= Solution
Write the <velocity field> as $u=U+u\prime$, $v=v\prime$. The uniform background has zero <relative vorticity>, and the perturbation has $\xi\prime=v\prime_x-u\prime_y$. Keeping only terms linear in the perturbations gives
$$
\boxed{(\partial_t+U\partial_x)\xi\prime+\beta v\prime=0.}
$$
The <incompressibility condition> permits a <streamfunction> $\psi$ with $u\prime=-\psi_y$, $v\prime=\psi_x$, hence $\xi\prime=\nabla_h^2\psi$. Thus the <Linear Rossby-wave equation> here is
$$
(\partial_t+U\partial_x)\nabla_h^2\psi+\beta\psi_x=0.
$$
Choose a <plane wave> $\psi=\operatorname{Re}\{\widehat\psi e^{i(kx+ly-\omega t)}\}$, with $K^2=k^2+l^2>0$. Substitution gives
$$
i(\omega-Uk)K^2\widehat\psi+i\beta k\widehat\psi=0,
$$
and therefore the <local Rossby-wave dispersion relation in a zonal jet>
$$
\boxed{\omega=Uk-\frac{\beta k}{k^2+l^2}.}
$$
The $Uk$ term is the background <Doppler shift>; the intrinsic <phase velocity> is westward. A real physical <wave> can always be represented using the conjugate <wavevector> if necessary to choose $\omega\geq0$.
The relevant ground-frame zonal <phase velocity> and <group velocity> are
$$
c_{p,x}=\frac\omega k=U-\frac\beta{K^2},\qquad
c_{g,x}=\frac{\partial\omega}{\partial k}
=U+\frac{\beta(k^2-l^2)}{K^4},\qquad
c_{g,y}=\frac{\partial\omega}{\partial l}=\frac{2\beta kl}{K^4}.
$$
\b[Time increases downward in the source diagram.] The individual phase contours lean towards increasing longitude as time increases, so their <phase velocity> is eastward. The envelope moves eastward faster: successive crests grow at the downstream edge and decay upstream. This illustrates <Rossby-wave downstream development>.
For the intended purely zonal <wave packet>, $l=0$, and
$$
c_p=U-\frac\beta{k^2}>0,\qquad c_g=U+\frac\beta{k^2}>U.
$$
Consequently, with $U,\beta>0$ and the positive-frequency choice $k>0$,
$$
\boxed{k>\sqrt{\beta/U}.}
$$
There is a small interpretation qualification: if “zonally propagating” specified only the plotted direction while allowing an unknown nonzero meridional <wavenumber>, eastward <phase velocity> alone would imply $k^2+l^2>\beta/U$, not the stronger bound on $k$ alone. In this uniform model a strictly zonal <group velocity> with $\beta k\ne0$ does imply $l=0$, which supplies the intended interpretation.
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