Example of smooth $f: \mathbb C \to \mathbb C$ holomorphic on annulus with non-vanishing line integral Is it possible to construct a smooth (when viewed as a map on $\mathbb R^2$) map $f: \mathbb C \to \mathbb C$ such that $f$ is holomorphic on an annulus $r < |z| < R$ however the contour integral $\int_{|z| = \epsilon} f(z) dz \neq 0$ where $r < \epsilon < R$?
If we take away some of the conditions, the construction is easy, e.g. $f= 1/z$ has non-vanishing contour integral on the unit circle however it is not defined on the entire domain. A bump function satisfies all the conditions except for non-vanishing contour integral. Intuitively I feel like there is a way to mash together these functions to get the desired example, however I'm not sure exactly how. 
 A: Yes, it is easy, and yes, you can mash bump functions, and by mash I mean multiply. Take any smooth function $\chi:\mathbb R^2 \to \mathbb R$ such that it is zero on an open ball of the origin, and takes the value 1 on the set $|z|\ge 1$. Then consider
$$ f:\mathbb C \to \mathbb C, \quad f(z) := \frac{\chi(z) }z$$
(continuously, not analytically, extended to $0$)
this function is pointwise equal to $1/z$ on $1\le |z| < \infty $, and therefore everything you know about $1/z$ as a function on $1\le |z|  $ carries over, in particular the contour integrals $\int_{|z|=1+\rho} f(z)dz$ do not vanish.
A: It is very possible to take the meromorphic function $f(z) = \frac1z$, and then change it only on the unit disc to become smooth as an $\Bbb R^2\to \Bbb R^2$ function in the entire complex plane. It will then be analytic on, say, the annulus $1<|z|<2$ and still have non-vanishing contour integral on that annulus.
More concretely, take a smooth function $g:\Bbb R\to \Bbb R$ which is $0$ on $(-\infty, \frac12]$ and $1$ on $[1, \infty)$, then the function
$$
h(z) = g(|z|)\cdot f(z)
$$
(with the obvious smooth extension $h(0) = 0$) will have the properties you're looking for.
