Let $\gamma$ parameterize the complex unit circle. We have the following holomorphic function $F$ on $\mathbb{C}\setminus \{\gamma\}$: $$F(z)=\frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}\frac{1}{\zeta-z}d\zeta$$
We are supposed to find all points $p \in \partial D_1(0)$ (all p in the complex unit circle) such that $\lim_{z\rightarrow p, |z|<1} F(z) = \frac{1}{p}$ or $\lim_{z\rightarrow p, |z|>1} F(z) = \frac{1}{p}$
I did the following for $0<|z|<1$ , using Cauchy's Integral Formula: $$F(z) =\frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}\frac{1}{\zeta-z}d\zeta =\frac{1}{2\pi i}\int_\gamma f(\zeta)\frac{1}{\zeta-z}d\zeta =f(z) =\frac{1}{z} $$
Which would mean $z\rightarrow p\implies F(z)\rightarrow p$
So all the points $p$ we are looking for would be all points on the complex unit circle. This seems so simple I'm afraid I missed something. Also we never needed the second condition from the or statement.
For $|z|<1,$
$$F(z)= \frac{1}{2\pi i}\int_\gamma\frac{d\zeta}{\zeta^2(1-z\zeta^{-1})}\,d\zeta$$ $$ = \frac{1}{2\pi i}\int_\gamma\zeta^{-2}\left (\sum_{n=0}^{\infty}(z\zeta^{-1})^n\right)\,d\zeta = \frac{1}{2\pi i}\cdot\sum_{n=0}^{\infty}\int_\gamma z^n\zeta^{-2-n}\,d\zeta =0.$$
Using a similar procedure, if $|z|>1,$ we find we find $F(z)=-1/z.$
The answer to the question thus seems to be that for no $p$ on the unit circle do we get the limit $1/p.$