Showing $1$ is not a branch point for $f(z) = z^2$?

Question

I can see geometrically why $1$ is not a branch point for $f(z) = z^\frac{1}{2}$ as if we take a a point $p$ on the Riemann surface for $z$, $\epsilon$ distance away from $1$ are able to rotate that point by $2\pi$ in a small closed circle around $1$, and when it returns to original starting point it has the same value. (As long as \epsilon < $1$).

But any point we take about $0$ will see it's value change as we rotate it by $2\pi$.

How I can I write it out mathematically that for $f(z) = z^2$, a point can be rotated about $1$ without changing its value, while if it's rotated around $0$, its value will change?

Answer 1

As $$f'(1)\ne 0,$$ $f$ is locally invertible near $z=1$.

See this notes p.13 or http://books.google.es/books?id=HoMhAwAAQBAJ&pg=PA320&lpg=PA320&dq=%22branch+point%22+%22locally+invertible%22&source=bl&ots=VW8ss3HEEx&sig=OtfSy4fSPRq6-2hwCTiAs0GxTyM&hl=es&sa=X&ei=COtuVJqjGtPjaITigsAO&ved=0CCgQ6AEwAQ#v=onepage&q=%22branch%20point%22%20%22locally%20invertible%22&f=false.