I am reading a text on branch points and branch cuts, and one of the exercises asks me to find the possible values of $\arccos (z)$ at $z = i$.
To do this, I understand that I should find the branch points of $\arccos (z)$, assign branch cuts, and then evaluate the function at $i$ for each branch. I see on Wikipedia that the branch points should be at $z = \pm 1$. I also see that the branch cuts extend to $\pm \infty$ on the real line, which seems to imply that $z = \pm \infty$ are also branch points, right?
I am having trouble squaring this with the instructions for finding branch points given in the text. Namely:
In most of the examples that we will encounter, the multivaluedness arises ultimately from the complex logarithm, and in such cases the branch points are the values of $z$ such that the input to the logarithm is $0$ or $\infty$.
Following along with Wikipedia's derivation of the logarithmic form of $\arcsin (z)$, I have shown that
$$\arccos(z) = -i \ln\left(z \pm \sqrt{z^2 -1}\right)$$
Since the branch points of $\ln(w)$ are at $0$ and $\infty$, I tried
$$\begin{align} z + \sqrt{z^2 -1} & = 0 \\ \implies z^2 & = z^2 -1& \text{never} \\\\ z - \sqrt{z^2 -1} & = 0 \\ \implies z^2 & = z^2 -1& \text{never} \\\\ z + \sqrt{z^2 -1} & \to \infty & \text{in the limit as }z\to\infty\\\\ z - \sqrt{z^2 -1} & \to \infty & \text{never(?)} \end{align}$$
(I put a question mark on the last one because the expression will go to $-\infty$ in the limit as $z\to-\infty$. Is this also considered an "infinity" in the sense that it is a branch point of complex $ln$?)
This seems to suggest that the only branch point of $\arccos(z)$ is $+\infty$, which I know is not true.
Meanwhile, if I plug in $z=\pm 1$ into the logarithmic expression for $\arccos(z)$, I get
$$\begin{align} \arccos(1) &= -i\ln(1 \pm 0) \\ &= -i\ln(e^{2\pi i n}), & n \in \mathbb{Z} \\ &= -i(1+ 2\pi i n), & n \in \mathbb{Z} \\ &= 2\pi n - i, & n \in \mathbb{Z} \\\\ \arccos(-1) &= -i\ln(-1 \pm 0) \\ &= -i\ln(e^\pi e^{2\pi i n}), & n \in \mathbb{Z} \\ &= -i(\pi+ 2\pi i n), & n \in \mathbb{Z} \\ &= 2\pi n - i \pi, & n \in \mathbb{Z} \\\\ \end{align}$$
In both cases, $\arccos(z)$ is multiple-valued, so $z=\pm 1$ does not seem like a branch point for this function.
Where have I gone wrong?