= Solution
{c}
Write $u=U+u'$, $v=v'$, and $\xi=\xi'$, with $U=\bar u$ constant. Discarding products of perturbations in the <vorticity equation> gives
$$
(\partial_t+U\partial_x)\xi'+\beta v'=0.
$$
Introduce the <streamfunction> convention
$$
u'=-\partial_y\psi',\qquad v'=\partial_x\psi',\qquad \xi'=\nabla_h^2\psi'.
$$
It enforces <incompressibility> identically. The linear <partial differential equation> is
$$
(\partial_t+U\partial_x)\nabla_h^2\psi'+\beta\partial_x\psi'=0.
$$
For the <plane wave> $\psi'=\operatorname{Re}\{\widehat\psi\exp[i(kx+ly-\omega t)]\}$, put $K^2=k^2+l^2>0$. Substitution yields
$$
i(\omega-Uk)K^2\widehat\psi+i\beta k\widehat\psi=0,\qquad
\boxed{\omega=Uk-\frac{\beta k}{k^2+l^2}.}
$$
The first term is the <Doppler shift> by the background flow; the second is the intrinsic <Rossby wave> <frequency>. The source's convention $\omega\geq0$ selects representatives of real waves, rather than making this expression nonnegative for arbitrary $k,l$. For $U=0$ and $\beta>0$, nonzero positive <frequency> requires $k<0$; the conjugate mode has opposite signs of $k,l,\omega$ and describes the same real field.
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