Prove that it's impossible to approximate $1/z$ with polynomials on an annulus I heard a nice problem, presumably from an old qual, that I thought I'd share.
Problem: Let A be the annulus (in the complex plane) $A=\{z: r_1 \leq |z|\leq r_2\}.$ Prove that $f(z) = 1/z$ cannot be approximated uniformly by polynomials on $A$.
 A: Here is another simple argument that only requires the maximum principle. Suppose that $1/z$ can be approximated uniformly on $A$ by polynomials. Let $m:=\operatorname{min} \{|1/z| : z \in A\}>0$. Then by assumption, there exists some polynomial $P$ such that
$$|1/z - P(z)|<m \qquad (z \in A).$$
Thus,
$$|1-zP(z)|<m|z|\leq1 \qquad (z \in A).$$
By the maximum principle, 
$$|1-zP(z)| < 1$$
for every $z$ in the disk whose boundary is the outer circle of $A$. But with $z=0$ we get a contradiction.
Essentially the same proof shows that for every compact set $K$ in the plane whose complement contains more than one component, there exists some function $f$ holomorphic on a neighborhood of $K$ that cannot be approximed uniformly on $K$ by polynomials.
A: If $f_n$ is a sequence of holomorphic functions converging uniformly to a holomorphic function $f$, then the contour integrals $\oint f_n$ of the $f_n$ around a circular contour converge to the contour integral $\oint f$. For $f = \frac{1}{z}$ the latter integral is nonzero, but it is zero for any sequence of polynomials $f_n$. 
