# Proof about complex holomorphic function

I have $f(z)$ which is holomorphic in $G$ open set, and there exist $g(z)$ such that is holomorphic in $G$ and $e^{g(z)}=f(z)$ then for every $n>1$ there exist $h(z)$ s.t. $h^n=f$.

I tried this:

I put $h(z)=e^{g(z)/n}$ then I have to proof that this sequence of function is holomorphic in $G$ but its obvious since $g(z)$ is holomorphic and its a composition over the exponential function, and the sequence $h_n(z)$ converges to $0$ as $n$ goes to $\infty$.

Should I define the $h$ as the limit of this sequence?

• If you stopped just before "and the sequence $h_n(z)$ converges to 0 as n goes to infiny", the solution would be complete. Why consider the limit when $n\to\infty$ at all? The last sentence "Should I define the h as the limit of this sequence?" is a mystery as well. – Did Nov 2 '17 at 16:09
• Ah okey so fixed n natural number not for the sequence of functions. But is it everything? So easy exercise? I Because the exercise is in a sheet of complex integrals and application of Cauchy's theorem and Deformation theorem. – energy Nov 2 '17 at 16:13
• Are you looking for the inverse which is also true? That is, If for every $n$, there exists a holomorphic nowhere vanishing $h$, such that $h^n=f$, then there exists a $g$, such that $e^g=f$. – Yiorgos S. Smyrlis Nov 2 '17 at 17:45

No. For each $n\in\mathbb N$, you just define $h_n(z)=e^{g(z)/n}$. That's all. It is clear that $h_n$ is holomorphic and that $(\forall z\in G):{h_n(z)}^n=f(z)$. What else could you possibly want?
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