Holomorphic functions reflected through segments that aren't on the real axis

My question concerns using the Schwarz Reflection principle (or symmetry principle) to reflect regions in the domain of a holomorphic function into a symmetric (with respect to a line segment) region in which $$f(z)$$ is already defined.

More specifically, if $$f: \mathbb{D} \rightarrow \mathbb{C}$$ is a holomorphic function that is real-valued along the line $$\arg z = \pi/3$$, we may take one "side" of a region whose boundary is along $$\arg z = \pi /3$$ and reflect through it via the formula $$\tilde{f}(\tilde{z})$$, where ~ denotes the reflection. Since this is an analytic continuation and $$f(z)$$ is already defined on the other "side" of $$\arg z = \pi /3$$, is it logical to conclude that $$f(z) = f(\tilde z)$$?

Furthermore, when the line segment in question is the real axis, the ~ map is really just the usual complex conjugation. I initially wanted to attack this question by rotating $$\arg z = \pi/3$$ clockwise by $$\pi /3$$ to bring the real-valued segment back to the axis, apply the symmetry principle here, and rotate back. Unless I am mistaken, I believe doing this shows that $$\tilde f (\tilde z) = \overline{f(\overline{z})}$$ ? That is to say, reflecting through line segments that are not on the real axis still follow the same reflection formula. This feels incorrect. Would love some insight about both of my (seemingly false) observations.. thanks!

Suppose we know the function $$f$$ is analytic on the values of $$\mathbb D$$ and is real (and at least continuous) on the line $$L = \{r, \theta \mid r \ge 0, \theta = \pi/3\}$$ intersected with $$\mathbb D.$$ By rotating, we conclude $$g(z) = f(e^{i\pi/3}z)$$ is analytic on the lower half plane $$\overline{\mathbb H}$$ intersected with $$\mathbb D$$ and is real on $$\mathbb R \cap \mathbb D.$$ By the Schwartz Reflection Formula, we can (uniquely) extend $$g$$ to an analytic function $$\hat g$$ on all of $$\mathbb D$$ by the rule $$g(\bar z) = \overline{g(z)}$$. We now rotate back by $$\pi/3$$ radians to conclude $$\hat f(z) =\hat g(e^{-i\pi/3}z)$$ is an analytic extension of $$f$$ to $$\mathbb D.$$

Note that for values on/below $$L \cap \mathbb D$$ we have $$\hat f(z) =\hat g(e^{-i\pi/3}z) = g(e^{-i\pi/3}z)=f(z)$$ and if I did the calculations right (please check), for values on/above $$L \cap \mathbb D$$ we have $$\hat f(z) = \hat g(e^{-i\pi/3}z) = \hat g\left(\overline{e^{i\pi/3}\bar{z}}\right) = \overline{\hat g\left(e^{i\pi/3}\bar{z}\right)} = \overline{ g\left(e^{i\pi/3}\bar{z}\right)} \\= \overline{ f\left(e^{i\pi/3}\overline{e^{-i \pi /3} z}\right)} = \overline{ f\left(e^{2i\pi/3} \bar z\right)}$$ So that we are extending by the rule $$f(\bar z) = \overline{f(e^{2i\pi/3} z)}.$$

Note that I assume you intend $$\mathbb D$$ to be the unit disk. Correct me if I misinterpreted you.