Question about correlation functions of 2d CFTs I have a question regarding equation (2.22) in Ginsparg's lecture notes on CFTs. Equation (2.22) is
$$
\langle T(z) \phi_1(w_1, {\bar w}_1) \cdots  \rangle = \sum_{i=1}^n \left( \frac{h_i}{(z-w_i)^2} + \frac{1}{z-w_i} \frac{\partial}{ \partial w_i} \right) \langle \phi_1(w_1, {\bar w}_1) \cdots  \rangle
$$
Here, $T(z)$ is the stress tensor of the CFT and $\phi_i$ is a primary operator of weight $(h_i,0)$ which transforms under conformal transformations as
$$
\delta_\epsilon \phi_i = \left( h_i \partial \epsilon + \epsilon \partial \right) \phi_i
$$
He derives (2.22) from (2.21) which reads
$$
\langle \oint \frac{dz}{2\pi i} \epsilon(z) T(z)\phi_1(w_1, {\bar w}_1) \cdots  \rangle = \sum_{i=1}^n \langle \phi_1(w_1, {\bar w}_1) \cdots \delta_\epsilon\phi_i(w_i, {\bar w}_i)  \cdots \rangle
$$
by setting $\epsilon(x) = \frac{1}{x-z}$.
My question is - Is (2.22) correct?
Here are my reasons to believe that it is not - 


*

*I believe he derives (2.22) from (2.21) by setting $\epsilon(x) = \frac{1}{x-z}$ in (2.21). (2.22) is then derived if the following holds
$$
\langle \oint \frac{dx}{2\pi i} \frac{T(x)}{x-z} \phi_1(w_1, {\bar w}_1) \cdots  \rangle = \langle   T(z)\phi_1(w_1, {\bar w}_1) \cdots  \rangle 
$$
This would be true if the integrand on the LHS had only a pole at $x-z$. However, it has also has poles at each $x = w_i$, but those contributions aren't considered.

*I can try and derive (2.22) in a different way - namely via contractions. I start with the LHS of (2.22) and contract $T(z)$ with each $\phi_i$. Each contraction is replaced with the operator product
$$
T(z) \phi_i(w_i {\bar w}_i) = \frac{h_i  \phi_i(w_i {\bar w}_i) }{ ( z - w_i )^2 }  + \frac{ \partial  \phi_i(w_i {\bar w}_i) }{ z - w_i } + : T(z) \phi_i(w_i {\bar w}_i)  :
$$
Again, if I only consider the singular terms, I reproduce the RHS of (2.22). But what about $: T(z) \phi_i(w_i {\bar w}_i)  :$?? In a general CFT, conformal normal ordering $:~:$ is not equivalent to creation-annihilation normal ordering ${}^\circ_\circ~{}^\circ_\circ$. The latter would vanish in a correlation function, but not the former. So, I believe in general there would be extra terms on the right of (2.22). 

What am I misunderstanding?

 A: The integral will give you the residue, which is the coefficient of the $\frac1{z-w}$ term. Nothing else will contribute (this is known form complex analysis). So we need to tag carefully the $\frac1{z-w}$ term of $\epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)$.
Using  $\epsilon(z)=\epsilon(w)+\epsilon'(w)(z-w)+\cdots$
and
$$T(z) \phi_i(w_i {\bar w}_i) = \frac{h_i  \phi_i(w_i {\bar w}_i) }{ ( z - w_i )^2 }  + \frac{ \partial  \phi_i(w_i {\bar w}_i) }{ z - w_i } + : T(z) \phi_i(w_i {\bar w}_i)  :$$
we see that
$$\epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)=\cdots+\frac{\epsilon(w) \partial  \phi_1(w_1, {\bar w}_1) }{ z - w_1 }+
\frac{\epsilon'(w) h_1  \phi_1(w_1, {\bar w}_1) }{  z - w_1  } 
+\cdots$$
where dots denote other terms that behave differently than $\frac1{z-w}$.
Then, doing the integral we obtain
$$\oint \frac{dz}{2\pi i} \epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)=\epsilon(w) \partial  \phi_1(w_1, {\bar w}_1)+\epsilon'(w) h_1  \phi_1(w_1, {\bar w}_1)  =\delta_\epsilon \phi_1.$$
"Putting the integral in brackets" and by re-instating the other (spectator) $\phi_i$'s we get (2.22).

Well in this last step we also have to use that the integral actually breaks as
$$\oint_{around\ everything}=\oint_{around\ w_1}+\cdots+\oint_{around\ w_n}$$
and when doing the i-th integral the only poles in that region will be $\frac{1}{z-w_i}$ and nothing more, (ie the other w's do not contribute. We can always choose the contours this way.)
