How can I calculate the F-terms for chiral superfields in a no-scale model?

[ChrisVer]

In case you have the Kahler and super- potential $K,W$:

$K(T,S,C) = -log (S +S^{*}) -3 log ( T+ T^{*} - C C^{*})$
$W(T,S,C)= C^{3} + d e^{-aS} +b$

with $T,S,C$ chiral super fields, $b,d$ complex numbers and $a>0$.
I tried to calculate the local F-terms arising from this. The local F-terms for the i-th chiral superfield are given by:

$F_{i}= D_{i}W = K_{i}W + W_{i}$

where in the rhs the index i denotes the derivative wrt to the i-th field. eg $W_{S}=\frac{\partial W}{\partial S}$

However I'm having a slight problem with the particular derivative. See what I mean...taking it:

$F_{S}= K_{S} W + W_{S} = - \frac{C^{3} + d e^{-aS} +b}{S+S^{*}} -d a e^{-aS}$

correct?
On the other hand, if I try to work with the covariant derivative wrt to the conjugate fields:

$F^{*}_{S}= D_{S^{*}} W^{*} = K_{S^{*}} W^{*} + W_{S^{*}}$
I don't get the complex conjugate of the above. Because in this case
$W_{S^{*}}=0$
and so:

$F^{*}_{S}= - \frac{(C^{3} + d e^{-aS} +b)^{*}}{S+S^{*}}$

what's the problem?

 

[ChrisVer]

Ah found the mistake... again by writing in LaTeX it became obvious- In the F* equation I needed the W* derivative as the second term...
However
In the case of $F$ let's say... How can I see if its module squared is simultaneously zero or not?

$|F_{T}|^{2}= \frac{9}{(T+T^{*} - CC^{*})^{2}} |C^{3}+ d e^{-aS} +b |^{2}$

$|F_{S}|^{2}= | \frac{C^{3} + d e^{-aS} +b}{S+S^{*}} + d a e^{-aS}|^{2}$

$|F_{C}|^{2}= | \frac{3 C^{*} (C^{3}+d e^{-aS} +b)}{T+T^{*}-CC^{*}} +3 C^{2}|^{2}$