Infinitesimal transformations in the Möbius subgroup In 2D a primary field of weight $(h,\bar{h})$ transforms as $\phi\rightarrow\phi '$ where,
$$
\phi'(f(z),\bar{f}(\bar{z}))=(\partial_zf)^{-h}(\partial_{\bar{z}}\bar{f})^{-\bar{h}}\phi(z,\bar{z})
$$
I want to show that 
an infinitesimal transformation of the form$$
z\mapsto z+\xi(z) \;\;\;\;\;\;\;\;\;\; \bar{z}\mapsto\bar{z}+\bar{\xi}(\bar{z})
$$
gives $$
\delta\phi=-h\frac{\partial\xi}{\partial z}\phi-\xi\frac{\partial\phi}{\partial z}-\bar{h}\frac{\partial\bar{\xi}}{\partial{\bar{z}}}\phi-\bar{\xi}\frac{\partial\phi}{\partial\bar{z}}
$$
My attempt:
$$
\phi'(f(z),\bar{f}(\bar{z}))=\phi'(z+\xi(z),\bar{z}+\bar{\xi}(\bar{z}))=
$$
$$
\left(\partial_z(z+\xi(z))^{-h}\right)\left(\partial_{\bar{z}}(\bar{z}+\bar{\xi}(\bar{z}))^{-\bar{h}}\right)\phi(z,\bar{z})
$$
I assume that I should take a Taylor expansion but I'm not quite sure how the Taylor expansion works here.
 A: You have done $\left(\partial_z(z+\xi(z))^{-h}\right)\left(\partial_{\bar{z}}(\bar{z}+\bar{\xi}(\bar{z}))^{-\bar{h}}\right)\phi(z,\bar{z})$ instead of writing $\left(\partial_z(z+\xi(z))\right)^{-h}\left(\partial_{\bar{z}}(\bar{z}+\bar{\xi}(\bar{z}))\right)^{-\bar{h}}\phi(z,\bar{z})$
Start from
$$\phi'(z+\xi(z),\bar{z}+\bar{\xi}(\bar{z}))=\left(\partial_z(z+\xi(z))\right)^{-h}\left(\partial_{\bar{z}}(\bar{z}+\bar{\xi}(\bar{z}))\right)^{-\bar{h}}\phi(z,\bar{z})$$
Expand the LHS by Taylor expansion:
$$\phi'(z+\xi(z),\bar{z}+\bar{\xi}(\bar{z}))=\phi '(z,\bar z)+\xi\frac{\partial\phi}{\partial z}+\bar{\xi}\frac{\partial\phi}{\partial\bar{z}}$$
RHS becomes(neglet the second order $\xi$ term):
$$\left(\partial_z(z+\xi(z))\right)^{-h}\left(\partial_{\bar{z}}(\bar{z}+\bar{\xi}(\bar{z}))\right)^{-\bar{h}}\phi(z,\bar{z})=\left(1+\frac{\partial{\xi(z)}}{\partial{z}}\right)^{-h}        \left(1+\frac{\partial{\bar\xi(\bar z)}}{\partial{\bar z}}\right)^{-\bar{h}}  \phi (z,\bar z)$$
$$=\left(1-h \frac{\partial{\xi(z)}}{\partial{z}}\right)  \left(1-\bar h \frac{\partial{\bar\xi(\bar z)}}{\partial{\bar z}}\right)  \phi (z,\bar z) $$
$$=\left(\phi(z,\bar z )-h \frac{\partial{\xi(z)}}{\partial{z}}\phi(z,\bar z )  -\bar h \frac{\partial{\bar\xi(\bar z)}}{\partial{\bar z}}\phi(z,\bar z )\right)$$
Equating LHS and RHS gives you :
$$\delta \phi=\phi '(z,\bar z)-\phi(z,\bar z)=-h\frac{\partial\xi}{\partial z}\phi-\xi\frac{\partial\phi}{\partial z}-\bar{h}\frac{\partial\bar{\xi}}{\partial{\bar{z}}}\phi-\bar{\xi}\frac{\partial\phi}{\partial\bar{z}}$$
