jmc

From $z+\frac{1}{z}=\frac{1+\sqrt{5}}{2},$ $$z+\frac{1}{z}-\frac{1}{2}=\frac{\sqrt{5}}{2}.$$Squaring both sides, we end up with $$z^2-z+\frac{9}{4}-\frac{1}{z}+\frac{1}{z^2}=\frac{5}{4}.$$Then $$z^2-z+1-\frac{1}{z}+\frac{1}{z^2}=0.$$Hence, $z^4-z^3+z^2-z+1=0.$ Then $$(z+1)(z^4-z^3+z^2-z+1)=0,$$which expands as $z^5+1=0.$ This gives us $z^5=-1.$

Therefore, $$z^{85}+\frac{1}{z^{85}}=(z^5{)}^{17}+\frac{1}{(z^5{)}^{17}}=(-1{)}^{17}+\frac{1}{(-1{)}^{17}}=\boxed{-2}.$$