= Solution
Use the <data augmentation> labels to separate the mixture into groups. Define $n_i=\sum_j\mathbf1_{\{z_j=i\}}$, $s_i=\sum_{z_j=i}x_j$ and $Q_i(\mu_i)=\sum_{z_j=i}(x_j-\mu_i)^2$. The intended joint <posterior distribution> is proportional to
$$
\prod_{j=1}^n\omega_{z_j}\varphi(x_j;\mu_{z_j},v_{z_j})\prod_i\exp[-\mu_i^2/(2\tau^2)]v_i^{-\alpha-1}e^{-\beta/v_i}\omega_i^{\epsilon_i-1}.
$$
Summing over all allocation labels recovers the marginal mixture posterior. The <Gaussian mixture Gibbs updates with independent priors> follow by collecting the terms in each parameter.
For $\mu_i$, the conditional exponent is
$$
-\frac12\left[\left(\frac1{\tau^2}+\frac{n_i}{v_i}\right)\mu_i^2-2\frac{s_i}{v_i}\mu_i\right]+\text{constant}.
$$
Completing the square gives
$$
\boxed{\mu_i\mid\mathbf v,\boldsymbol\omega,\mathbf z,\mathbf x\sim N(m_i,V_i),\quad V_i=\left(\tau^{-2}+n_i/v_i\right)^{-1},\quad m_i=V_i s_i/v_i.}
$$
For $v_i$, the conditional density is proportional to $v_i^{-(\alpha+n_i/2)-1}\exp[-(\beta+Q_i(\mu_i)/2)/v_i]$, hence
$$
\boxed{v_i\mid\boldsymbol\mu,\boldsymbol\omega,\mathbf z,\mathbf x\sim\operatorname{InvGamma}\left(\alpha+\frac{n_i}2,\ \beta+\frac{Q_i(\mu_i)}2\right).}
$$
The independent prior on $\mu_i$ adds neither a $v_i^{-1/2}$ factor nor a $\mu_i^2/v_i$ penalty. Finally, combining the allocation counts with the <Dirichlet distribution> prior gives
$$
\boxed{\boldsymbol\omega\mid\boldsymbol\mu,\mathbf v,\mathbf z,\mathbf x\sim\operatorname{Dirichlet}(\epsilon_1+n_1,\ldots,\epsilon_k+n_k).}
$$
A <Gibbs sampler> successively draws from these conditionals, using the most recently updated values. An empty component has $n_i=s_i=Q_i=0$, so its mean and variance conditionals reduce to their proper priors without dividing by an undefined group sample mean.
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