= Solution
Use periodic boundary conditions on the square crystal. Each allowed <wavevector> occupies area $(2\pi)^2/A$ in two-dimensional $k$-space. Including the two <phonon> polarizations, the number of modes in the disk $|k|\leq K$ is
$$
\mathcal N(K)
=2\frac{A}{(2\pi)^2}\pi K^2
=\frac{AK^2}{2\pi}.
$$
The <phonon dispersion relation> $\omega=C|k|^\alpha$ gives $K=(\omega/C)^{1/\alpha}$, so
$$
\mathcal N(\omega)=\frac{A}{2\pi}
\left(\frac\omega C\right)^{2/\alpha}.
$$
Differentiating the cumulative mode count produces the <density of states for a two-dimensional power-law phonon dispersion>:
$$
\boxed{
g(\omega)=B\omega^b,
\qquad
B=\frac{A}{\pi\alpha C^{2/\alpha}},
\qquad
b=\frac2\alpha-1.}
$$
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