Analytic continuation of $f(z) = \sum_{n=1}^{\infty} \frac{1}{2^n (z-e^{2\pi i r_n})}$

Question

Let $\{r_k:k\in\mathbb{N}\}$ be a counting set of rational numbers in $[0,1]$. Show that $$f(z) = \sum_{k=1}^{\infty} \frac{1}{2^k (z-e^{2\pi i r_k})}$$ is holomorphic on $U_1=\{z\in \mathbb{C} : |z| < 1\}$ and $U_2 = \{z\in \mathbb{C} : |z|>1\}$. Are both functions analytic continuations from each other?

I've already learned that if $(f_n)_{n\in \mathbb{N}}$ is a compactly convergent sequence of holomorphic functions and $f$ its limit, then $f$ is also holomorphic.

Taking $$f_n(z) = \sum_{k=1}^{n} \frac{1}{2^k (z-e^{2\pi i r_k})}$$ it would make sense that it's enough to prove that this series is uniformly convergent on $U_1, U_2$.

For $k\in \mathbb{N}$: $\frac{1}{2^k(z-e^{2\pi i r_k})}$ is holomorphic because we know that $\frac{1}{z-\exp(2\pi ir_k)}$ is holomorphic in $z\in \mathbb{C}\setminus \{\exp(2\pi i r_k)\}$ and $\exp(2\pi i r_k)$ is neither in $U_1$ nor in $U_2$ $\forall k$.

Let $z\in U_1$ and $R = \inf\{|z-\exp(2\pi i r_k)|: k\in \mathbb{N}\}$:

\begin{align*} |f_n(z)-f(z)| = |\sum_{k=n}^{\infty} \frac{1}{2^k(z-e^{2\pi i r_k})}| \leq \sum_{k=n}^{\infty} \frac{1}{2^k|(z-e^{2\pi i r_k})|} \leq \sum_{k=n}^{\infty} \frac{1}{2^k R} \overset{n\rightarrow \infty}{\longrightarrow} 0 \end{align*}

Analogous with $z\in U_2$ and $R = \inf\{|z-\exp(2\pi i r_k)|: k\in \mathbb{N}\}$.

With that I should have proven that $f(z)$ is holomorphic on $U_1$ and $U_2$, right?

Now to my main question: What does "are both functions analytic continuations from each other" mean in this context? $U_1\cap U_2 = \emptyset$ and they're essentially the same function, so yes?

Answer 1

If $\{r_k,k\ge 1\}$ is a dense subset of $[0,1]$ then $f(z)=\sum_{k\ge 1} \frac{2^{-k}}{z-e^{2i\pi r_k}}$ is analytic for $|z|<1$ and it has a natural boundary on $|z|=1$.

This is because for $x\in [0,1)$, $\sum_{k\ge 1,|e^{2i\pi r_k}-e^{2i\pi r_l}|\ge \epsilon} \frac{2^{-k}}{x e^{2i\pi r_l }-e^{2i\pi r_k}}$ is bounded while $\lim_{x\to 1^-} \Re(\sum_{k\ge 1, |e^{2i\pi r_k}-e^{2i\pi r_l}|< \epsilon} \frac{2^{-k}}{e^{2i\pi (r_k-r_l)}-x})\ge \frac{2^{-k}}{1-x} =+\infty$

So $f(z)$ isn't continuous at $e^{2i\pi r_l}$.