# Show that for $g:G'(\subset f(G))\to \Bbb{C}$ continuous, with $f:G\to \Bbb{C}$ holomorphic, $g$ is holomorphic

Let $G$ be a domain and let $f:G\to \Bbb{C}$ be holomorphic. Now let $G'\subset f(G)$, $g:G'\to \Bbb{C}$ continuous such that $f(g(z))=z$ for all $z\in G'$ (Meaning, $g$ is some branch of $f^{-1}$). Show that $g(z)$ is holomorphic and that $g'(z)={1\over f'(g(z)}$ given that $f'\ne 0$. Hint: Look carefully at the definition for the derivative of $g$ in $z_0\in G'$. Use the continuity of $g$ to justify the transitions from one limit to another: $$\lim_{z\to z_0}{g(z)-g(z_0)\over z-z_0}=\lim_{g(z)\to g(z_0)}{g(z)-g(z_0)\over z-z_0}={1\over \lim_{g(z)\to g(z_0)}{z-z_0\over g(z)-g(z_0)}}$$

I am quite clueless: do I show that $g$ is holomorphic and than find its derivative? Are those two question a single question? And why is the first transition legitimate? Never have I ever done such a thing before. Could you elaborate?

• @ctst: You need to justify that $g$ is differentiable in the first place. – copper.hat Dec 21 '16 at 15:51
• I am so sorry there is a shameful mistake in the hint... I've been awake for ages and this is another result. Let me change it... – Meitar Dec 21 '16 at 16:01

I presume that $g(G') \subset G$ so $f \circ g$ is well defined.

Pick some $z_0 \in G'$ and let $w_0 = g(z_0)$.

I am assuming that $f'(w_0) \neq 0$.

Since $f$ is analytic at $w_0$, we have some $R>0$ such that if $|w-w_0| < R$ then $f(w) = f(w_0) + f'(w_0)(w-w_0) + (w-w_0)^2\sum_{k \ge 2} a_k (w-w_0)^{k-2}$, and the function $\phi(w) = \sum_{k \ge 2} a_k w^{k-2}$ is continuous (in fact analytic, but that is not used here).

Now choose $\delta>0$ such that $B(z_0,\delta) \subset G'$ and if $|z-z_0| < \delta$, then $|g(z)-g(z_0)| < R$ (this is where continuity of $g$ is important), then we have $z = z_0 + f'(g(z_0))(g(z)-g(z_0)) + (g(z)-g(z_0))^2 \phi(g(z)-g(z_0))$.

Note that $g$ is injective on $B(z_0,\delta)$, hence $g(z) \neq g(z_0)$ so we can rearrange and divide across by $g(z)-g(z_0)$ to get ${z -z_0 \over g(z)-g(z_0)} = f'(g(z_0)) + (g(z)-g(z_0)) \phi(g(z)-g(z_0))$ from which we get $\lim_{z \to z_0} {z -z_0 \over g(z)-g(z_0)} = f'(g(z_0))$.

It follows from this that $\lim_{z \to z_0} {g(z)-g(z_0) \over z -z_0}$ exists and is equal to ${1 \over f'(g(z_0))}$.