= Solution
Include temporal dependence $e^{\sigma t}$ and put $p=k^2+\ell^2$. The two Fourier amplitudes obey
$$
(\sigma+\eta p)A=\alpha B,\qquad
(\sigma+\eta p)B=(ik\Omega+\alpha p)A.
$$
Consequently the <dispersion relation> and its branches are
$$
\boxed{(\sigma+\eta p)^2=\alpha^2p+i\alpha\Omega k,\qquad
\sigma=-\eta p\pm\sqrt{\alpha^2p+i\alpha\Omega k}.}
$$
For $\eta>0$ and $\ell\ne0$, the branch with positive real square root is the potentially unstable one. At marginal stability set $\sigma=i\omega$; comparing real and imaginary parts gives
$$
\eta^2p^2-\omega^2=\alpha^2p,\qquad 2\eta p\omega=\alpha\Omega k.
$$
Eliminating $\omega$ produces the <alpha-squared Omega dynamo threshold>
$$
\boxed{\alpha_m^2(k)=\frac{4\eta^4(k^2+\ell^2)^4}{4\eta^2(k^2+\ell^2)^3+\Omega^2k^2},\qquad
\omega=\frac{\alpha_m\Omega k}{2\eta(k^2+\ell^2)}.}
$$
Either sign of $\alpha_m$ is allowed; it changes the wave's propagation sign. Larger $|\alpha|$ increases the real part of the growing branch through zero.
To minimize the threshold differentiate with respect to $x=k^2$. The sign of this derivative is the sign of
$$
F(x)=4\eta^2(x+\ell^2)^3+\Omega^2(3x-\ell^2).
$$
Its derivative is strictly positive for $x\geq0$, so there is at most one interior minimum. Its value at zero changes sign at
$$
\boxed{\Omega_0^2=4\eta^2\ell^4.}
$$
For $\Omega^2\leq\Omega_0^2$, $F(x)\geq0$ and the minimum is $\boxed{k_c=0}$, with $|\alpha_c|=\eta|\ell|$. For stronger shear, the minimizing positive wavenumber is the unique root
$$
\boxed{4\eta^2(k_c^2+\ell^2)^3+\Omega^2(3k_c^2-\ell^2)=0,\qquad 0<k_c^2<\ell^2/3.}
$$
An explicit form follows by setting $s=\Omega^2/(4\eta^2\ell^4)>1$ and $y=1+k_c^2/\ell^2$. The cubic becomes $y^3+3sy-4s=0$, with real solution
$$
\boxed{k_c=|\ell|\sqrt{2\sqrt s\,\sinh\left[\frac13\operatorname{arsinh}\left(\frac2{\sqrt s}\right)\right]-1}.}
$$
The negative wavenumber $-k_c$ has the same threshold. In the strong-shear limit $k_c\to|\ell|/\sqrt3$.
The threshold curve is even in $k$ and behaves as $\eta|k|$ at large $|k|$. Below the shear threshold it increases monotonically away from its central minimum. Above the threshold, zero is a local maximum flanked by the two nonzero minima. At equality the central minimum has vanishing quadratic curvature but remains a minimum. These two cases are sketched from the derived formula below. If $\ell=0$, the zero-wavenumber uniform/gauge mode is degenerate and the positive-wavenumber threshold tends to zero as $k\to0$; the nonzero-$\ell$ minimum formula must not be used as a nondegenerate onset criterion there.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-65-thresholds.png]
{title=Marginal alpha-squared Omega dynamo thresholds below and above the critical shear, with the minimizing wavenumbers}
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