= Solution
Put $a=c/(4\pi en)$ and retain terms linear in $\delta\mathbf B$. The <electron magnetohydrodynamics> induction equation becomes
$$
\partial_t\delta\mathbf B=-aB_0\partial_z(\nabla\times\delta\mathbf B).
$$
Here the <curl> of $(\nabla\times\delta\mathbf B)\times B_0\hat{\mathbf z}$ reduces to $B_0\partial_z\nabla\times\delta\mathbf B$, since the background <magnetic field> is constant and the <divergence> of a curl vanishes. For a <plane wave> $\delta\mathbf B=\mathbf b\,e^{i(\mathbf k\cdot\mathbf r-\omega t)}$, the linear equation and the <solenoidal> constraint are
$$
\omega\mathbf b=i aB_0k_\parallel\,\mathbf k\times\mathbf b,\qquad \mathbf k\cdot\mathbf b=0.
$$
On the transverse plane, the operator $i\mathbf k\times$ has <eigenvalues> $\pm k$ by the <helicity decomposition of a transverse Fourier mode>. Equivalently, squaring the equation and using the <vector triple product> gives $\omega^2=a^2B_0^2k_\parallel^2k^2$. Since $aB_0=v_Ad_i$, where $v_A$ is the <Alfvén speed> and $d_i$ the <ion skin depth>,
$$
\boxed{\omega_\pm(\mathbf k)=\pm v_Ad_i k_\parallel k.}
$$
The two branches have opposite <circular polarizations>. The specified electron-only induction model describes a <whistler wave>; the paper calls these branches <kinetic Alfvén waves>. The following cascade calculation uses the specified model and its <dispersion relation>, without adding the pressure response needed for the usual kinetic Alfvén interpretation.
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