Gauge invariant supersymmetric transformations Given the following action
$$
\mathcal{S}=\int{d^4x\;d^2\theta\;d^2\bar{\theta}\left(\bar{Q}_+e^{2V}Q_++\bar{Q}_-e^{-2V}Q_--2\xi V\right)}+\int{d^4x\;d^2\theta\left(mQ_-Q_++\frac{\tau}{16\pi i}W^\alpha W_\alpha\right)}+h.c.
$$
Where $\bar{Q}$ is the hermitian conjugated $Q$, $\tau = \frac{\theta}{2\pi}+i\frac{4\pi}{g^2}$ the gauge coupling, $\xi$ the Fayet-Iliopoulos parameter, $W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V$ is the gaugino superfield and $V$ the abelian vector superfield.
How can I show that the action is invariant under 
$$
Q_+\rightarrow e^{i\Lambda}Q_+ \;\\
Q_-\rightarrow e^{-i\Lambda}Q_- \;\\
V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}
$$
When I try the transformation on the first term I get the following:
$$
\bar{Q}_+e^{2V}Q_+\rightarrow \bar{Q}_+e^{2V}Q_+ e^{Im(\Lambda)}
$$
I have a similar problem with the next term
Edit:
$Q_+$ is a chiral superfield with charge 1
$Q_-$ is a chiral superfield with charge -1
 A: Let's check invariant of first term.
Using transformation laws and fact, that transformation are abelian:
$$Q_+\rightarrow e^{i\Lambda}Q_+ \Rightarrow \bar{Q}_+\rightarrow \bar{Q}_+ e^{-i\bar{\Lambda}}$$
$$
V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}
\Rightarrow
e^{2V}\rightarrow e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda} $$
We obtain:
$$
\bar{Q}_+ e^{2V} Q_+ \to  \bar{Q}_+ e^{-i\bar{\Lambda}} e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda} e^{i\Lambda}Q_+ = \bar{Q}_+ e^{2V} Q_+
$$
For gauge multiplet term:
$$
W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V
\to
-\frac{1}{4}\bar{D}^2D_\alpha V 
-\frac{i}{8}\bar{D}^2D_\alpha\Lambda +\frac{i}{8}\bar{D}^2D_\alpha\bar{\Lambda}
$$
Now using that $\Lambda$ is chiral, $\bar{\Lambda}$ antichiral and $[\bar{D}^2,D_\alpha]=0$:
$$
D_\alpha\bar{\Lambda} =0
$$
$$
\bar{D}^2D_\alpha\Lambda = D_\alpha\bar{D}^2\Lambda = 0
$$
So $W_\alpha$ is gauge invariant quantity.
Fayet-Iliopoulos term:
$$
\int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi V\rightarrow \int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi \left(V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}\right)
$$
Using that $\int d\theta_\alpha = D_\alpha$  and $\int d\theta_{\dot{\alpha}} = \bar{D}_{\dot{\alpha}}$ (up to the boundary term) and chirality properties,  we immediately see, that FI term is gauge invariant.
The same idea works with other terms.
