Let $T$ be a bounded linear operator on a real Hilbert space $H$ such that $\|T\|< \sin\left(\frac{\pi}{2n}\right)$ for some $n\in\mathbb{N}$. I have proved that $\langle x,(I-T)^kx\rangle\neq 0$ for all $x\in H\setminus\{0\}$ and $1\leq k\leq n$ using the law of cosines.
The proof of this statetment is based on the fact that the angle between $x$ and $(I-T)x$ is less than $\frac{\pi}{2n}$. In paricular, the angle between $x$ and $(I-T)^kx$ is less than $\frac{k\pi}{2n}\leq\frac{\pi}{2}$.
Now, if $H$ is a complex Hilbert space, I do not how to extend this proof because I used "the law of cosines" and this is true only for the real-case.
Can someone give me an idea? Thank you.
Every complex Hilbert space is a real Hilbert space, by defining the real inner product to just be the real part of the complex inner product. Note that in particular the real Hilbert space then has the same norm, so the operator norm of $T$ is the same whether you consider it as an operator on the real Hilbert space or on the complex Hilbert space. So, from the real Hilbert space case, you can deduce that $\operatorname{Re}\langle x,(I-T)^kx\rangle\neq 0$ and thus in particular $\langle x,(I-T)^kx\rangle\neq 0$.