= Solution
Take the <curl> of the linear momentum equation. For constant $\boldsymbol\Omega$ and zero <divergence>, $\nabla\times(\boldsymbol\Omega\times\mathbf u)=-(\boldsymbol\Omega\cdot\nabla)\mathbf u$. With $\boldsymbol\Omega=\Omega\mathbf e_z$, this gives
$$
\partial_t\boldsymbol\omega=2\Omega\partial_z\mathbf u.
$$
Take another <curl> and use part (a): $-\partial_t\nabla^2\mathbf u=2\Omega\partial_z\boldsymbol\omega$. Differentiate in time and substitute the previous equation:
$$
-\partial_t^2\nabla^2\mathbf u=4\Omega^2\partial_z^2\mathbf u.
$$
Consequently
$$
\boxed{\partial_t^2\nabla^2\mathbf u+4\Omega^2\partial_z^2\mathbf u=\mathbf0.}
$$
The two curl operations remove the pressure gradient while retaining the rotational restoring force.
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