= Solution
Include the <Z boson> amplitude and its <quantum interference> with the <photon> amplitude. Define $s_W=\sin\theta_W$, $c_W=\cos\theta_W$, $e=gs_W$, $g_L^f=T_3^f-Q_fs_W^2$, $g_R^f=-Q_fs_W^2$, and
$$
C_{ij}(s)=\frac{e^2Q_eQ_\mu}{s}+\frac{g^2}{c_W^2}\frac{g_i^eg_j^\mu}{s-M_Z^2+iM_Z\Gamma_Z},\qquad i,j=L,R.
$$
For the same unpolarized and massless approximations as <2/b>, the complete tree-level <angular distribution> is
$$
\frac{d\sigma}{dc_*}=\frac{s}{128\pi}\left[(|C_{LL}|^2+|C_{RR}|^2)(1+c_*)^2+(|C_{LR}|^2+|C_{RL}|^2)(1-c_*)^2\right].
$$
The absolute squares include the pure terms and their interference. Put $A=|C_{LL}|^2+|C_{RR}|^2$ and $B=|C_{LR}|^2+|C_{RL}|^2$. The even part is $(A+B)(1+c_*^2)$ and the odd part is $2(A-B)c_*$. Thus the <forward-backward asymmetry> is $A_{\rm FB}=3(A-B)/(4(A+B))$.
At the <Z boson> pole, neglect the small <photon> contribution and use real tree-level couplings. Then
$$
\frac1\sigma\frac{d\sigma}{dc_*}=\frac38\left[1+c_*^2+2\mathcal A_e\mathcal A_\mu c_*\right],\quad
\mathcal A_f=\frac{(g_L^f)^2-(g_R^f)^2}{(g_L^f)^2+(g_R^f)^2}=\frac{2g_V^fg_A^f}{(g_V^f)^2+(g_A^f)^2},\quad
A_{\rm FB}^0=\frac34\mathcal A_e\mathcal A_\mu,
$$
where $g_V=g_L+g_R$, $g_A=g_L-g_R$. For charged <leptons>, $g_V/g_A=1-4\sin^2\theta_W$. Fitting the forward/backward counts, after detector and radiation corrections, therefore measures the <weak mixing angle>, conventionally an effective leptonic angle once electroweak corrections are included. For illustration $\sin^2\theta_W=0.231$ gives $\mathcal A_\ell\simeq0.151$ and $A_{\rm FB}^0\simeq0.0171$. This illustrative parameter choice is not extracted from this paper's data. The real tree-level <photon>-<Z boson> interference vanishes exactly at the pole but matters away from it.
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