If $A$ is a C\*-algebra and $n$ is a normal element of $A$, then we have: (By Gelfand duality for example.)


$\operatorname{spec}( |N| ) = | \operatorname{spec}(N) | := \left\{ | \lambda | ; \lambda \in \operatorname{spec}(N) \right\}$


where we define: $|n|:=(n^\*n)^{1/2}$. My question is, does the converse also hold?


That is if $a\in A$ and for $r>0$:


$\left\{ r e^{it} : t \in [0, 2\pi[ \right\} \cap \operatorname{spec}(a)$ is not empty if and only if $r\in \operatorname{spec}(|a|)$


implies that $a$ is normal. (Possibly some exceptions made for the zero-element) Or bluntly speaking if the mapping $a$ to $a^\*a$ does not create any "new" (or removes any "old") elements in the spectrum then $a$ is normal.


For example if $e$ is an idempotent in $A$, then $e$ is a projection if and only if $||e||=1$. Hence if $e$ is a non-projection idempotent we have $\left\{0,1\right\} = \operatorname{spec}(e) \subsetneq \operatorname{spec}(|e|)$, since $||e||>1$ and by the spectral radius $\operatorname{spec}(e^\*e)$ contains an element strictly bigger that one. 


Clearly if p is a projection, then p=|p|.



Normal operators and it's spectrum in C*-algebras
You can take a normal operator $N$ with spectrum filling the unit disk and take its tensor product with any operator $T$ of norm less than $1$ thus effectively hiding the non-normal component's contribution to the spectrum in both $A$ and $|A|$.


