Exercise II.6.3. $f(z) = (z - 1)/(z -2)$ at $z = l.$ (Shakarchi) Determine if this function is a local analytic isomorphism at the given point. Give the reason for your answer.
Solution. We write $ζ — 2 = —(1 — (ζ — 1))$ so that for all $ζ$ near 1 we have $f(z) = -(z - 1) [1 + (z - 1) + (z - l)2 4- · · ·]$ so that $f'(1) = -1$. This proves that $f$ is a local analytic isomorphism at $ζ = 1$
In this context, what does local analytic isomorphism mean? Does it just mean that $ f'(z) /= 0 $ at the given $z$?
Like you said, $f$ is a local analytic isomorphism at some $z_0$ if $f$ is analytic at $z_0$ and $f'(z_0) \neq 0$. This comes from the local inverse theorem for analytic maps.