Does the series $\sum\limits_{n=1}^\infty \frac{e^{inz}}{\bar zn^3+|z|^3}$ converge uniformly on $[-2,0)\cup(0,2]$ or $[-2)\cup(-2,\infty)$?

I attempted to use the Weierstrass M-test:

Since $z=\bar z$ on all $A\subseteq \mathbb R$, we have, bounding the term from above, $$\left| \frac{e^{inz}}{\bar zn^3+|z|^3}\right |\leq \frac{1}{|z|n^3+|z|^3}\leq \frac1{|z|n^3}$$I do not see how I could've procured a different upper bound that would've been independent of $z$ terms on $[-2,0)\cup (0,\infty]$ since $|z|<2$, but I'd presume to say that the Weierstrass M-test fails here. What next? I am aware that the failure of the M-test does not imply non-uniform convergence.

Edit: As @Duong Ngo pointed out, there was a typo in the domains—they should should exclude zero. Following this, I believe that I can rescue the M-test strat since if $z\in [-2,2]\setminus\{0\}$, then $\exists \epsilon >0$ s.t. $\epsilon < |z|<2 \iff \frac 1\epsilon >\frac 1{|z|}$, so the M-test still applies. But this upper bound seems to be independent of the domain, i.e., it seems like it should hold for all $z\in \mathbb R$?