No Sequence of polynomials converging to $f(z)= \frac{1}{z}$ on given compact set $K$

Question

let $K=\left \{z \in \mathbb{C}: \frac{1}{2}\leq |z|\leq2 \right \}$. Then $f(z)=\frac{1}{z}$ is holomorphic on an open neighbourhood of $K$.

use Maximum principal to show that there is no sequence $P_n$ of polynomials such that $P_n \to f$ uniformaly on $K$.

first please tell what it mean to say a function $f$ is holomorphic in open neighbourhood of compact set $K$.

Now in the question $f(z)=\frac{1}{z}$ has maximum value of $2$ on $K$. let $P_n \to f$

then it mean $\text{sup}_{z\in K}|P_n(z)-f(z)| \to 0$. from this how can i get contraduction . also i donot understand how can i use that sequence $P_n$ are polynomials.

any suggestion ??

Answer 1

Hint: the maximum value of $P_n$ on $K$ occurs on $\{z: |z|=2\}$.