= Solution
Put $K=(k^2+m^2)^{1/2}$ and $P=p'/\rho_0$. Acting on the <Fourier mode>, $D_0=-i(\hat\omega-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The <buoyancy> equation gives $g\rho'/\rho_0=iN^2w/\omega$. Substitution into vertical momentum yields
$$
\boxed{\omega^2=\frac{N^2k^2}{k^2+m^2},\qquad
\hat\omega=Uk\pm\frac{Nk}{K}.}
$$
The two signs describe the two frequency branches; when the <intrinsic frequency> is required positive, $\omega=N|k|/K$. For a stationary disturbance $\hat\omega=0$, so $\omega=-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat\omega=Uk-Nk/K$. The stationary dispersion condition is
$$
\boxed{K=\frac NU,\qquad m^2=\frac{N^2}{U^2}-k^2.}
$$
Differentiate the \i[unconstrained] dispersion before imposing stationarity. The <stationary mountain-wave group velocity> is obtained from the ground-frame dispersion; before imposing stationarity it is
$$
\hat c_{gx}=U-\frac{Nm^2}{K^3},\qquad
\hat c_{gz}=\frac{Nkm}{K^3}.
$$
On the stationary locus this becomes
$$
\boxed{\widehat{\mathbf c_g}=\frac{Uk}{K^2}(k,m),\qquad
\hat c_{gx}=\frac{Uk^2}{K^2},\quad
\hat c_{gz}=\frac{Ukm}{K^2}.}
$$
Differentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.
For upward <energy> propagation from the lower boundary, choose $m<0$, since $k<0$ and $\hat c_{gz}>0$. It requires real nonzero $m$, hence
$$
\boxed{0<|k|<N/U\quad\Longleftrightarrow\quad\lambda_x>2\pi U/N.}
$$
The downstream ray inclination above horizontal is
$$
\boxed{\theta=\tan^{-1}\frac{|m|}{|k|}
=\cos^{-1}\frac{U|k|}{N}=\cos^{-1}\frac\omega N.}
$$
It is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.
\b[Frame qualification for the requested phase angle:] the normal intrinsic <phase velocity> is $\mathbf c_p=\omega(k,m)/K^2$. Because $\omega=-Uk$, the boxed stationary <group velocity> is exactly $-\mathbf c_p$: its angle to \i[intrinsic] <phase velocity>[phase propagation] is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic <group velocity> $\mathbf c_g=\widehat{\mathbf c_g}-U\hat{\mathbf x}$ satisfies $\mathbf c_g\cdot(k,m)=0$, so its angle to intrinsic <phase velocity>[phase propagation] is $\pi/2$. In the ground frame the <phase velocity> of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{\mathbf c_g}=(0.36U,0.48U)$ and $\mathbf c_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.
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