Does there exists a sequence of polynomials and rational functions approximating an analytic function uniformly? This question was asked in my complex analysis quiz and I was absolutely confused on which result to use.

Consider the function $f(z)=1/z$ on the annulus $A=[{z \in \mathbb{C} : 1/2 < |z|<2}]$.  Then

(a) Does there exists a sequence ${p_n(z)}$ of polynomials that approximate f(z) uniformly on compact subsets of A.
(b) DOes there exists a sequence ${r_n(z)}$ of rational functions , whose poles are contained in $\mathbb{C}/A$ and which approximate  f(z) uniformly on compact subsets of A.
Attempt: $1/z$ is analytic on $A$ so there will exist a sequence of analytic functions which converge uniformly to $1/z$ on compact subsets of $A$ but why should they be polynomials or rational functions specifically?
 A: To summarize the comments above :

(a)
Let $p_n(z)$ be polynomials uniformly approximating $f(z)$ on $A$. Consider the contour $\gamma = \{|z| = 1\}$.
By Cauchy's theorem, since polynomials are holomorphic everywhere in the interior of the circle $|z| = 2$, we get that $\int_{\gamma} p_n(z)dz = 0$ for all $n$.
However, it is well known that $\int_{\gamma} \frac 1z dz = 2 \pi i$.
Now, if $p_n(z) \to \frac 1z$ uniformly on $A$, in then particular we must have $\int_{\gamma} p_n(z)dz \to \int_{\gamma} \frac 1zdz$, which does not hold. Consequently, no sequence of polynomials can uniformly approximate $\frac 1z$ in this annulus (or for that matter, in any annulus centered at the origin by the same argument).

(b)
This is far more obvious : the function $f(z)$ has only one pole, which is outside $A$. Consequently, we can just take $r_n = f$ for all $n$.

The theorem that generalizes this kind of approximation, is Runge's theorem. It states :

Let $K \subset \mathbb C$ be compact and $f$ be a holomorphic on an open set containing $K$. Let $D$ be any set, which has at least one element from every bounded component of $\mathbb C \setminus K$. Then, we can find rational functions $r_n$ such that :

*

*For all $n$, the poles of $r_n$ are contained in $D $.

*$r_n \to f$ uniformly on $K$.


We also have the stronger Mergelyan's theorem, which removes the holomorphicity assumption on the boundary of $K$.

If $K$ is a subset of $\mathbb C$ such that $\mathbb C \setminus K$ is connected, then every continuous function $f:K \to \mathbb C$ with $f$ holomorphic on $K^{\circ}$ is approximable by polynomials uniformly on $K$.

