In the book "Conformal field theory" by Di Francesco et al, or the BPZ paper and many other places, the infinitesimal conformal transformation in the (compactified) complex plane/Riemann sphere, supposed to be a local holomorphic function, is represented by a Laurent series around $z=0$ \begin{align} \epsilon(z) = \sum_{-\infty}^\infty c_n z^{n+1} \end{align}
It looks suspicious, why Laurent series not Taylor series, why around a specific point $z=0$, if we think about the symmetry algebra of the system, why local not global?
With almost no knowledge in CFT and Riemann surface, I don't know the answer and would appreciate an authoritative one. But I just suspect what we are considering is actually a global meromorphic vector generating the infinitesimal conformal transformation. I want to argue $z^n\partial_z$ span the infinite dimensional linear space of meromorphic vectors on a Riemann sphere. For simplicity of argument, consider meromorphic 1-forms instead. Away from the poles, a meromorphic 1-form $\eta$ must be with holomorphic coefficient in an annulus $r_1<|z|<r_2$, in which it can be represented by a Laurent series \begin{align} \eta = (\sum_{-\infty}^\infty a_n z^{n+1}) dz \end{align} It then suffices to prove this expansion holds on the whole Riemann sphere for $\eta$, or equivalently, we need to prove a meromorphic 1-form $\omega$ that vanishes in an annulus should vanish on the whole Riemann sphere. By taking the contour integral \begin{align} \int z^n \omega = 0 \end{align} in the annulus, we see no poles can exist either in $|z|\leq r_1$ or $|z|\geq r_2$, so $\omega$ does vanish on the Riemann sphere.
Even if my suspicion is correct, it remains a question to me why we consider meromorphic vectors. Although it makes sense in Minkowski signature, $z^n\partial_z$ clearly correspond to the Fourier expansion of $Diff(\mathbb{S}^1)$, together with the barred sector, they span the whole conformal group $Diff(\mathbb{S}^1) \times Diff(\mathbb{S}^1)$ of the compactified two dimensional Minkowski space.
