Consider the operator $T_{c}=-\frac{d}{dx^{2}}+ c x^{2}$ with $c\in C^{*}$, $Re(c)>0$ defined on its domain $D_{c}=\{u\in L^{2}; T_{c}(u)\in L^{2}\}$.

Put $c=a+ib$ avec $a>0$ et $b\in R$.

And $\Omega_{a+ib}=\{(2n+1)(a+ib), n\in N\}$. I have the following situation:

1. If $\xi\not\in\Omega_{a}$ then $ (T_{a}-\xi) : D_{a}\to L^2$ Is bijective , so $Im(T_{a}-\xi)=L^{2}$

1. If $\xi\not\in\Omega_{a+ib}$ then $ (T_{a+ib}-\xi) : D_{a+ib}\to L^2$ is one to one and $\overline{Im(T_{a+ib}-\xi)}=L^{2}$

How can I do to show that if $\xi\not\in\Omega_{a+ib}$ then $Im(T_{a+ib}-\xi)=L^{2}$.

Thank you very much.