Solution
= Solution
The equation implies $|z|\geq1$. Its intersections with $\{z>0\}$ and $\{z<0\}$ are disjoint, nonempty subsets open in the <subspace topology>, and together cover the entire surface. They therefore form a separation, proving \b[the surface is disconnected]. More precisely its two sheets are $z=\pm\sqrt{1+x^2+y^2}$; each is homeomorphic to the <connected> plane by projection, so these are exactly its two <connected components>.