Let $f$ be a holomorphic function in the open subset $G$ of $C$. Let the point $z_0$ of $G$ be a zero of $f$ of order $m$. I want to prove that there is a branch of $f^{1/m}$ in some open disk centered at $z_0$.
I tried to prove this way but I am not sure about my solution:
Since $z_0$ is a zero of $f$ of order $m$, there exist an analytic function $g$ s.t. $f(z)=(z-z_0)^mg(z)$; $g(z_0)\neq0$. If there is a branch $h$ of log$f$, then $e^{h/m}$ is a branch of $f^{1/m}$. My problem is how to prove such $h$ exists becuase $f(z_0)=0$.
I appreciate any help to solve this problem.
Because every analytic function $g$ without zeros defined on an open disk $D(z_0.r)$ has some analytic logarithm; just take a primitive $G$ of $\dfrac{g'}g$ such that $\exp\bigl(G(z_0)\bigr)=g(z_0)$.