= Solution
The <BRST operator> $Q_B$ is Grassmann odd and represents the gauge symmetry on the gauge-fixed state space. Requiring two successive BRST transformations to vanish means
$$
Q_B^2=\frac12\{Q_B,Q_B\}=0.
$$
This nilpotence makes physical states a <BRST cohomology>. If $[Q_B,S_0]=0$ and $\Psi$ is the <gauge-fixing fermion>, the graded Jacobi identity gives
$$
[Q_B,S]=[Q_B,S_0]+[Q_B,\{Q_B,\Psi\}]
=\frac12[\{Q_B,Q_B\},\Psi]=0,
$$
so $S=S_0+\{Q_B,\Psi\}$ is BRST invariant.
A holomorphic field of <conformal weight> $h$ has the Laurent expansion
$$
\phi(z)=\sum_{n\in\mathbb Z}\phi_nz^{-n-h}.
$$
Under the <state–operator correspondence>, $\phi(z)|0\rangle$ must be regular at the origin. Terms with $n>-h$ have negative powers, so
$$
\boxed{\phi_n|0\rangle=0\quad(n>-h)}.
$$
For the anticommuting <bc system>,
$$
b(z)=\sum_nb_nz^{-n-2},\qquad c(z)=\sum_nc_nz^{-n+1},
\qquad\{b_m,c_n\}=\delta_{m+n,0}.
$$
Separating creation and annihilation modes and summing the geometric series for $|z|>|w|$ gives the <bc ghost operator-product expansion>
$$
\boxed{b(z)c(w)\sim\frac1{z-w}}.
$$
The <BRST current> built from the matter and ghost stress tensors has an operator-product expansion with $c$ whose residue gives
$$
\boxed{\{Q_B,c(z)\}=c(z)\partial c(z)}.
$$
For a matter <Virasoro primary operator> $\Phi(z,\bar z)$ of weights $(1,1)$, the two standard <string vertex operators> are
$$
\boxed{U(z,\bar z)=c(z)\bar c(\bar z)\Phi(z,\bar z)},
\qquad
\boxed{V=\int_\Sigma d^2z\,\Phi(z,\bar z)}.
$$
The first is the local, unintegrated vertex. Using $Q_Bc=c\partial c$, $Q_B\bar c=\bar c\bar\partial\bar c$, and the weight-$(1,1)$ transformation of $\Phi$, the terms cancel pairwise and $Q_BU=0$. For the integrated vertex,
$$
\{Q_B,\Phi\}=\partial(c\Phi)+\bar\partial(\bar c\Phi),
$$
so $\{Q_B,V\}=0$ on a closed worldsheet because the variation is a total derivative. Both vertices therefore represent the same <BRST cohomology> class.
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