Here is a question from a past qual exam:
For $0 < r < s$, let $A_{r,s} = \{z \in \mathbb{C} \colon r < |z| < s\}$.
a: Are there polynomials $p_j$ which converge uniformly to $z + \frac{1}{z}$ on $A_{2,4}$?
b: Let $f$ be holomorphic on $A_{1,5}$. How can we know if there exist $p_j$ which converge uniformly to $f$ on $A_{2,4}$?
c: Now let $R_{r,s} = \{z = x + iy \in \mathbb{C} \colon r < \text{max}(|x|,|y|) < s\}$ (essentially a square annulus). If $f$ is holomorphic on $R_{1,5}$, how can we know if there exist polynomial $p_j$ which converge uniformly to $f$ on $R_{2,4}$?
My attempt: Letting $\gamma$ be a loop around $D_2(0)$, since $\int_\gamma p_j dz = 0$ for any poly but $\int_\gamma z + \frac{1}{z} dz = 2\pi i$, it becomes clear that the answer to (a) is no. Thus, for (b) and (c), a necessary condition is that the poles of $f$ "cancel" themselves out in that $\int_\gamma f(z) dz = 0$ for any closed loop $\gamma$ in $A_{2,4}$. I believe the answer is actually that $f$ must be holomorphic on all of $D_5(0)$ as well (and similarly the entire $10$-by-$10$ box). However, I am struggling to think of a rigorous proof and am perplexed by how part c seems the same as part b.
More thoughts: By Runge's theorem, we can approximate $f$ by rational functions whose denominator is $z^n$. This gives us $|p_j(z) - \frac{q_i(z)}{z^n}| < \epsilon$ for specific values of $i$ and $j$ which perhaps leads to a contradiction, but I can't see exactly how.
Thanks!