= Solution
To find the <threefold phase-locked equilibria>, write $C=Re^{i\theta}$ with $R>0$; separating real and imaginary parts gives
$$
\boxed{R_T=R(\mu-R^2)+R^2\cos3\theta,\qquad
\theta_T=-\omega-R\sin3\theta.}
$$
A steady state satisfies
$$
\cos3\theta_0=\frac{R_0^2-\mu}{R_0},\qquad
\sin3\theta_0=-\frac{\omega}{R_0},
$$
so
$$
\boxed{(\mu-R_0^2)^2+\omega^2=R_0^2.}
$$
Writing $y=R_0^2$ and $D=1+4\mu-4\omega^2$, the amplitude branches are
$$
\boxed{y_\pm=\mu+\frac12\pm\frac12\sqrt D.}
$$
There are two distinct positive roots for \b[$\mu>\omega^2-1/4$], except at $(\mu,\omega)=(0,0)$, where $y_-=0$ and only the upper root is nonzero. Indeed their sum is positive in this range and their product is $\mu^2+\omega^2$. Below the fold condition there are none. At equality, there is one positive repeated root $y=\omega^2+1/4$, the <saddle-node bifurcation> limit.
For each positive amplitude, the sine and cosine determine $3\theta_0$ modulo $2\pi$, giving three phases separated by $2\pi/3$. Thus the source's “two states” means \b[two amplitude branches modulo the threefold spatial symmetry]. Generically there are six nonzero complex equilibria, three on each branch, not literally two.
The polar <Jacobian matrix> at an equilibrium is
$$
J=\begin{pmatrix}
-\mu-y&3\omega R_0\\
\omega/R_0&3\mu-3y
\end{pmatrix},
\quad
\operatorname{tr}J=2\mu-4y,\quad
\det J=3y(2y-2\mu-1).
$$
On the lower branch, $\det J=-3y_-\sqrt D<0$, so it is a <saddle equilibrium> and unstable. On the upper branch, $\det J=3y_+\sqrt D>0$ and
$$
\operatorname{tr}J=-2(\mu+1+\sqrt D)<0,
$$
because existence implies $\mu>-1/4$. Therefore \b[every upper-branch equilibrium is <asymptotically stable>, and every lower-branch equilibrium is a <saddle equilibrium>], away from the degenerate endpoints. The <eigenvalues> are unchanged by the smooth polar coordinate transformation at $R_0>0$. The origin, not covered by those coordinates, has <eigenvalues> $\mu\pm i\omega$ and is stable for $\mu<0$ and unstable for $\mu>0$.
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