 Finite zero-energy vacuum for a single exponential superpotential

For $K=-\log (S+\bar{S})-3\log (T+\bar{T})$ and $W=a{e}^{-\alpha S}+b$ with $\alpha >0$, the potential is $|b+[1+\alpha (S+\bar{S})]a{e}^{-\alpha S}{|}^2/[(S+\bar{S})(T+\bar{T}{)}^3]$. With nonzero $a,b$, a finite zero-energy vacuum exists exactly when $0<|b/a|\le 2e^{-1/2}$. To prove this, set $x=\alpha \mathrm{Re}S>0$; the magnitude equation is $|b/a|=(1+2x)e^{-x}$, whose derivative changes sign at $x=1/2$ and whose maximum is $2e^{-1/2}$. The phase fixes the imaginary part of $S$. At a zero-energy solution the $T$ auxiliary field remains nonzero, so supersymmetry is broken and both real components of $T$ are flat. If $a=b=0$, the family is instead supersymmetric.