I am trying to do the proof of the following statement:
We work with a unital $C^\ast$-algebra $\mathscr{A}$ where $B\in\mathscr{A}$ is a normal element.
Statement: Show that $B$ is unitary if and only if $σ(B) \subseteq S^1$ where $S^1=\{z\in\mathbb{C}:|z|= 1\}$.
I managed to show the direction where we say $B$ is unitary. However how would one show the other direction?
Attempt: We want to show $\sigma(B)\subseteq \{z\in\mathbb{C}:|z|= 1\}$. Let $B-\lambda I$ then we want to show that this is invertible for all $\lambda$, where $|\lambda|<1$. However one have that $B$ is invertible so we will now see that $B-\lambda I$ is invertible for $0<|\lambda|<1$ That is, $-\lambda B(B^{-1}-\frac{1}{\lambda}I)=(B-\lambda I)$. From there one see that $\lambda\neq 0$. Now since $B^{-1}-\frac{1}{\lambda}I$ is invertible so is $(B-\lambda I)$. This gives us that $\sigma(B)\subseteq\{z\in\mathbb{C}:|\lambda|=1\}$.
But the other way is not clear for me.
The usual trick is to assume first that $\mathcal A$ is commutative. You can then apply that to the commutative sub-aglebra $C^*(B)$, using the fact that $\sigma(B)= \sigma_{C^*(B)}(B)$.
Suppose that $\mathcal A$ is commutative, and $$\Gamma \colon \mathcal A \to C(Δ) $$ is the Gelfand transform, where $\Delta$ is the maximal ideal space of $\mathcal A$. If $B \in \mathcal A$ is unitary, that is, $B^{-1}=B^*$, then $$ |\Gamma(B)| = \Gamma(B) \Gamma(B)^* = \Gamma(B) \Gamma(B)^{-1} =1$$ and so $\sigma(x) = range (\Gamma(x)) \subset S^1$. Conversely, if $ range(\Gamma(B))=\sigma(B) \subset S^1 $ then $$\Gamma(B) \Gamma(B^*) = \Gamma(B^*) \Gamma(B) =1 $$ and so $B^*B=BB^*=1$.