If $\,f^{7} $ is holomorphic, then $f$ is also holomorphic. I need some help with this problem:

Let $ \Omega $ be a complex domain, i.e., a connected and open non-empty subset of $ \mathbb{C} $. If $ f: \Omega \to \mathbb{C} $ is a continuous function and $ f^{7} $ is holomorphic on $ \Omega $, then $ f $ is also holomorphic on $ \Omega $.

Thanks in advance.

This is a topological variant of Christian’s elegant solution.

Let $ g \stackrel{\text{df}}{=} f^{7} $. As $ g $ is holomorphic on $ \Omega $, either (i) it is $ 0 $ everywhere on $ \Omega $ or (ii) its set of roots is an isolated subset of $ \Omega $. If Case (i) occurs, then we are done. Hence, suppose that we are in Case (ii).
Let $ \mathcal{Z} $ denote the (closed) set of roots of $ g $. For any point $ p \in \Omega \setminus \mathcal{Z} \neq \varnothing $, there exist


*

*a (sufficiently small) open disk $ D $ centered at $ p $ and contained in $ \Omega \setminus \mathcal{Z} $, and

*a holomorphic function $ h: D \to \mathbb{C} $ such that $ g|_{D} = e^{h} $.


Now, for each $ z \in D $, we must have
$$
f(z) \in \left\{ e^{2 \pi i k / 7} ~ e^{h(z) / 7} ~ \Big| ~ k \in \{ 0,\ldots,6 \} \right\}.
$$
This is because the right-hand side is precisely the set of all complex $ 7 $-th roots of $ g(z) $. This means that we have a function $ k: D \to \{ 0,\ldots,6 \} $ such that
$$
\forall z \in D: \quad
f(z) = e^{2 \pi i \cdot k(z) / 7} ~ e^{h(z) / 7}.
$$
We claim that $ k $ is locally constant on $ D $. Assume the contrary. Then there exists a $ q \in D $ such that $ k $ is not constant on any open neighborhood of $ q $. We can thus find a sequence $ (z_{n})_{n \in \mathbb{N}} $ in $ D \setminus \{ q \} $ converging to $ q $ such that $ k(z_{n}) \neq k(q) $ for all $ n \in \mathbb{N} $. By the Pigeonhole Principle, we can extract a subsequence $ (z_{n_{j}})_{j \in \mathbb{N}} $ such that all the $ k(z_{n_{j}}) $’s are equal to a single integer $ m \in \{ 0,\ldots,6 \} $. Then
\begin{align}
       0
& =    \lim_{j \to \infty} \left[ f(q) - f(z_{n_{j}}) \right] \quad
       (\text{As $ f $ is continuous.}) \\
& =    \lim_{j \to \infty} \left[
                           e^{2 \pi i \cdot k(q) / 7}         ~ e^{h(q) / 7} -
                           e^{2 \pi i \cdot k(z_{n_{j}}) / 7} ~ e^{h(z_{n_{j}}) / 7}
                           \right] \\
& =    e^{2 \pi i \cdot k(q) / 7} ~ e^{h(q) / 7} - e^{2 \pi i m / 7} ~ e^{h(q) / 7} \quad
       (\text{As $ h $ is continuous.}) \\
& =    \left[ e^{2 \pi i \cdot k(q) / 7} - e^{2 \pi i m / 7} \right] e^{h(q) / 7} \\
& \neq 0. \quad (\text{As $ m \neq k(q) $.})
\end{align}
We have a contradiction, so $ k $ must be locally constant on $ D $. As $ D $ is connected, $ k $ must be constant on $ D $. Consequently, $ f $ is a holomorphic function on $ D $. As $ p \in \Omega \setminus \mathcal{Z} $ is arbitrary, $ f $ is holomorphic on $ \Omega \setminus \mathcal{Z} $. (Note: Although holomorphicity has global consequences, it is a local property.) Finally, $ f $ is holomorphic on all of $ \Omega $ as it is continuous at each (isolated) root.
