mathcal{A}^{c \bar a}(w) \right. \cr
& \qquad + \left. j^b_{L,z}(x)
 \left( \frac{c_+}{c_++c_-} \frac{i{f^{ac}}_d 
\mathcal{A}^{d \bar a}(w)}{z-w} + ... \right) \right] \cr
%
& = -\frac{c_-}{c_+} \left[ \frac{c_1 \mathcal{A}^{a \bar a}(w)}{(z-w)^2} + \left(-c_2 + \frac{i c_+}{c_++c_-} \right) \frac{{f^a}_{bc}:j^b_{L,z}\mathcal{A}^{c \bar a}:(w)}{z-w} \right. \cr
& \qquad \left. - (c_2-g) \frac{{f^a}_{bc}:j^b_{L,\bar z}\mathcal{A}^{c \bar a}:(w)(\bar z - \bar w)}{(z-w)^2}  + ... \right]\end{aligned}$$ In principle the second- and first-order poles that we obtain in the last line may receive corrections from the lower-order terms that we neglected in the penultimate line. We will now argue that it is not the case. Let us consider the first term in the last line (the second-order pole). This term may receive corrections of the form ${T^a}_b \mathcal{A}^{b \bar a}$, where the tensor ${T^a}_b$ contains at least one structure constant. Such a tensor vanishes by using properties of the Lie super algebras under consideration [@Bershadsky:1999hk]. Let us now consider the second term (the holomorphic simple pole). It could receive corrections of the form ${T^a}_{bc}:j^b_{L,z}\mathcal{A}^{c \bar a}$, where ${T^a}_{bc}$ contains at least two structure constants. Again, according to [@Bershadsky:1999hk], this tensor vanishes because traceless four-tensors invariantly contracted with structure constants over two indices vanish. The third term receives no higher order corrections for the same reason. Thus the terms written in the last line of are not corrected. Using equations and we finally obtain: \[jLjR1\] j\^a\_[L,z]{}(z) j\^[|a]{}\_[R,z]{}(w) = ( + + ) + ... where the ellipses refer to terms of order zero or more in the distance between the two operators. Similarly we can compute: $$\begin{aligned}
 \label{jLjR2}
j^a_{L,\bar z}(z) j^{\bar a}_{R,\bar z}(w) &=  
\frac{c_+ c_-}{c_++c_-}\left( \frac{\mathcal{A}^{a \bar a}(w)}{(\bar z-\bar w)^2}
+ \frac{c_+}{c_++c_-} \frac{\partial \mathcal{A}^{a \bar a}(w)(z-w)}{(\bar z-\bar w