jmc

We start by trying to express $z$ in a more convenient form. We are given that $z+z^{-1}=\sqrt{3}=\frac{2\sqrt{3}}{2}=2\cos \frac{\pi }{6}$ So we know that $z$ is $\text{cis}\frac{\pi }{6}$ or $\text{cis}-\frac{\pi }{6}$

Say that $z=\text{cis}\frac{\pi }{6}$. Then, $$z^{2010}={\left(\text{cis}\frac{\pi }{6}\right)}^{2010}=\text{cis}\frac{2010\pi }{6}=\text{cis}335\pi =\text{cis}\pi =-1.$$Then $z^{-1}=-1^{-1}=-1$. So $$z^{2010}+z^{-2010}=-1+(-1)=\boxed{-2}.$$Similarly, if $z=\text{cis}-\frac{\pi }{6}$. Then, $$z^{2010}={\left(\text{cis}-\frac{\pi }{6}\right)}^{2010}=\text{cis}-\frac{2010\pi }{6}=\text{cis}-335\pi =\text{cis}-\pi =-1.$$Then $z^{-1}=-1^{-1}=-1$. So $$z^{2010}+z^{-2010}=-1+(-1)=\boxed{-2}.$$