Talking to the physics advisor is not a bad idea, but I would definitely recommend stopping by room 236 to talk to the students, if you haven't already.
The fixed point theorem requires f to be bounded. Otherwise, take f(x) = x + 1
See http://en.wikipedia.org/wiki/Smooth_function" [Broken] for a differentiable but not C^1 function.
Define g_y : \mathbb{R} \rightarrow \mathbb{C} so that g_y(x) = f(x+iy). Cauchy's theorem plus continuity of f at the boundary imply that
\int_{-a}^a (g_y(x)+g_y(-x))dx = 0
(taking a symmetric rectangular contour with base arbitrarily close to the real line). The continuity of g_y gives that...
Neither. Both poles and essential singularities require the relevant function to be holomorphic on a deleted neighborhood of the singularity. z^{-\frac{1}{2}} isn't even continuous on one of these neighborhoods. In descriptive terms, however, it would look like half of a simple pole stretched...