In the article Rational Solutions of the Fifth Painleve Equation (see here) proposition 2.3 derives the orders and residues of singularities of rational solutions to a particular differential equation (the fifth Painlevé equation). I personally don't see how they arrive at the result. For example, in the previous propositions it is shown that if $\alpha = 0$, then the rational solution must have a Laurent series around $z= \infty$ of the form \begin{equation} y = -1 + az^{-1} + bz^{-2} + cz^{-3} + \mathcal{O}(z^{-4}) \\ \text{or} \\ y = az^{-1} + bz^{-2} + cz^{-3} + \mathcal{O}(z^{-4}). \end{equation}
We can turn these into Laurent series around $0$ which won't have any terms with negative exponents using $z \mapsto 1/z$. I would think this implies that zero can't be a pole at all. Whereas if $\alpha \neq 0$ it is possible for zero to be a pole because the solution can have a Laurent series around $\infty$ of the following form:
\begin{equation} y = az + b + cz^{-1} + \mathcal{O}(z^{-2}), \, a \neq 0. \end{equation}
Which would make zero a pole, again using $z \mapsto 1/z$. However the theorem claims that when $\alpha \neq 0$ that $y$ can't have zero as a pole and when $\alpha = 0$ then zero is (or at least could be) a pole. What exactly am I doing wrong?
I also don't really see how they can use the Laurent series around $z =\infty$ to say anything about non-zero poles, unless of course they simply wrote $y$ as a Laurent series $\sum_{k = k_0}^{\infty} a_k(z - z_0)^k$ and substituted that back into the differential equation (P'.5).