Which options are true for $f(z)=\frac{1}{z}$

Question

Consider $f(z)=\frac{1}{z}$ on the annulus $A= \{z\in \Bbb{C} | \frac{1}{2}<|z|<2\}$. Which of the following are true?

1. There is a sequence $\{p_n(z)\}$ of polynomials that approximate $f(z)$ uniformly on compact subsets of $A$.

2. There is a sequence $\{r_n(z)\}$ of rational functions, whose poles are contained in $\Bbb{C}\setminus A$ and which approximates $f(z)$ uniformly on compact subsets of $A$.

3. No sequence $\{p_n(z)\}$ of polynomials approximate $f(z)$ uniformly on compact subsets of $A$.

4. No sequence $\{r_n(z)\}$ of rational functions whose poles are contained in $\Bbb{C}\setminus A$ approximate $f(z)$ uniformly on compact subsets of $A$.

I think options 2 and 3 are correct, 2 follows from Runge's theorem, but I don't know how to prove 3.

Answer 1

Let $K = \{ z \in \mathbb{C} \, | \, |z| = 1 \}$ and assume that there exists a sequence $\{p_n(z)\}_{n=1}^{\infty}$ of polynomials such that

$$ \max_{z \in K} |p_n(z) - f(z)| \xrightarrow[n \to \infty]{} 0. $$

Choose $n \in \mathbb{N}$ such that $\max_{|z| = 1} |p_n(z) - f(z)| < \frac{1}{2}$. Hence,

$$ |p_n(z) - f(z)| = \left| \frac{z \cdot p_n(z) - 1}{z} \right| = \left|z \cdot p_n(z) - 1 \right| < \frac{1}{2} $$

for all $|z| = 1$. By the maximum modulus principle, we have

$$ 1 = |0 \cdot p_n(0) - 1| \leq \max_{|z| \leq 1} |z \cdot p_n(z) - 1| = \max_{|z| = 1} |z \cdot p_n(z) - 1| < \frac{1}{2} $$

and we have obtained a contradiction.