OPE's from Spacetime Lorentz Invariance of the Polyakov action How to explicitly determine the other singular terms Polchinski (2.4.14) using wick's theorem
$$
T(z)A(0,0) = ...+\frac{h}{z^2} A(0,0)+ \frac{1}{z} \partial A(0,0)+...
$$
if $$z \rightarrow z' = az+b $$
then we have
$$
A'(z', \bar z') = a^{-h}\bar a^{-\tilde h} A(z, \bar z)
$$
where $A$ is a primary field.
I suppose at first we should break the problem into 2 parts
$$A'(z', \bar z') = A(z, \bar z);  \,\,for \,\,  z' = z+b
$$
$$ A'(z', \bar z') = a^{-h}\bar a^{-\tilde h} A(z, \bar z); \,\, for \,\, z' = az
$$
The first part is exactly what we had for OPE's from spacetime translation invariance but I do not have any idea how to evaluate the second part.
 A: If nothing else about the operator $A$ is given then you can't determine them.
One case you might be interested in is when $A$ is a Virasoro primary. In that case
\begin{equation}
A^\prime(z^\prime, \bar{z}^\prime) = \left ( \frac{\partial z^\prime}{\partial z} \right )^{-h} \left ( \frac{\partial \bar{z}^\prime}{\partial \bar{z}} \right )^{-\bar{h}} A(z, \bar{z})
\end{equation}
holds not just for $z^\prime = az + b$ but for $z^\prime$ being any holomorphic function of $z$. In that case, it is a standard exercise to show that the more singular terms are zero. It follows from the fact that the infinitesimal form of the finite transformation above must be generated by $z^\prime(z) T(z)$.
Another case, showing that you can't set them to zero always is $A(z, \bar{z}) = T(z)$ itself. This is a quasiprimary so it still has the above transformation (for $h = 2, \bar{h} = 0$) when $z^\prime = az + b$ but not for more general $z^\prime$. Non $SL(2)$ transformations modify the above to
\begin{equation}
T^\prime(z^\prime) = \left ( \frac{\partial z^\prime}{\partial z} \right )^{-2} \left [ T(z) - \frac{c}{12} \{ z^\prime, z \} \right ]
\end{equation}
where the new term is the Schwarzian derivative. This leads to the more singular term $\frac{c / 2}{z^4}$ as the only one.
Edit: This is true in any 2d CFT, in particular the one you're referring to. The Polyakov action (whether or not you set $c = 26$ to get spacetime Lorentz invariance) has local operators of all types (Virasoro primary, quasiprimary, descendant) so you need to know what Verma module your operator is in and how far down it is before the other terms can be worked out.
