\section{Unitary Elements}
\label{section:unitary-c-star}

\begin{definition}[Unitary]
\label{definition:unitary-element}
    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$. 
\end{definition}

\begin{lemma}
\label{lemma:unitary-unit}
    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$. 
\end{lemma}
\begin{proof}
    $\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$. 
\end{proof}

\begin{definition}[Unitarily Equivalent]
\label{definition:unitary-equivalent}
    Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$. 
\end{definition}

\begin{lemma}
\label{lemma:unitary-equivalent-same-stuff}
    Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then:
    \begin{enumerate}
        \item $\norm{x}_A = \norm{y}_A$. 
        \item $\sigma_A(x) = \sigma_A(y)$. 
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): Let $u \in A$ be unitary such that $x = uyu^*$, then $\norm{x}_A \le \norm{u}_A\norm{x}_A\normn{u^*}_A$. By \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, so $\norm{x}_A \le \norm{y}_A$. As the argument is symmetric, $\norm{x}_A = \norm{y}_A$. 

    (2): Let $\lambda \in \complex$, then
    \[
    u(y - \lambda)u^* = uyu^* - \lambda uu^* = uyu^* - \lambda = x - \lambda
    \]

    Since $u, u^* \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_A(x) = \sigma_A(y)$. 
\end{proof}

\begin{proposition}
\label{proposition:unitary-spectrum}
    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 8.2]{Zhu}}}. ]
    By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus
    \[
    \bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)}
    \] 

    by (4) of \autoref{proposition:c-star-algebra-gymnastics}. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
\end{proof}



