Center of the unit group $R^\times$ of a ring. Let $R$ be a (unital, noncommutative) ring. 
If $R^\times$ is the unit group of $R$ (the set of elements with a two-sided inverse), it is clear that $R^\times \cap Z(R) \subset Z(R^\times)$, where $Z$ means center (in the LHS, $Z(R)$ is the center of the ring $R$, in the RHS, $Z(R^\times)$ is the center of the group $R^\times$) : if an invertible element commutes with everyone, it commutes with every invertible element.
I don't see any reason why the reverse inclusion should be true, but I'm unable to find an example. So here's the question:
Is there a ring $R$ such that $R^\times \cap Z(R) \varsubsetneq Z(R^\times)$ ?
 A: OK, I hope that's not out of line, but I asked this question to a collegue of mine, and she gave me an answer worth posting here.
Consider the twisted polynomial ring $\mathbb C\langle X \rangle$, which is the ring of formal finite combinations
$$ \sum_{i=0}^n a_i X^i, \qquad (a_i)_{i=0}^n \in \mathbb C^{n+1}$$
with the multiplication
$$\left( \sum_{i=0}^n a_i X^i\right) \cdot \left(\sum_{j=0}^{m} b_j X^j \right) = \sum_{k=0}^{n+m} \left( \sum_{i+j=k} a_i \sigma^i(b_j)   \right) X^k,$$
(where $\sigma^i$ is the complex conjugation if $i$ is odd, and the identity if $i$ is even).
(That's example (1.7) in Lam's First Course in Noncommutative Rings)
One then checks easily that :


*

*The center of $\mathbb C\langle X \rangle$ is the (classical) real polynomial ring $\mathbb R[X]$;

*The unit group of $\mathbb C\langle X \rangle$ is simply $\mathbb C^\times$.
In brief, 
$$ Z(\mathbb C\langle X \rangle) \cap \mathbb C\langle X \rangle^\times = \mathbb R[X] \cap \mathbb C^\times = \mathbb R^\times \varsubsetneq Z(\mathbb C\langle X \rangle^\times) = Z(\mathbb C^\times) = \mathbb C^\times.$$
A follow-up to Alex's comment: it seems to me that something similar could be said on the group ring $\mathbb C[D_{\infty}]$ of the infinite dihedral group, bit I haven't been able to it precise (to be honest, I have no idea what the unit group is in that case—but it's certainly bigger than $\mathbb C^\times$).
