Let
$$g(z):=\frac{f(z)(1-z\overline{a})}{z-a}$$
It's not hard to prove that $g:U\to U$ (hint: use the maximum modulus principle) and it is easy to see that $g(0)=0$. Thus, by Schwarz' lemma,
$$\begin{align*}1\ge &|g'(0)|=\left|\frac{f'(0)}{a}\right|\\
|a|\ge&|f'(0)|\end{align*}$$
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 $K$.
Best Answer
False statement in case $z_n=z_0$ for all $n$. True if this case is excluded.
A pole is an isolated singularity. This means if $f$ has pole at $z_0$ it must be analytic throughout some disk containing $z_0$ except for the point $z_0$. So the conclusion follows just by definition!.