Proposition 6.4.4.label Let $A$ be a unital $C^{*}$-algebra then
$\fall u\in A_{u}$, $\spec{A}{u}\suf\pd\cl{D_1(0)}=\curl{\lam\in\com:\abs{\lam}=1}$
$\fall h\in A_{h}$, $\spec{A}{h}\suf\R$
Proof. Let $u\in A_{u}$ then
\begin{align*}1=\norm{1}{}=\norm{u^*u}{}=\norm{u}{}^{2}\end{align*}
hence $\norm{u}{}=\norm{u^*}{}=1$. By Proposition 6.4.3, $\norm{u}{sp}=\norm{u^*}{sp}=1$. Since $u^{*}=u\inv$ we have
\begin{align*}\spec{A}{u\inv}=\curl{\lam\inv:\lam\in\spec{A}{u}}=\curl{\cl{\lam}:\lam\in\spec{A}{u}}\suf\cl{D_1(0)}\end{align*}
so
\begin{align*}\spec{A}{u}\suf\pd\cl{D_1(0)}\end{align*}
Let $h\in A_{h}$ and set $u\define \exp(ih)$ then $u^{*}=\exp(-ih)$ so $u\in A_{u}$. By Proposition 6.2.11
\begin{align*}\curl{\exp(i\lam):\lam\in\spec{A}{h}}=\spec{A}{u}\suf\pd\cl{D_1(0)}\end{align*}
which implies that $\spec{A}{h}\suf\R$$\square$