jmc

The solutions of the equation $z^{1997}=1$ are the $1997$th roots of unity and are equal to $\cos \left(\frac{2\pi k}{1997}\right)+i\sin \left(\frac{2\pi k}{1997}\right)$ for $k=0,1,\dots ,1996.$ They are also located at the vertices of a regular $1997$-gon that is centered at the origin in the complex plane.

By rotating around the origin, we can assume that $v=1.$ Then $$\begin{array}{rl}|v+w{|}^2& ={\left|\cos \left(\frac{2\pi k}{1997}\right)+i\sin \left(\frac{2\pi k}{1997}\right)+1\right|}^2\\ & ={\left|\left[\cos \left(\frac{2\pi k}{1997}\right)+1\right]+i\sin \left(\frac{2\pi k}{1997}\right)\right|}^2\\ & ={\cos }^2\left(\frac{2\pi k}{1997}\right)+2\cos \left(\frac{2\pi k}{1997}\right)+1+{\sin }^2\left(\frac{2\pi k}{1997}\right)\\ & =2+2\cos \left(\frac{2\pi k}{1997}\right).\end{array}$$We want $|v+w{|}^2\ge 2+\sqrt{3}.$ From what we just obtained, this is equivalent to $\cos \left(\frac{2\pi k}{1997}\right)\ge \frac{\sqrt{3}}{2}.$ This occurs when $\frac{\pi }{6}\ge \frac{2\pi k}{1997}\ge -\frac{\pi }{6}$ which is satisfied by $k=166,165,\dots ,-165,-166$ (we don't include 0 because that corresponds to $v$). So out of the $1996$ possible $k$, $332$ work. Thus, the desired probability is $\frac{332}{1996}=\boxed{\frac{83}{499}}.$