Example of holomorphic function $f$ such that $\lim_{|\zeta|\rightarrow z}\text{Im}f(\zeta) = arg(z)$

Question

This question came up in a recent qualifying exam.

Prove that there exists a holomorphic function $f : D(0,1) \rightarrow \mathbb{C}$ such that $$\lim_{\zeta \rightarrow z}\text{Im}\ f(\zeta) = \text{Arg}(z)$$ for all $z \in \{\omega : |\omega| = 1\} \setminus \{-1\}$, where $\text{Arg}(z) \in (-\pi, \pi)$.

I'm really stuck on this one, so any help would be appreciated.

UPDATE: Following votes and comments that this question is "off-topic" or just "not a good question", let me be more specific about what I am asking:

Can you provide an explicit construction of such a function?

That is my question.

Answer 1

Following the comment by @user1952009...

If we define $g(z) := \text{Arg}(z)$ for $z \in \partial D(0,1)$, where $\text{Arg}(z) \in (-\pi, \pi]$, then $g$ is continouous for all $z \in \partial D(0,1)\setminus \{-1\}$. It follows that $$ u(z) := \left\{ \begin{array}{cl} \frac{1}{2\pi} \int_{-\pi}^{\pi} g(e^{i\theta})\cdot \dfrac{1 - |z|^{2}}{|z - e^{i\theta}|^{2}} d\theta & \text{ if } z \in D(0,1) \\ g(z) & \text{ if } z \in \partial D(0,1) \end{array} \right. $$ is harmonic on $D(0,1)$ and continuous on $\overline{D(0,1)}\setminus \{-1\}$.

For a proof of this claim, see Function Theory of One Complex Variable by Krantz & Green, page 213. Note that in the proof they assume the function $g$ is continuous for all $z \in \partial D(0,1)$, but the same steps apply.

So, we have that $u$ is harmonic on $D(0,1)$ and therefore there exists a holomorphic function $f$ on $D(0,1)$ such that $\text{Im}\ f = u$.