I have a reasoning that is wrong but I don't understand why.
Suppose we have a unital $C^*$-algebra $A$ and $v \in A$ is a proper isometry, so $v^*v = 1$ but $v v^* \neq 1$. Take the function $f: \mathbb{C} \to \mathbb{C}: y \mapsto \overline{y}y$. Then by functional calculus the $f(v) = v^* v = 1$. This means that by the spectral mapping theorem we get $\sigma(f(v)) = \sigma(1) = 1$ and $1 = \sigma(f(v)) = f(\sigma(v))$. This shows that for every element in the spectrum of $v$ we have $$ \forall y \in \sigma(v): f(y) \in f(\sigma(v)) = 1, $$ so $\forall y \in \sigma(v): |y| = 1$. This shows that $\sigma(v)$ is a subset of the circle.
Of course, this calculation is wrong, because this would show that every isometry is a unitary. However, I can't seem to find the mistake.
The reason this does not work is because a proper isometry is not a normal element, so functional calculus is not applicable.