= Solution
Write the vortex aspect ratio as $\chi$ to distinguish it from cylindrical radius. The <Kida vortex> core flow is
$$
u_x=\frac{3\Omega}{2\chi(\chi-1)}y,
\qquad
u_y=-\frac{3\Omega\chi}{2(\chi-1)}x.
$$
As $\chi\to\infty$, $u_x\to0$ and $u_y\to-(3/2)\Omega x$, recovering <Keplerian shear>.
For a fluid particle,
$$
\ddot x=-\left[\frac{3\Omega}{2(\chi-1)}\right]^2x.
$$
It therefore circulates around an ellipse with angular frequency $3\Omega/[2(\chi-1)]$ and period
$$
\boxed{T_{\rm vort}=\frac{4\pi(\chi-1)}{3\Omega}}.
$$
For a perturbation depending only on $z,t$, horizontal pressure gradients vanish. Linearization gives
$$
\boxed{\partial_tu_x'
=\Omega\left[2-\frac{3}{2\chi(\chi-1)}\right]u_y'},
$$
$$
\boxed{\partial_tu_y'
=-\Omega\left[2-\frac{3\chi}{2(\chi-1)}\right]u_x'}.
$$
Taking $\mathbf u'\propto e^{-i\omega t}$ yields
$$
\boxed{\omega^2=\Omega^2
\left[2-\frac{3}{2\chi(\chi-1)}\right]
\left[2-\frac{3\chi}{2(\chi-1)}\right]}.
$$
The first bracket is positive for $\chi>3/2$, while the second is negative for $1<\chi<4$. Their product is therefore negative, so $\omega$ is imaginary and the mode grows precisely when
$$
\boxed{\frac32<\chi<4}.
$$
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