The set-up is as follows:
Given $\alpha \in \mathbb{R}$, set $f_\alpha (z) = \sqrt{r}e^{i(arg_\alpha (z) /2)}$ where $arg_\alpha(z)$ is the unique $\theta \in (\alpha, \alpha+2\pi]$ such that $z=re^{i\theta}$. Setting $L_\alpha = \{z \in \mathbb{C} : z = te^{i\alpha}, t \geq 0\}$, show that $f_\alpha$ is not continuous at any $z \in L_\alpha$.
Initially, my idea was to simply create the following single-valued function:
$\phi(k) = f_\alpha(te^{ik})$ and evaluate the following limits:
$\lim_{k \to \alpha^+} \phi(k)$ and $\lim_{k \to (\alpha + 2\pi)^-} \phi(k)$ and show that they are not equal to each other.
I found that
$\lim_{k \to \alpha^+} \phi(k) = \lim_{k \to \alpha^+} \sqrt{t}e^{i(arg_\alpha(te^{ik})/{2})}$.
Would I be able to say that this limit does not exist since $arg_\alpha(te^{ik})$ is not defined when $k = \alpha$? If this is not correct, how would I evaluate this limit? Furthermore, is my approach even correct for trying to show what is being asked for?
your idea is correct : $\lim_{k \rightarrow \alpha^+} e^{ik} = lim_{k \rightarrow (\alpha + 2\pi)^-} e^{ik}$ so for $f_\alpha$ to be continuous, you would need to have $\lim_{k \to \alpha^+} \phi(k) = \lim_{k \to (\alpha + 2\pi)^-} \phi(k)$.
However, it is not correct to say that the limit in $\alpha^+$ does not exist because the function is not defined when $k = \alpha$.
For starters, $arg_\alpha (te^{i\alpha}) = \alpha + 2 \pi$ according so your definition, so it is in fact defined. But the limit $\lim_{k \rightarrow \alpha^+} e^{ik}$ is equal to $\alpha$. This is because if $k \in (\alpha;\alpha + 2\pi],\ arg_\alpha(e^{ik}) = k$.
This shows that $arg_\alpha$ is not continuous at any $z \in L_\alpha$ and it is what you need to use to show that $\lim_{k \to \alpha^+} \phi(k) \neq \lim_{k \to (\alpha + 2\pi)^-} \phi(k)$.