Is there a continuous branch in the neighborhood of function $z$ around $0.$ In my opinion no, because as we take a circle and go around it, the argument gains $2\pi$ and the value of logarithm changes by $2i\pi$. But do I understand it well?
My problem was to determine if there is a continuous branch of a function $f(z)=(z^2-1)^{1/2}$ in infinity, I changed it into $g(z)=1/f(1/z)$ in the 0 neighborhood, which makes it $g(z)=z/(1-z)^{1/2}$, the denominator has continuous branch ofc as I can take the Arg z, but does the numerator have it too?
I think you are confusing $z$ with $\log z$. To the second part, yes there is, provided you cut the plane between the branch points at $\pm1$. Circling around $\infty$ causes the argument to clock up $\pi i$ twice, which cancels. As you say, it looks like $z$ at $\infty$. Note, $\log \sqrt{z^2-1}$ does not have a cts branch around $\infty$.