$T = A+i B$ is normal operator, if $\Vert T x \Vert^2 = \Vert Ax \Vert^2 + \Vert Bx \Vert^2$.

Question

Let $T = A+iB$, where $A$ and $B$ are self-adjoint and bounded. Prove that if $\Vert T x\Vert^2 = \Vert Ax\Vert^2 + \Vert Bx\Vert^2$, then $T$ is normal operator.

So it's obvious that $\Vert T^* x \Vert^2 = \Vert Tx \Vert^2$, but how do I prove that $T^* T = T T^*$. I tried to use this equation $$\langle Tx, Ty \rangle = \frac{1}{4} \left( \Vert T(x+y) \Vert^2 - \Vert T(x-y) \Vert^2 + i \Vert T(x+iy) \Vert - i \Vert T(x-iy) \Vert^2\right),$$ but I'm stuck on the calculations. Is this equation even correct?

Answer 1

$\langle (T^{*}T-TT^{*})x, x \rangle =\|Tx||^{2}-\|T^{*}x||^{2}=0$. In a complex Hilbert space this is enough to conclude that $T^{*}T-TT^{*}=0$.