Center of finitely generated $C^\ast$-algebra Let $\mathcal{A}$ be a $C^\ast$-algebra finitely generated by $n$ elements $\{a_1,\dots,a_n\}$.
Elements in the center $Z(\mathcal{A})$ commute with each of the generators
$$Z(\mathcal{A})\subset\{a\in\mathcal{A}:aa_i=a_ia\}$$
What about vice versa:
$$\{a\in\mathcal{A}:aa_i=a_ia\}\subset Z(\mathcal{A})?$$
If $a\in \mathcal{A}$ is non-central does  $aa_i\neq a_ia$ for some generator $a_i$?
It would seem that a proof by contradiction might do it.
 A: Let $a \in \mathcal A$ be an element commuting with all the generators. The centralizer of $a$ is defined as :
$$Z_\mathcal A(a) = \{b\in A | ab = ba\}$$
It a sub-$C^*$-algebra  of $\mathcal A$ (in fact it is the largest sub-$C^*$-algebra containing $a$ in it's center). Since $Z_\mathcal A(a)$ contains the generators of $\mathcal A$, it is equal to $\mathcal A$ and $a \in Z(\mathcal A)$.
Therefore :
$$Z (\mathcal A) = \{ a\in \mathcal A|\forall i\in \{1,\ldots,n \}, a_i a= aa_i\}$$
A: I have accepted another answer while I was writing this answer. I am not sure about this polynomial *-algebra $B$ being dense in $\mathcal{A}$.
The generators generate a polynomial *-algebra $B$ and if $a$ commutes with all the generators of $\mathcal{A}$ then it commutes with everything in $B$.
Take $(b_\lambda)\subset B$ converging to $b\in \mathcal{A}$ such that $ba-ab\neq 0$. But as a Banach algebra the multiplication and addition in $\mathcal{A}$ is continuous and so:
$$ab_\lambda-b_\lambda a\to ab-ba.$$
But $(ab_\lambda-b_\lambda a)$ is constant equal to zero.
