Selfadjoint and continuous operator on a complex Hilbert space Let $T\colon H\to H$ be a selfadjoint continuous operator on a complex Hilbert space. Show: $$ \lVert (T\pm i\mbox{Id})x\rVert^2=\lVert Tx\rVert^2+\lVert x\rVert^2~\forall~x\in H. $$ \-- How can I show that? I started with $$ \lVert (T\pm i\mbox{Id})x\rVert^2=\langle Tx\pm ix,Tx\pm ix\rangle. $$

You have $$\| T(x) + ix\|^2 = \|T(x)\|^2 + \|x\|^2 + \langle T(x), ix\rangle + \langle ix, T(x)\rangle = \|T(x)\|^2 + \|x\|^2 - i\langle T(x), x\rangle + i \langle x, T(x)\rangle$$

Now we know that $\langle x, T(x)\rangle = \langle T(x), x\rangle$, so our conclusion follows forthwith.