I have $V$ an inner product space above $\mathbb{C}$ and a linear operator $T$ such as $T=T^{*}$ on $V$ (surjective).
I need to prove that the operators $I+iT$, $I-iT$ are invertible and that $U=\left(I-iT\right)\left(I+iT\right)^{-1}$ satisfies the condition $U^{*}=U^{-1}$, by $*$ I mean adjoint operator.
I know that: $$\left\|v+iT(v)\right\|=\left\|v-iT(v)\right\|$$ $$v=0 \iff v+iT(v)=0$$
Will be thankful for your help, thank you.
Edit:
I shouldn't use the rule $(ST)^*=T^*S^*$ or any other rules without proving it, except the definition: $<T(u),v>=<u,T^*(v)>$ and also I found that $(I+iT)^*=I-iT$.
Since $T^* = T$ for any $v \in V$ we have $\langle Tv,v\rangle \in \Bbb{R}$ so $$\|(I+iT)v\|^2\|v\|^2 \ge |\langle (I+iT)v,v\rangle|^2 = |\|v\|^2 + i\langle Tv,v\rangle|^2 =\|v\|^4 + |\langle Tv,v\rangle|^2 \ge \|v\|^4$$ and therefore $\|(I+iT)v\| \ge \|v\|$ so the operator $I+iT$ is bounded from below. In particular it is injective and has closed range. Similarly we show that $I+iT$ is bounded from below. We now have $$V = \overline{R(I+iT)} \oplus N((I+iT)^*) = \overline{R(I+iT)} \oplus N(I-iT) = R(I+iT) \oplus \{0\} = R(I+iT)$$ so $I+iT$ is surjective and injective, hence invertible.
As for the second claim, I believe it is false unless $T=0$. Namely, if $U = (I+iT)(I-iT) = I+T^2$ is unitary, then its spectrum is contained in the unit sphere so
$$1+\sigma(T)^2 = \sigma(U) \subseteq \mathbb{S}^1$$ and $\sigma(T)^2 \subseteq [0, +\infty\rangle$ so it follows $\sigma(T) = 0$ and from $T^*=T$ we get $T = 0$.