Application of Ward identities for OPE under scaling and rotations

Question

I think this is a very straightforward question but I don't see it right now. In Tong's notes on String theory (http://www.damtp.cam.ac.uk/user/tong/string/four.pdf) section 4.2.3 he defines the weight of an operator under $\delta z=\epsilon z$, $\delta \bar{z}=\bar{\epsilon}\bar{z}$ in equation 4.16. Then at the end of the next page he uses the Ward identity 4.12 for that transformation. That is $$\delta O=-\mathrm{Res}[\epsilon zT(z)O(w)]=-\epsilon( hO(w)+w\partial O(w))$$ So $$\epsilon zT(z)O(w)=...+\frac{\epsilon( hO+z\partial O)}{z-w}+...$$ But the I don't see how he gets the term $(z-w)^{-2}$ which gives $$T(z)O(w)=...+\frac{ hO}{(z-w)^{2}}+\frac{\partial O}{z-w}+...$$ The naive thing is to divide by $\epsilon z$ but that doesn't give the result.

Answer 1

For a pole of order $n+1$, $$\mathrm{Res}[f,w]=\frac{1}{n!}\lim_{z\to w}\frac{d^n}{dz^n}(z-w)^{n+1}f(z)$$ So for a function of the type $$f(z)=\frac{g(z)}{(z-w)^{n+1}}$$ $$\mathrm{Res}[f,w]=\frac{1}{n!}g^{(n)}(w)$$ where $g$ is regular. Now the important part is to notice that $O$ depends on $w$ and it is not the result of evaluating $z$ at $w$.

Also because $\epsilon z$ is regular, we expect that $\epsilon z T(z)O(w)$ is of the form $\epsilon zh(z,w)$, where $h(z,w)$ has the singularities.

With this it becomes clear that the term $\epsilon h O(w)$ cannot arise from a simple pole as otherwise we won't get the factor $\epsilon z$. But for a pole of order 2, $$g^{(1)}(w)=\epsilon h O(w) \Rightarrow g(z)=\epsilon z h O(w)$$ On the other hand, the term $\epsilon w\partial O(w)$ can arise from a single pole: $$g(w)=\epsilon w \partial O(w)\Rightarrow g(z)=\epsilon z \partial O(w)$$ Therefore $$\epsilon z T(z)O(w)=...+\frac{\epsilon z h O(w)}{(z-w)^2}+\frac{\epsilon z \partial O(w)}{(z-w)}+...$$ Thus we get the result $$ T(z)O(w)=...+\frac{ h O(w)}{(z-w)^2}+\frac{ \partial O(w)}{(z-w)}+...$$