Let $u, v$ be unitaries in a unital $C^*$-algebra satisfying $uv=e^{2\pi i \theta}vu$ where $\theta$ is irrational (so $\{e^{2 \pi i n \theta} : n \in \mathbb{Z} \}$ is dense in $\mathbb{T}$). For fixed $z \in \mathbb{T}$, finitely supported $a: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{C}$, and $\varepsilon >0$, show that
$||\sum_{m,n} a_{m,n} z^m v^m u^n|| \leq ||\sum_{m,n} a_{m,n} v^m u^n|| + \varepsilon$
There was a hint about being able to find $k$ such that $|e^{2 \pi i k \theta} - z|$ can be made small as desired.
So I thought to do something like
$||\sum_{m,n} a_{m,n} ((z^m - e^{2 \pi i k_m \theta}) + e^{2 \pi i k_m \theta}) v^m u^n|| $
$\leq \sum_{m,n} |a_{m,n}| |z^m - e^{2 \pi i k_m \theta}| + ||\sum_{m,n} a_{m,n} e^{2 \pi i k_m \theta} v^m u^n||$
choosing the $k_m$ such that the left sum is less than $\varepsilon$. Then if the right sum can be shown to be less than or equal to $||\sum_{m,n} a_{m,n} v^m u^n||$, that would be it. To do that I was thinking of using the commuting relation for $u, v$ to cancel out the $e^{2 \pi i k_m \theta}$, but then the $v, u$ are permuted from the desired $v^m u^n$ form...
With $u, v$ as above, consider $B = C^*(u,v)$. Note that $u, zv$ are also unitaries, and still $u(zv) = e^{2 \pi i \theta} (zv) u$. By universality of the irrational rotation algebra, $B$ is *-isomorphic to $C^*(u,zv)$. In particular there is a *-isomorphism $\pi$ satisfying $u \mapsto u$ and $v \mapsto zv$. Hence, as an isometry,
$||\sum_{m,n}a_{m,n} z^m v^m u^n||= ||\pi (\sum_{m,n}a_{m,n} v^m u^n)|| = ||\sum_{m,n}a_{m,n} v^m u^n|| $