Show that there exists $g: \mathbb{\Omega} \rightarrow \mathbb{C}$ analytic such that $g(z)^n = f(z)$ for all $z \in \mathbb{\Omega}$

Question

I'm learning about complex analysis and need some help with this problem:

Let $\mathbb{\Omega}$ open, simply connected, $f: \mathbb{\Omega} \rightarrow \mathbb{C}$ analytic without zeros in $\Omega$ and $n \in \mathbb{N}$. Show that there exists $g: \mathbb{\Omega} \rightarrow \mathbb{C}$ analytic such that $g(z)^n = f(z)$ for all $z \in \mathbb{\Omega}$.

If I understand the problem correctly we want to show that there exists $g(z) = \sqrt[n]{f(z)}$ analytic in $\Omega$ but I have no idea how to tackle it.

Answer 1

Since the domain is simply connected, there is an analytic branch of $\log f(z)$ on $\Omega$ given by $L(z)$, where $$L(z) = \int_{z_0}^z {f'(z) \over f(z)}\,dz$$ Here $z_0$ is any fixed point in $\Omega$, and the integral is over any curve starting at $z_0$ and ending at $z$. Since the domain is simply connected, by Cauchy's theorem this definition of $L(z)$ does not depend on which curve is chosen. It is not hard to show that $e^{L(z)} = f(z)$ for all $z \in \Omega$. Therefore ${\displaystyle g(z) = e^{L(z) \over n}}$ will satisfy $$g(z)^n = e^{L(z)} = f(z)$$

Answer 2

Remember we deal with branching through the exponential. $$ \sqrt[n] { f(z) } = \exp \left ( \frac{ \ln f(z) } {n} \right) = \exp \left ( \frac{ \ln | f(z) | + i \arg f(z) } { n} \right ) = \exp \left ( \frac{ \ln | f(z) | }{n} \right ) \exp \left ( \frac{ i (\text{Arg } f(z) + 2\pi k) } {n} \right) $$ where $\ln | f(z) | $ is well defined since $|f(z)| >0$ on $\Omega$ and $k \in [0, n-1]$, thus we've found $n$ possible $g(z)$'s with the property $g(z)^n = f(z)$

Answer 3

The statement is also valid if instead of analytic functions we consider smooth, or continuous. The important thing: We have an $n$-fold covering map $p \colon \mathbb{C}\backslash\{0\}\to \mathbb{C}\backslash\{0\}$, $z \mapsto p(z) = z^n$. Given now a continuous map $f$ from a simply connected open set $\Omega$ ( manifold, etc) to (the bottom) $\mathbb{C}\backslash\{0\}$ ( the top one), $\omega_{0} \in \Omega$ and $y_0$ so that $p(y_0) = f(\omega_0)$, there exists a (unique) continuous map $g \colon \Omega \to \mathbb{C}\backslash\{0\}$ so that $p\circ g = f$ and $g(\omega_0) = y_0$ ( the existence and uniqueness of liftings). Since the covering $p$ is locally an analytic isomorphism, $g$ will be analytic (smooth) whenever $f$ will be so.