= Solution
In the <Polonyi model>, the <Kähler metric> is $K_{z\bar z}=1$. The relevant <Kähler covariant derivative of a superpotential> is
$$
D_zW=m^2\{1+\bar z(z+\beta)\}.
$$
The <supergravity auxiliary field> is $F^z=-e^{|z|^2/2}\overline{D_zW}$, up to an irrelevant common phase convention. Thus a constant vacuum preserves <supersymmetry> exactly when this <auxiliary field> vanishes. For \b[$m=0$], the <superpotential> and <scalar potential> vanish identically, and every constant scalar value is a <supersymmetric vacuum>.
For $m\ne0$, put $\langle z\rangle=x+iy$. Its <supersymmetry> condition becomes
$$
1+x^2+y^2+\beta x-i\beta y=0.
$$
Since $\beta>0$, it requires $y=0$ and $x^2+\beta x+1=0$. Hence the <Polonyi supersymmetry branches> are
$$
\boxed{\text{unbroken supersymmetry: }\beta\geq2,\quad
\langle z\rangle=\frac{-\beta\pm\sqrt{\beta^2-4}}2,\quad m\ne0.}
$$
At these points $W\ne0$, so the <supergravity F-term potential> is negative, $V=-3e^K|W|^2$: these are supersymmetric <Anti-de Sitter spacetime> vacua, not zero-energy ones. The condition $D_zW=0$ also makes them stationary, as follows by differentiating the <supergravity F-term potential>.
For $0<\beta<2$ and $m\ne0$, the <auxiliary field> cannot vanish anywhere, so any vacuum has <supersymmetry breaking>. For $\beta\geq2$, a stationary vacuum at any other scalar value still breaks <supersymmetry>; the parameter condition alone does not determine which vacuum is selected. In particular, a nontrivial zero-energy vacuum cannot preserve <supersymmetry>: $D_zW=0$ and $V=0$ would also imply $W=0$, whereas $W=0$ gives $z=-\beta$ and $D_zW=m^2\ne0$.
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