= Solution
{c}
In a parabolic conduction valley, $\mathbf v=\hbar M^{-1}\mathbf k$. The <Lorentz force> on an <electron> gives $M\dot{\mathbf v}=-e\mathbf v\times\mathbf B=e\mathbf B\times\mathbf v$. In the principal axes of the stated <effective-mass tensor>,
$$
\begin{pmatrix}\dot v_x\\\dot v_y\\\dot v_z\end{pmatrix}
=e\begin{pmatrix}0&-B_z/m_1&B_y/m_1\\B_z/m_1&0&-B_x/m_1\\-B_y/m_2&B_x/m_2&0\end{pmatrix}
\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}.
$$
Call the matrix on the right $A$. Its characteristic polynomial is
$$
\det(\lambda I-A)=\lambda\left[\lambda^2+e^2\left(\frac{B_z^2}{m_1^2}+\frac{B_x^2+B_y^2}{m_1m_2}\right)\right].
$$
One mode has zero frequency, corresponding to constant drift along the field; the other two have eigenvalues $\pm i\omega$. Thus
$$
\omega^2=e^2\frac{m_1(B_x^2+B_y^2)+m_2B_z^2}{m_1^2m_2},
\qquad
\boxed{\omega=\frac{e|\mathbf B|}{m_{\mathrm{cr}}},\quad m_{\mathrm{cr}}=|\mathbf B|\sqrt{\frac{m_1^2m_2}{m_1B_x^2+m_1B_y^2+m_2B_z^2}}.}
$$
More generally the <ellipsoidal cyclotron mass> is $m_{\mathrm{cr}}^2=\det M/(\widehat{\mathbf B}^TM\widehat{\mathbf B})$. The tensor axes must be rotated for the differently oriented valleys.
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