Let $T: \operatorname{dom}(T) \rightarrow H$ be self-adjoint, then $U(T):=(T+i)(T-i)^{-1}$ is defined and unitary( this is clear to me). Furthermore, we have that $\sigma(U(T)):= \overline{\{ t; \exists y \in \sigma(T): t = \frac{y+i}{y-i}\}}$, i.e. $1 \notin \sigma(U(T)).$ I guess that there is a nice trick to show this that I currently just don't see. I hope that anybody here knows how to do this.
If anything is unclear, please let me know.
This is the brute force approach: $$ \begin{align} (T+iI)(T-iI)^{-1}-\lambda I & = (T+iI)(T-iI)^{-1}-\lambda(T-iI)(T-iI)^{-1} \\ & = (T+iI-\lambda T+\lambda iI)(T-iI)^{-1} \\ & = ((1-\lambda)T+i(1+\lambda)I)(T-iI)^{-1} \\ & = (1-\lambda)(T+i\frac{1+\lambda}{1-\lambda})(T-iI)^{-1} \end{align} $$ The above is invertible iff $-i(1+\lambda)/(1-\lambda) \notin\sigma(T)$. If $\mu \in \sigma(T)$, then $$ -i\frac{1+\lambda}{1-\lambda} = \mu \iff \lambda = \frac{\mu+i}{\mu-i}. $$ There is a special case to be considered, but I'll leave that.