the nonnegative selfadjoint relation $H_r$ corresponding to the real part $(\st_H)_{\r}$ of the form $\st$ is given by $$\label{henilll}
 H_\r=T^*T.$$

It is straightforward to check that $\st$ in is a closed sectorial form as indicated, since $$\st_{r}[h,k]=(Th,Tk), \quad \st_{\I}[h,k]=(B Th,Tk).$$ Therefore, $|\st_{i}[h]|=|(B Th,Th)| \leq \|B\| \|Th\|^{2}=\|B\| \st_{r}[h]$, so that $\st$ is closed and sectorial with vertex at the origin and semi-angle $\alpha \leq \arctan \|B\|$. Moreover, since $T$ is closed, it is clear that $\st_{r}$ and hence $\st$ is closed.

Now let $\{h,h'\} \in T^*(I+iB)T$, then there exists $\varphi \in \sK$ such that $$\{h,\varphi\} \in T, \quad \{(I+iB)\varphi, h'\} \in T^*,$$ from which it follows that $$(h',h)=(\varphi, \varphi)+i(B \varphi, \varphi).$$ Consequently, one sees that $$| \IM (h'h) |=|(B\varphi, \varphi)| \leq \|B\|\,\|\varphi\|^2= \|B\| \,\RE (h',h),$$ which implies that $T^*(I+iB)T$ is a sectorial relation with vertex at the origin and semi-angle $\alpha \leq \arctan \|B\|$. Furthermore, observe that the above calculation also shows that $\mul T^*(I+iB)T=\mul T^*$.

To see that $T^*(I+iB)T$ is closed, let $\{h_n, h_n'\} \in T^*(I+iB)T$ converge to $\{h,h'\}$. Then there exist $\varphi_n \in \sK$ such that $$\{h_n,\varphi_n\} \in T, \quad \{(I+iB)\varphi_n, h_n'\} \in T^*,$$ and the identity $\RE(h_n',h_n)=\|\varphi_n\|^2$ shows that $(\varphi_n)$ is a Cauchy sequence in $\sK$, so that 