I am trying to get better intuition for branch points and a branch cut. I understand the textbook definitions:
Branch point (of an analytic function $f$): Point in the complex plane whose value in the domain of $f$ takes on multiple values in the range of $f$. (Taken from here)
Branch cut (of an analytic function $f$): Points in the complex plane along which the function $f$ is discontinuous. (Taken from here.)
So, now when I look at the function $f(z)=log(z)$, and I see that for any value along the negative real axis in the complex plane $z$, I get two distinctly different points in the plane for $f(z)$. This tells me that all of the points $z\in(-\infty,0]$ are branch points, and they also form a set of points corresponding to a branch cut: $z\in(-\infty,0]$.
From what I am seeing, $z=-1$ is also a branch point because it corresponds to two distinctly different points on the $f(z)$ plane: In polar, $z=-1=-e^{i\pi}=-e^{-i\pi}$. and when I map these to $f(z)=log(z)$, I'm getting the following: $f(-e^{i\pi})=log(-e^{i\pi})=ln|-e^{i\pi}|+i*arg(-e^{i\pi})$. and since $ln|-e^{i\pi}|=0$, and $arg(-e^{i\pi})=i\pi$, then we get $f(-e^{i\pi})=i\pi$. Similarly,
$$f(-e^{-i\pi})=log(-e^{-i\pi})=ln|-e^{-i\pi}|+i*arg(-e^{-i\pi})$$.
and since $ln|-e^{i\pi}|=0$, and $arg(-e^{-i\pi})=-i\pi$, then we get $f(-e^{-i\pi})=-i\pi$.
So, for one $z$ value $z=-1$, I am getting two different values for $f(z)$: $f(-e^{-i\pi})=-i\pi$ $f(-e^{i\pi})=i\pi$
Can someone please help me understand why only the point $z=0$ is considered a branch point, and not, for example $z=-1$?