= Solution
Eliminate <pressure> by multiplying the radial equation by $k_z$ and subtracting $k_x$ times the vertical equation. For $k_z\ne0$, incompressibility gives $\tilde u_z=-(k_x/k_z)\tilde u_x$, and the pressure-free relation is $s(k_x^2+k_z^2)\tilde u_x/k_z=2\Omega k_z\tilde u_y$. Substitution in the azimuthal equation yields
$$
\boxed{s^2=\Omega^2\frac{2qk_xk_z-k_z^2}{k_x^2+k_z^2}=-\Omega^2\frac{k_z^2}{k^2}\left(1-2q\frac{k_x}{k_z}\right).}
$$
Thus the <vertical shear instability> occurs when $2qk_xk_z>k_z^2$, or $2q(k_x/k_z)>1$ for $k_z\ne0$. Otherwise the nonzero modes oscillate; equality is marginal in the exponential-growth sense. With $q>0$, growing disturbances have $k_x/k_z>1/(2q)$ and are strongly inclined in <wavevector> space.
Put $R=k_x/k_z$. Maximizing $(2qR-1)/(1+R^2)$ gives $qR^2-R-q=0$. For $q>0$, the unstable maximizing root is
$$
R_{\max}=\frac{1+\sqrt{1+4q^2}}{2q},\qquad \boxed{s_{\max}=\Omega\sqrt{\frac{\sqrt{1+4q^2}-1}{2}}=\Omega q\,[1+O(q^2)].}
$$
More precisely its bracket is $1-q^2/2+O(q^4)$, and $R_{\max}\sim q^{-1}$. This is the <maximum growth rate of the vertical shear instability>. For negative $q$, the maximum uses the corresponding negative root and is asymptotically $\Omega|q|$. At $q=0$ there is no exponentially growing mode. The undivided formula handles $k_z=0$: $s^2=0$, not an exponential instability; an allowed vertical velocity can instead force a secular azimuthal change. The excluded $\mathbf k=0$ is a spatially uniform disturbance and does not belong to the stipulated nonconstant one-phase family.
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