= Solution
Linearize about zero bulk flow and the uniform guide field. Put $\mathbf b=\delta\mathbf B/\sqrt{4\pi\rho}$, and take <plane waves> proportional to $e^{i(\mathbf k\cdot\mathbf x-\omega t)}$. Let $k_\parallel=\mathbf k\cdot\hat{\mathbf z}$, $k=|\mathbf k|$, $q=k_\parallel v_A$, and
$$
d_i=\frac c{\omega_{pi}},\qquad v_Ad_i=\frac{cB_0}{4\pi en}.
$$
Both $\mathbf u$ and $\mathbf b$ are perpendicular to $\mathbf k$. Projecting momentum onto that plane eliminates total <pressure>. The remaining magnetic-tension term gives
$$
-i\omega\mathbf u=iq\mathbf b,\qquad \omega\mathbf u=-q\mathbf b.
$$
In the induction equation, the Hall contribution is $-c(B_0\partial_z)(\nabla\times\delta\mathbf B)/(4\pi en)$; its two derivative factors of $i$ in the <Fourier mode> produce a positive coefficient multiplying $\mathbf k\times\mathbf b$. Thus
$$
-i\omega\mathbf b=iq\mathbf u+qd_i\mathbf k\times\mathbf b,
\qquad\omega\mathbf b=-q\mathbf u+i qd_i\mathbf k\times\mathbf b.
$$
Eliminating $\mathbf u$ gives
$$
(\omega^2-q^2)\mathbf b=i\omega qd_i\mathbf k\times\mathbf b.
$$
On the transverse plane, $(i\mathbf k\times)^2=k^2$, so applying that operator again proves the <incompressible Hall-MHD wave dispersion>:
$$
\boxed{(\omega^2-k_\parallel^2v_A^2)^2=\omega^2 k_\parallel^2v_A^2k^2\frac{c^2}{\omega_{pi}^2}.}
$$
Equivalently, in a <helicity decomposition of a transverse Fourier mode> with $i\mathbf k\times\mathbf b_h=hk\mathbf b_h$, $h=\pm1$, each polarization obeys $\omega^2-hqkd_i\omega-q^2=0$. For $k_\parallel=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_\parallel\ne0$.
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