# Normal ordering by contour integral in CFT

In chapter 6 of Di Francesco, they introduce the normal ordering $$(AB)(w) = \oint_w \frac{ dz }{ 2\pi i (z-w) }A(z) B(w)\ .\tag{6.130}$$

So far so good. But then starting eq (6.139) $$\oint_w \frac{ dz }{ z-w }A(z)B(w) = \oint_{|z|>|w|} \frac{ dz }{ z-w }A(z)B(w)-\oint_{|z|<|w|} \frac{ dz }{ z-w }B(w)A(z) \tag{6.139}$$ and $$A(z) = \sum_n (z-x)^{-n - h_A}A_n(x), \quad B(z) = \sum_m (w-x)^{-m - h_B}B_n(x) \ , \tag{6.140}$$ they try to convert the integral into sum of modes, which confuses me a lot.

The expansion coefficients/modes $$A_n(x), B_n(x)$$ in eq (6.140) clearly depend on the intermediate $$x$$. Moreover, when treating the two integrals in (6.139), two different intermediate $$x$$'s are needed. That makes the final conclusion eq (6.144) mysterious, $$(AB)_m = \sum_{n \le -h_A}A_n B_{m - n }+ \sum_{n > -h_A} B_{m-n}A_n, \tag{6.144}$$ since there the dependence on $$x$$ is removed some how: how are the $$A_n$$ and $$B_n$$ defined concretely?

(By making the $$z$$ contours to be $$|z| = |w| \pm \epsilon$$ and take the $$\epsilon \to 0$$ seems to indicate that the intermediate $$x \to w$$: but that will render the expansion $$B(w) = \sum_m (w-x)^{-m - h_B}B_m(x)$$ funny.)

OP has a point. The text in Ref. 1 above eq. (6.140) (which claims that $$|x|$$ should be between $$|z|$$ and $$|w|$$) is wrong. Here is a hopefully better argument: Let there be given a fixed $$w\in\mathbb{C}\backslash\{0\}$$. Now choose a fixed $$x\in\mathbb{C}$$ with $$|x|<|w|$$. One can for simplicity choose $$x=0$$, but it is enough if the following is satisfied.
• On the outer contour $$|z| > |w|$$ we demand that $$|z-x|>|w-x|$$ for running $$z$$. Then we may perturbatively expand the following geometric series $$\frac{z-w}{z-x}~=~1-\frac{w-x}{z-x}\quad\Rightarrow\quad \frac{z-x}{z-w}~=~\left(1-\frac{w-x}{z-x}\right)^{-1}~=~\sum_{\ell\geq 0}\left( \frac{w-x}{z-x}\right)^{\ell}. \tag{6.141}$$
• On the inner contour $$|z| < |w|$$ we demand that $$|z-x|<|w-x|$$ for running $$z$$. Then we may perturbatively expand the following geometric series $$\frac{z-w}{x-w}~=~1-\frac{z-x}{w-x}\quad\Rightarrow\quad \frac{x-w}{z-w}~=~\left(1-\frac{z-x}{w-x}\right)^{-1}~=~\sum_{\ell\geq 0}\left( \frac{z-x}{w-x}\right)^{\ell}.$$
Now the rest of the proof and the eqs. (6.142)-(6.145) in Ref. 1 are correct. [Note that eqs. (6.144)-(6.145) refer to $$x=0$$ but they can easily be generalized to non-zero $$x$$.] $$\Box$$