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