 Solution

Set $x=T+\bar{T}>0$, the domain in which the Kähler potential is defined, and use Planck units. The nonzero entries of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areDenote $W=C^3+B$ and $S=3C^2+\bar{C}W$. The supergravity F-term potential isHere the $T$ contribution is exactly $3|W{|}^2$, cancelling the negative term. This is the no-scale supergravity identity, givingThe cancellation is important: replacing the covariant derivatives by ordinary derivatives would miss both the matter dependence and the possibility of zero vacuum energy with supersymmetry breaking.

In the convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$, the chiral supergravity auxiliary fields areAn overall auxiliary phase or sign convention does not change their vanishing conditions. These are upper-index fields, including the inverse Kähler metric, rather than just the covariant derivatives.

Since $V$ is nonnegative and vanishes at $C=0$, its minimum energy isThere are additional minima that should not be discarded by examining only real $C$. For $C=\rho e^{i\alpha }\ne 0$, the condition $S=0$ becomesIf $B\ne 0$, the magnitude equationhas exactly one positive root: its left side is strictly increasing from zero to infinity. There are three phases, differing by $2\pi /3$, determined by $e^{3i\alpha }=-B/|B|$. Thus for nonzero $B$ the finite zero-energy matter vacua consist of $C=0$ and these three nonzero phase-related values, each with arbitrary $T$ in the allowed half-plane. At a nonzero matter vacuum,At $C=0$, $W=B$.

For $B\ne 0$, every such minimum has $F^C=0$ but $F^T\ne 0$, so supersymmetry is spontaneously broken despite zero vacuum energy. Indeed no finite point anywhere is fully supersymmetric: $D_{T}W=0$ would force $W=0$, after which $D_{C}W=3C^2=0$ forces $C=0$ and then $B=0$. This proves breaking for the specified nonzero-constant case without relying just on the positivity of the potential.

For the allowed exceptional value (B=0), the requested breaking statement is false. Then $S=C^2(3+|C{|}^2)$ vanishes only at $C=0$, where $W=F^T=F^C=0$. Every finite allowed $T$ gives an unbroken supersymmetric vacuum with zero energy. This is an explicit counterexample to unqualified breaking for an arbitrary complex $B$.

Both real components of $T$ are flat directions of a scalar potential along every minimum: the imaginary component never appears in $V$, and when $S=0$ the entire $x$ dependence is multiplied by zero. The matter vacua are isolated in the $C$ plane; there is no continuous matter vacuum direction. At $B=0$ the potential begins quartically in $C$, so zero quadratic mass is not a flat vacuum manifold. Away from $S=0$, the real $T$ direction instead has $V\propto x^{-3}$ and a decompactification-type runaway towards zero as $x\to \infty$, not a further finite positive-energy minimum.

The gravitino mass from a superpotential isAt the $C=0$ branch it is $|B|/x^{3/2}$; at each nonzero branch it is $3\rho e^{{\rho }^2/2}/x^{3/2}$. Its value is not fixed because the $T$ modulus is flat. For $B=0$ it vanishes at the supersymmetric minimum. For nonzero $B$, the super-Higgs mechanism absorbs the Goldstino associated with $F^T$ into the massive gravitino. The zero-energy relation is explicitly satisfied:which balances the auxiliary contribution against the gravitational negative term.