How can we show that if $f$ is holomorphic on a region $D$, and has no root in $D$ ($D$ is simply connected), then $g:D\rightarrow \mathbb{C}$ is holomorphic such that for every $z\in D$, $ g^n(z)=f(z)$
Alot of the exercises in my notes use this result, but I was just wondering why is this true?
Since $f'/f$ is holomorphic on the simply connected domain, let $\phi$ be its primitive, then $\dfrac{d}{dz}(fe^{-\phi})=0$, and hence $fe^{-\phi}=c$ for some constant $c\ne 0$. And then $f=e^{\phi+c'}$. Now we let $g=e^{(\phi+c')/n}$.
\begin{align*} (fe^{-\phi})'=f'\cdot e^{-\phi}-f\cdot e^{-\phi}\phi'=f'e^{-\phi}-f\cdot e^{-\phi}\cdot\dfrac{f'}{f}=0. \end{align*}