= Solution
For a vertical plane wave $\chi=e^{imz}$, the vertical-mode equation gives
$$
\boxed{c=\frac N{|m|}}.
$$
Consequently the Boussinesq <Equatorial Kelvin wave> and $n=1$ <Equatorial Rossby wave> have
$$
\boxed{\omega_K=\frac{Nk}{|m|}},
\qquad
\boxed{\omega_R=-\frac{Nk}{3|m|}}.
$$
Their meridional structures are those in part b with $L_e=[N/(\beta|m|)]^{1/2}$.
Let the positive forcing frequency be $\omega=N\sin\theta$ with $\theta\ll1$. Upward radiation into $z>0$ requires $m<0$ for both responses. The Kelvin wave has $k>0$ and
$$
\frac{k}{|m|}=\sin\theta.
$$
Its group velocity points up and right along a ray making angle approximately $\theta$ above the horizontal, so it occupies the region to the right of the localized source. Its phase propagates down and right.
The $n=1$ Rossby wave has $k<0$ and
$$
\frac{|k|}{|m|}=3\sin\theta.
$$
Its group velocity points up and left along a ray making angle approximately $3\theta$ above the negative horizontal direction, so it occupies the region to the left. Its phase propagates down and left. Thus energy radiates upward and away from the source on both sides, while the vertical phase propagation is downward in both wave beams.
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