Uniform convergence of sequence of complex polynomials

Question

Show that there does not exist any sequence of complex polynomials that converges to $\frac{1}{z^2}$ uniformly on the annulus $A=\{z\in\mathbb{C}:1<|z|<2\}$.

I don't know how to even think in this problem. Can someone please give me hint or way ?

Answer 1

Let $(p_n) \subseteq \mathbb{C}[X]$ such that $p_n \to f$ in $A$, with $f(z) = \frac{1}{z^2}$. Now, this implies $q_n := z^2p_n \to 1$ in $A$. By the maximum modulus principle,

$$ \sup_{D(0,2)}|1 - q_n(z)| = \sup_{\partial D(0,2)}|1 - q_n(z)| \leq \sup_{A}|1 - q_n(z)| \to 0 $$

so we have that $q_n \to 1$ in $D(0,2).$ However, this would imply

$$ 1 = \lim_{n \to \infty}q_n(0) = 0 $$

which is absurd. Note that the only thing we've used about $(p_n)_{n \geq 1}$ is that we can apply the maximum modulus principle to $1 - q_n$ for each $n \in \mathbb{N}$, so the result extends to any sequence of holomorphic functions.

Answer 2

Here is one approach:

Note that if $\gamma$ is the circle of radius ${3 \over 2}$ (so it lies in $A$) then $\int_\gamma {1 \over z} dz = 2 \pi i$, and since $\int_\gamma p(z) dz = 0$ for all polynomials, we see that no sequence of polynomials can be uniformly convergent to $z \mapsto {1\over z}$ on $A$.

Let $f(z) = {1 \over z^2}$ and suppose $p_n$ is a sequence of polynomials uniformly convergent to $f$ on $A$. Then the sequence $z \mapsto z p_n(z)$ converges uniformly to $z \mapsto {1\over z}$ on $A$ which is a contradiction.