= Solution
Now let the basic buoyancy be
$$
B=M^2x+N^2z.
$$
<Thermal-wind balance> requires the basic along-front velocity to have vertical shear $V_z=M^2/f$, so
$$
\boldsymbol U=\left(0,\Lambda x+\frac{M^2}{f}z,0\right).
$$
The basic absolute vorticity and buoyancy gradient are
$$
\boldsymbol\omega_a
=\left(-\frac{M^2}{f},0,f+\Lambda\right),
\qquad
\boldsymbol\nabla B=(M^2,0,N^2),
$$
and hence
$$
Q=\boldsymbol\omega_a\mathbin{\cdot}\boldsymbol\nabla B
=(f+\Lambda)N^2-\frac{M^4}{f}.
$$
For the prescribed disturbance the <buoyancy perturbation> vanishes, so the linearized buoyancy equation and <incompressibility> give
$$
M^2u+N^2w=0,\qquad ku+mw=0.
$$
The wavevector must therefore satisfy
$$
\frac{m}{k}=\frac{N^2}{M^2};
$$
the disturbance velocity lies along a basic <isopycnal>. The along-front momentum equation becomes
$$
-i\omega v+
\left(f+\Lambda-\frac{M^4}{fN^2}\right)u=0
=-i\omega v+\frac{Q}{N^2}u.
$$
Projecting the remaining momentum equations onto the divergence-free direction eliminates the pressure and yields
$$
\omega^2=\frac{fQ}{N^2}\frac{m^2}{k^2+m^2}
=\frac{fQN^2}{N^4+M^4}.
$$
Thus this isopycnal disturbance grows if and only if $fQ<0$. Geometrically, $Q$ is the component of the absolute vorticity along the buoyancy gradient, multiplied by $|\boldsymbol\nabla B|$; the horizontal buoyancy gradient reduces that component through the term $-M^4/f$. When $M=0$, the result reduces to the most unstable, nearly horizontal-wavevector limit of part i.
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