\pi)}$?
(*) Responding to the Maxim's comment sent just before:
Yes, you are right. I seem to have forgotten dropping off the fraction command and should have multiplied instead during the copy and paste procedure. It should be as you pointed out,
\begin{equation} (z-1)z^2\frac{d^2y}{dz^2}+[-(c\,z^2-(a+b+1)z)+2\,z\,(z-1)]\frac{dy}{dz}-a\,b\,y=0 \end{equation}
I have corrected the equation above. Very sorry for the confusion.