= Solution
Let $L_\perp=k_\perp^{-1}$ and let $U$ be the common size of horizontal and vertical <velocity> perturbations, as found in part <1/b>. The linear <inertial wave> time and nonlinear advection time are
$$
\tau_w^{-1}\sim2\Omega\frac{|k_\parallel|}{k_\perp},\qquad
\tau_{\rm nl}^{-1}\sim k_\perp U.
$$
The <critical balance in a rapidly rotating fluid> is therefore
$$
\boxed{\epsilon=\frac{|k_\parallel|}{k_\perp}
\sim\frac{k_\perp U}{2\Omega}\ll1,\qquad
\omega\sim k_\perp U\sim2\Omega\epsilon.}
$$
The <Rossby number> is of order $\epsilon$. Horizontal <geostrophic flow> balances the leading <pressure gradient> against <Coriolis acceleration>, so
$$
\boxed{u_\perp\sim u_\parallel\sim U,\quad
\delta p\sim2\Omega L_\perp U\sim U^2/\epsilon,\quad
k_\parallel\sim\epsilon k_\perp.}
$$
These are dimensionally comparable statements, rather than comparisons of a dimensional <frequency> with a dimensional <velocity>. Equivalently, use $L_\perp$ and $(2\Omega)^{-1}$ as units. Then
$$
k_\perp L_\perp=O(1),\quad
k_\parallel L_\perp=O(\epsilon),\quad
\frac{\omega}{2\Omega}=O(\epsilon),\quad
\frac{u_\perp}{2\Omega L_\perp},\frac{u_\parallel}{2\Omega L_\perp}=O(\epsilon),\quad
\frac{\delta p}{(2\Omega L_\perp)^2}=O(\epsilon).
$$
Thus rotation stays $O(1)$ in these units, while the motions evolve on the slower $O(\epsilon^{-1})$ time. Vertical advection $u_\parallel\partial_z$ is smaller than horizontal advection by $\epsilon$.
Write $\mathbf u_\perp=\mathbf u_\perp^{(0)}+\mathbf u_\perp^{(1)}+\cdots$, with $\mathbf u_\perp^{(1)}=O(\epsilon U)$, and $p=p_0+p_1+\cdots$. At leading horizontal order,
$$
0=-\nabla_\perp p_0+2\Omega\mathbf u_\perp^{(0)}\times\widehat{\mathbf z}.
$$
Define the <stream function> $\Phi=p_0/(2\Omega)$, including its horizontally uniform part in this convention. Then
$$
\boxed{\mathbf u_\perp^{(0)}=\widehat{\mathbf z}\times\nabla_\perp\Phi
=(-\partial_y\Phi,\partial_x\Phi),\qquad
\nabla_\perp\cdot\mathbf u_\perp^{(0)}=0.}
$$
The next-order <incompressibility> condition is not zero horizontal divergence:
$$
\boxed{\nabla_\perp\cdot\mathbf u_\perp^{(1)}=-\partial_z u_\parallel.}
$$
This term has size $\epsilon k_\perp U$ and is essential: its product with the large rotation rate is a leading contribution to the vertical <vorticity> equation.
To see that contribution explicitly, write $D_\perp=\partial_t+\mathbf u_\perp^{(0)}\cdot\nabla_\perp$. At the first horizontal inertial order,
$$
D_\perp\mathbf u_\perp^{(0)}=-\nabla_\perp p_1
+2\Omega\mathbf u_\perp^{(1)}\times\widehat{\mathbf z}.
$$
Taking its vertical <curl> eliminates $p_1$. The two-dimensional advection identity and the two divergence orders give
$$
D_\perp\zeta=-2\Omega\nabla_\perp\cdot\mathbf u_\perp^{(1)}
=2\Omega\partial_z u_\parallel,\qquad
\zeta=(\nabla_\perp\times\mathbf u_\perp^{(0)})_z=\nabla_\perp^2\Phi.
$$
Equivalently the <ageostrophic flow> satisfies
$$
\mathbf u_\perp^{(1)}=\frac{\widehat{\mathbf z}\times[D_\perp\mathbf u_\perp^{(0)}+\nabla_\perp p_1]}{2\Omega},
$$
whose divergence is $-D_\perp\zeta/(2\Omega)$. This explicitly verifies that the next-order divergence has the required sign.
The vertical momentum equation, at the same slow order, is
$$
D_\perp u_\parallel=-\partial_zp_0=-2\Omega\partial_z\Phi.
$$
Finally $\mathbf u_\perp^{(0)}\cdot\nabla_\perp f=\{\Phi,f\}$, where the canonical two-dimensional <Poisson bracket> is $\{f,g\}=f_xg_y-f_yg_x$. The <reduced equations for an anisotropic rotating fluid> are
$$
\boxed{\begin{aligned}
\partial_t\nabla_\perp^2\Phi+\{\Phi,\nabla_\perp^2\Phi\}
&=2\Omega\partial_z u_\parallel,\\
\partial_tu_\parallel+\{\Phi,u_\parallel\}
&=-2\Omega\partial_z\Phi.
\end{aligned}}
$$
The omitted vertical-advection, ageostrophic-advection and relative-vorticity stretching terms are smaller by $O(\epsilon)$ in the equations retained here. Linearizing these equations recovers $\omega^2=4\Omega^2k_\parallel^2/k_\perp^2$, the anisotropic limit of part <1/a>.
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