Given notion of “commutativity” in $C^{\ast}-$ algebra same as usual notion of commutativity? Let $A$ be a $C^{\ast}$-algebra. We say $A$ is “commutative“ if $ab^*c=cb^*a$ for all $a,b,c \in A$ and define “center” of $A$ as $$Z(A)= \{ v \in A : av^*c=cv^*a \forall a,c \in A \}$$

Are these notions of “commutativity“ and “center” same as usual notions of commutativity and center in $C^{\ast}$-algebras?

I tried by using that fact that every element in $C^{\ast}$-algebra is product of two elements but I dint get success. Any ideas?
 A: I think the answer is yes (EDIT: see below), but my experience with C* algebras is very very limited so I could be making a silly mistake! Hopefully someone will correct me if so. Anyway:
Suppose the C* algebra is commutative (in the weird sense). Choose an approximate identity $\{e_\lambda\}_{\lambda \in L}$. Then the net $\{a e_\lambda^* c\}_{\lambda \in L} = \{a e_\lambda c\}_{\lambda \in L}$ converges to $ac$. At the same time, this net equals $\{c e_\lambda^* a\} = \{c e_\lambda a\}$, which converges to $ca$. Since limits of nets are unique in any Banach algebra, $ac = ca$. Of course, if the C* algebra is commutative in the normal sense, then it is commutative in the weird sense. Thus, the two definitions of "commutative" agree.
Similarly, if $v$ is in the (weird) center, then $\{e_\lambda v^* c^*\}_{\lambda \in L} = \{c^* v^* e_\lambda\}_{\lambda \in L}$ converges to $(cv)^* = v^* c^* = c^* v^* = (vc)^*$, so $cv = vc$ for all $c$.
Edit: oops, in the second argument here I just showed that the weird center is a subset of the normal center! I don't see how to get the other containment.
