On some lecture notes that I am working on there is an exercise to prove that $-1,1$ are branch points of the multi-function $\sqrt{z^2-1}$.
I know that the branch $f=\sqrt{rs}e^{i\frac{1}{2}(\theta_1+\theta_2)}$, obtained by letting $z=1+re^{i\theta_1}=-1+se^{i\theta_2}$,$\theta_1,\theta_2\in (-\pi,\pi]$ encodes the the branch cut $[-1,1]$.
The notes I have, seem to imply that if we suppose $g$ is a branch on $B(1,r),r\lt1$ (open ball about $1$), then $g=\pm f$, with $f$ as above giving a contradiction.
Could an explanation be given as follows?
For a branch $g$ we can write $z=1+re^{i\phi_1+2n\pi}=-1+se^{i\phi_2+2m\pi}$ for some integers $n,m$ giving $$g(z)=\pm \sqrt{rs}e^{\frac{1}{2}i(\phi_1+\phi_2)}e^{i\pi(n+m)}\\=\pm \sqrt{rs}e^{\frac{1}{2}i(\phi_1+\phi_2)}$$ so whether $g$ consistently takes the positive or negative root, the freedom of $n+m$ to be even or odd will add a $\pm$.
By the way the arguments are measured from the positive real axis, this is then equivalent to $\pm f$.
My feelings are that is not correct since I don’t seem to be using at any point the open ball itself.
What am I missing?
The open ball would imply that $\phi_1$ matches $\theta_1$ on the full range $(-\pi,\pi]$