Showing $\frac{1}{z}$ cannot be approximated by any polynomial on $\partial{\mathbb{D}}$.

Question

I am trying to understand why $f(z)=\frac{1}{z}$ cannot be approximated by any polynomials uniformly on $\partial\mathbb{D}$.

I have the following idea, but I am really not sure if it is valid.

Given $\epsilon>0$, I assume that there exists a sequence of polynomial $P_n(z)$ such that $$\|P_n-f\|_{\partial\mathbb{D}}\leq \epsilon.$$

I think to get a contradiction, I should prove $(P_n)$ is convergent to $f$ uniformly in $\mathbb{D}$, which is a contradiction, since the uniformly convergent limit of analytic functions must be analytic.

Answer 1

Hint: polynomials satisfy the Cauchy Integral Theorem.