In a unit C*algebra $A$, let $x\in A$ is normal, show that
(1) if $\sigma(x)\subset \mathbb{T}=\{\lambda \in \mathbb{C}:|\lambda|= 1\}$, then $x$ is an untiary element.
(2) if $\sigma(x)\subset \mathbb{R}$, then $x$ is a self-adjoint element.
(3) give examples to show that if $x$ is not normal, then (1) and (2) is not right.
We all know theorem the sepctrum of unitary element in circle, and the sepctrum of self-adjoint element in real number set, but i couldn't find any help in proof of this theorem.
My attempt: for (1), only to show $xx^*=1$, because $xx^*$ is a positive element, so $xx^*\le 1 \Leftrightarrow \|xx^*\|\le 1$, as the sepctrum of $x$ in circle , then i got $xx^*\le 1$, but i can't prove the other side.
For(2) and (3) i have no idea.
Any help will be appreciated.
So this really comes down to using functional calculus. The element is normal, so we can identify $C^*(x) \simeq C(\sigma(x))$, where the element $x$ is the identity map of $\sigma(x)$: $x(z) = z$. But in $C(\sigma(x))$, the spectrum of a function $f$ is the image of $f$. From here you should be able to show the results (a function is self-adjoint if and only if its real-valued, and unitary if and only if it is $\mathbb{T}$-valued). Note this also gives that a normal element is positive if and only if its spectrum is contained in $[0,\infty)$.
For an example of a non-normal element, consider the upper triangular matrix: $$ T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. $$ Its clearly not self-adjoint nor unitary, and you can check that the spectrum is $\{1\}$.