Let $F_{0}(z)=\frac{1}{(1-z)^{i}i}$.
a. Verify that $|F_{0}(z)| \leq e^{\frac{\pi}{2}}$ in the unit disc, but that $\lim_{r \rightarrow 1} F_{0}(r)$ does not exist. (Hint: Note that $|F_{0}(r)|=1$ and $F_{0}(r)$ oscillates between $\pm$ infinitely often as $r \rightarrow 1$).
b. Let $\{a_{n}\}$ be an enumeration of the rationals, and let $F(z)=\sum_{j=1}^{\infty} \delta ^{j}F_{0}(ze^{-ia_{j}})$, where $\delta$ is sufficiently small. Show that $\lim_{r \rightarrow \infty} F(re^{i \theta})$ fails to exist whenever $\theta = a_{j}$, and hence $F$ fails to have radial limit for a dense set of points of the unit circle.
Firstly, in solving this problem, I don't understand what is "$r$"; is $r$ just representing the same a $z$ but notated differently because it is approaching infinity? And, does this function have a "known" pattern or does it have a name so that I can learn more about it?
$r \to 1$ simply stands for restricting z to the intersection of the unit disk with the real axis and letting z approach 1. For the question to make sense you have to first know how $(1-z)^{i}$ is defined. I will assume that it is defined as $e^{ilog(1-z)}$ where log stands for the principal branch of logarithm. If $|z| <1$ then the argument of $1-z$ lies between $-\pi /2$ and $\pi /2$ so $|F_0 (z)| =|e^{-i log (1-z)}| = e^{- \Re \{i log (1-z)\}} =e^{arg (1-z)} < e^{ \pi /2}$. Now take z to be a real number $r<1$ and take limit as $r \to 1$. $e^{-i log (1-r)}$ does not have a limit as $r \to 1$ because its imaginary part is $- \sin (\ln (1-r))$ and limit of $\sin x$ as $x \to -\infty$ does not exist.