= Solution
Every forced propagating mode must have the imposed zonal frequency
$$
\omega=kC.
$$
For $C>0$, only the <Equatorial Kelvin wave> has the correct sign. Upward group velocity in the fluid $z>0$ selects
$$
m_K=-\frac NC.
$$
With $K(y)=\exp[-\beta y^2/(2C)]$, its pressure field can be written
$$
\boxed{
\phi_K=\operatorname{Re}
\left\{
a_KK(y)e^{i[k(x-Ct)+m_Kz]}
\right\}}.
$$
Here $v_K=0$, $u_K=\phi_K/C$, and the complex vertical-velocity amplitude is
$$
\hat w_K=-\frac{kCm_K}{N^2}a_KK
=\frac kN a_KK.
$$
The lower boundary is matched only if
$$
\boxed{
W(y)=\frac kN a_K
\exp\left(-\frac{\beta y^2}{2C}\right)}
$$
after choosing the phase of $a_K$ to make the prescribed cosine amplitude real.
For $C<0$, the upward-radiating <Equatorial Rossby waves> have
$$
\boxed{
m_n=-\frac{N}{(2n+1)C}>0},
\qquad n=1,2,\ldots.
$$
Choose each pair $(\hat v_n,\hat\phi_n)$ as in part c for this $m_n$. The propagating sum is
$$
\boxed{
v=\operatorname{Re}
\sum_{n=1}^{\infty}
a_n\hat v_n(y)
e^{i[k(x-Ct)+m_nz]}},
$$
$$
\boxed{
\phi=\operatorname{Re}
\sum_{n=1}^{\infty}
a_n\hat\phi_n(y)
e^{i[k(x-Ct)+m_nz]}}.
$$
The other fields follow mode by mode from part c and $\hat w_n=-kCm_n\hat\phi_n/N^2$. Thus the boundary matching condition is
$$
\boxed{
W(y)=
\sum_{n=1}^{\infty}
\frac{k\,a_n}{N(2n+1)}
\hat\phi_n(y)}.
$$
For $C>0$, the space of upward-radiating wave profiles is only the one-dimensional Gaussian Kelvin profile, so a generic $W(y)$ cannot be matched by propagating waves. Its Kelvin projection radiates upward; the remaining forcing produces a balanced, vertically evanescent response trapped near the lower boundary. This is the equatorial analogue of the fact that quasi-geostrophic <Rossby waves> have westward rather than eastward phase propagation.
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