= Solution
Take the outward <wavenumber> $k>0$; the absolute value in the source makes the two propagation directions equivalent. The <phase velocity> of the <gravity-capillary wave> obeys
$$
c_p^2=\frac{\omega^2}{k^2}=\frac gk+\frac\sigma\rho k.
$$
Differentiate with respect to $k$:
$$
\frac{d(c_p^2)}{dk}=-\frac g{k^2}+\frac\sigma\rho=0
\quad\Rightarrow\quad k_*=\sqrt{\frac{\rho g}{\sigma}}.
$$
The second derivative is positive, and $c_p\to\infty$ at both ends of the positive-$k$ range, so this is the global minimum. The <minimum phase speed of a gravity-capillary wave> is
$$
\boxed{c_{p,\min}=\left(\frac{4g\sigma}{\rho}\right)^{1/4}.}
$$
For $g=9.81\ \mathrm{m\,s^{-2}}$ and the supplied <surface tension> and <mass density>,
$$
\boxed{c_{p,\min}=0.231\ \mathrm{m\,s^{-1}}.}
$$
The corresponding <wavenumber> is $k_*\simeq369\ \mathrm{m^{-1}}$, giving <wavelength> $\lambda_*=2\pi/k_*\simeq1.70\ \mathrm{cm}$. At this minimum, $c_g=d\omega/dk=c_p+k\,dc_p/dk=c_p$. Thus the <group velocity> and <phase velocity> agree for this particular ripple, although they differ for general <gravity-capillary waves>.
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