Need help with a proof concerning zero-free holomorphic functions.

Question

Suppose $f(z)$ is holomorphic and zero-free in a simply connected domain, and that $\exists g(z)$ for which $f(z) =$ exp$(g(z))$.

The question I am answering is the following:

Let $t\neq 0$ be a complex number. Prove that $\exists h(z)$ holomorphic such that $f(z) = (h(z))^t$.

I see that the idea makes sense, but a nudge in the right direction would be appreciated.

Answer 1

If $f$ is zero free and it is defined in a simply-connected domain, you can define a logarithm of $f$ as $$ r(z)=\int_{z_0}^z \frac{f'(w)\,dw}{f(w)}. $$ This integral is univalent, as $f'/f$ is holomorphic in a simply-connected domain.

Clearly $f(z)=f(z_0)\exp\big(r(z)\big)$.

Then $h(z)=w_0\exp\big(r(z)/t\big)$, where $w_0^t=f(z_0)$.

Answer 2

Nudge: Remember that for real numbers $r>0$, you can define $r^t = \exp(t\ln r)$. Maybe you can do something similar for holomorphic functions?