Are squares dense in C*-algebras?

Question

Suppose that we have a (complex) C*-algebra $A$. Let us say that $x\in A$ is square if $x=y^2$ for some $y\in A$.

Is the set of squares dense in $A$? What if $A$ is commutative?

Answer 1

Not in general, not even if $A$ is commutative. For instance, let $A=C(S^1)$ and let $z\in A$ be the inclusion map of the unit circle $S^1\to \mathbb{C}$. If $f\in A$ and $\|z-f\|<1$, then $f$ never vanishes, so can be considered as a map $f:S^1\to\mathbb{C}\setminus\{0\}$. The winding number of a map $S^1\to\mathbb{C}\setminus\{0\}$ is locally constant (with respect to the sup norm), so any $f$ sufficiently lose to $z$ has winding number $1$. Since the winding number of a product of two functions $S^1\to\mathbb{C}\setminus\{0\}$ is the sum of their winding numbers, the winding number of a square is always even. Thus no square can be close to $z$.

Answer 2

$B(H)$ is also a counterexample. Halmos and Lumer showed that squares are not dense in the invertible operators on an infinite dimensional Hilbert space in "Square roots of operators. II", 1954.

(In a commutative von Neumann algebra, all elements are squares.)

You might find interesting "On C*-algebras with the approximate $n$-th root property" by A. Chigogidze et al., 2005.