Ward identities and operator product expansions Polchinski's (2.3.11) gives the Ward Identity
$$i\epsilon[Res_{z\rightarrow z_0}j(z)\mathcal A(z_0,\bar z_0)+\bar {Res}_{\bar z\rightarrow \bar z_0}\tilde j(\bar z)\mathcal A(z_0,\bar z_0)]=\delta\mathcal A(z_0,\bar z_0).$$
Then he says on p46 that by this identity, $T(z)\mathcal A(0,0)$ contains the terms
$$\frac{h}{z^2}\mathcal A(0,0)~~~~~\text{and}~~~~~\frac1z\partial\mathcal A(0,0),$$
the former due to $\mathcal A(z,\bar z)\rightarrow\mathcal \zeta^{-h}\bar\zeta^{-\bar h}\mathcal A(z,\bar z)$, and the later due to $\mathcal A\rightarrow\mathcal A-\epsilon v^a\partial_a\mathcal A.$
Could someone please tell me how the two terms are obtained from the Ward Identity? I'm very confused about why they're the residues. Also, just by staring at it, why don't I see $\bar h$ in the first term given that things look symmetrical?
 A: There are a few different ingredients going into this:
Firstly, the (holomorphic) current generating the infinitesimal conformal transformation $\delta z=\epsilon v(z)$ is $j(z)=i v(z) T(z)$ (there's a similar antiholomorphic one too). The general Ward identity (at $z=0$) for this current gives
$$
\frac{1}{i\epsilon}\delta\mathcal A(0,0)=Res_{z\rightarrow 0}j(z)\mathcal A(0,0)+ \text{antiholomorphic}\\
=Res_{z\rightarrow 0}i v(z)T(z)\mathcal A(0,0)+ \text{antiholomorphic}\\
=Res_{z\rightarrow 0}i v(z)\sum_{n=0}^\infty\frac{1}{z^{n+1}}\mathcal A^{(n)}(0,0)+ \text{antiholomorphic}
$$
using the definitions in the $T\mathcal{A}$ OPE (2.4.11). We can drop regular terms because they won't contribute to the residue. In the $n$th term of the sum, we need to find $Res_{z\rightarrow 0}z^{-n-1}v(z)$, which is $\partial^n v(0)/n!$, the nth term in the Taylor expansion of $v$. This finally gives us the transformation property of $\mathcal{A}$ under a general conformal tranformation:
$$
\delta\mathcal A(0,0)=-\epsilon\sum_{n=0}^\infty\left[\frac{1}{n!}\partial^n v(0)\mathcal A^{(n)}(0,0)+ \text{antiholomorphic}\right].
$$
(There's nothing special about 0, so shift it to wherever else to get (2.4.12)).
Now to actually find out what some of those $\mathcal A^{(n)}(0,0)$s are, we can use what we know about how $\mathcal{A}$ transforms under some specific conformal transformations. We can work out the left hand side of the Ward identity and learn about the right.
Firstly, translations, where $\delta z=\epsilon v$, where $v$ is actually constant. We have $$\delta\mathcal{A}=-\epsilon v^a\partial_a\mathcal{A}=-\epsilon (v\partial\mathcal{A}+v^*\bar\partial\mathcal{A})$$ for the left hand side, and $$-\epsilon(v\mathcal{A}^{(0)}+v^*\tilde{\mathcal{A}}^{(0)})$$ for the right. Match coefficients and you're away: you've got $\mathcal{A}^{(0)}$ and $\tilde{\mathcal{A}}^{(0)}$.
Secondly, we're talking about a quasi-primary operator, so we specify its behaviour under dilations and rotations, the infinitesimal form of which is $\delta z = \epsilon \xi z$ for some $\xi$. So $v(z)=\xi z$ here. Translating to the finite form given, this is $z'=\zeta z=(1+\epsilon \xi)z$, and $$\mathcal{A}'(z',\bar{z}')=(1+\epsilon \xi)^{-h}(1+\epsilon \bar\xi)^{-\bar{h}}\mathcal{A}(z,\bar{z})=(1-\epsilon h\xi-\epsilon\bar{h}\bar{\xi})\mathcal{A}(z,\bar{z}).$$ So $$\delta\mathcal{A}(0,0)=-\epsilon(h\xi+\bar{h}\bar{\xi}) \mathcal{A}(0,0)$$
is the left hand side, and
$$-\epsilon(\xi\mathcal{A}^{(1)}+\bar{\xi}\tilde{\mathcal{A}}^{(1)})$$
is the right.
The upshot: you now have the $z^{-1}$ and $z^{-2}$ coefficients of the $T(z)\mathcal{A}(0,0)$ OPE (where $h$, the holomorphic weight, appears) and of the $\tilde{T}(\bar z)\mathcal{A}(0,0)$ OPE (where $\bar h$, the antiholomorphic weight, appears).
