It is well known that the 2pt function of 2D free massless boson is given by \begin{equation} \label{s} \langle \phi(z,\bar{z})\phi(y,\bar{y})\rangle=-\frac{1}{4\pi g} \ln |z-y|^2\tag1 \end{equation} which is the eqn(2.10) in the CFT yellow book ("Conformal field theory" by P. Francesco et al). And this result is derived by Green function method.
I am trying to produce the above 2pt function via an alternative way, i.e. the mode expansion. For scalar field, the mode expansion is (6.54) in that book, which is \begin{equation} \phi(z,\bar{z})=\phi_0-\frac{i}{4\pi g}\pi_0 \ln(z\bar{z})+\frac{i}{\sqrt{4\pi g}}\sum_{n\neq 0}\frac{1}{n}(a_nz^{-n}+\bar{a}_n\bar{z}^{-n}) \end{equation} with the coefficients satisfying the commutators \begin{equation} [a_n,a_m]=n\delta_{m+n},~~[\bar{a}_n,\bar{a}_m]=n\delta_{m+n},~~a_0\equiv \bar{a}_0\equiv \frac{\pi_0}{\sqrt{4\pi g}} \end{equation} For vacuum state, we have \begin{equation} a_n |0\rangle=\bar{a}_n|0\rangle=0, ~(n\geq 0) \end{equation} Then using above commutators, we can compute the vacuum 2pt function as follows \begin{eqnarray} \langle\phi(z,\bar{z})\phi(y,\bar{y})\rangle&=&\phi^2_0-\frac{1}{4\pi g}\sum_{n>0}\frac{-1}{n^2}\langle(a_nz^{-n}+\bar{a}_n\bar{z}^{-n})(a_{-n}y^{n}+\bar{a}_{-n}\bar{y}^{n})\rangle\\ &=&\phi^2_0-\frac{1}{4\pi g}\sum_{n>0}\frac{-1}{n^2}n((y/z)^n+(\bar{y}/\bar{z})^n)\\ &=&\phi^2_0-\frac{1}{4\pi g}(\log(1-y/z)+\log(1-\bar{y}/\bar{z})),\tag2 \end{eqnarray} where we used \begin{equation} \sum_{n=1}\frac{x^n}{n}=-\log(1-x) \end{equation} However equation $(2)$ is not the standard form as in equation $(1)$. They are different by terms $\log z$ and $\log \bar{z}$. My question is what is wrong with equation $(2)$?
