How does one get Di Francesco Conformal Field Theory equation 6.65:
$$ V_\alpha(z,\bar{z})V_\beta(w,\bar{w}) \sim |z-w|^{\frac{2\alpha\beta}{4\pi g}} V_{\alpha+\beta}(w,\bar{w})+\ldots~?\tag{6.65}$$
By definition, $$V_\alpha(z,\bar{z}) =\, :e^{i\alpha \phi(z,\bar{z})}:\tag{6.58}$$
We can use 6.64 which says: $$ :e^{a\phi_1}:\, :e^{b\phi_2}:\ =\ :e^{a\phi_1+b\phi_2}:e^{ab\langle\phi_1\phi_2\rangle}\tag{6.64}$$
Also we have from 5.75: $$ \langle\phi(z,\bar{z})\phi(w,\bar{w})\rangle = -\frac{1}{4\pi g}\ln(|z-w|^2)$$
Here we go: $$ V_\alpha(z,\bar{z})V_\beta(w,\bar{w}) \overset{\text{def}}{=} \ :e^{i\alpha \phi(z,\bar{z})}: :e^{i\beta \phi(w,\bar{w})}:\ \overset{6.64}{=}\ :e^{i\alpha \phi(z,\bar{z})+ i\beta \phi(w,\bar{w})}:e^{-\alpha\beta\langle\phi(z,\bar{z})\phi(w,\bar{w})\rangle}$$ $$\overset{5.75}{=}|z-w|^{\frac{2\alpha\beta}{4\pi g}}:e^{i\alpha \phi(z,\bar{z})+ i\beta \phi(w,\bar{w})}: $$
This is very close to 6.65 except that we still have a $z$ and a $w$. Normally we would now Taylor expand $z$ around $w$. But I think this might give additional terms that might be singular, since I'm assuming that we can have ${\frac{2\alpha\beta}{4\pi g}} < -1$. How to proceed? (Or can it not be that ${\frac{2\alpha\beta}{4\pi g}} < -1$?)
