Polynomial approximation obstruction on a punctured circle
= Polynomial approximation obstruction on a punctured circle
Polynomials cannot converge uniformly to $1/z$ on $\{z:|z|=1,z\ne1\}$. Uniform convergence there makes the polynomials uniformly Cauchy on the full circle because deleting one point does not change the supremum of a continuous function. They would therefore converge uniformly to $1/z$ on the full circle, contradicting
$$
\oint p(z)\,dz=0,
\qquad
\oint\frac{dz}{z}=2\pi i.
$$