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Suppose $\Omega$ is a domain of the complex plane (i.e. an open and connected subset of the plane). Suppose $f$ is holomorphic on $\Omega$, and $f$ is not identically zero.

Suppose $f$ has a holomorphic logarithm on $\Omega$, which means that there is a function $g$ holomorphic on $\Omega$ such that $e^g=f$. Then it is easy to show that $f$ has holomorphic $n$-th roots on $\Omega$ for each $n$, which means that for each integer $n$, there exist a function $g_n$ holomorphic on $\Omega$ such that $(g_n)^n = f$.

Is the converse true? i.e. if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then can we find a function $g$ holomorphic on $\Omega$ such that $f=e^g$?

A few remarks :

One can prove that if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then $f$ does not vanish on $\Omega$. Therefore, we can define a holomorphic logarithm locally, but is it possible to find a global holomorphic logarithm?

Furthermore, notice that $\Omega$ is not supposed simply connected, in which case the answer to my question is yes.

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The condition that $f$ have a holomorphic logarithm is equivalent to $df/f=f'(z)dz/f(z)$ being an exact differential. This is equivalent to the integral of $df/f$ over all closed curves in $\Omega$ vanishing. Let $C$ be a closed curve in $\Omega$.

If $f=g^n$ is an $n$-th power in $\Omega$ of a holomorphic $g$ then $\int_C df/f=n\int_C dg/g$. But $\int_C dg/g$ is an integer multiple of $2\pi i$. Hence $\int_C df/f$ is an integer multiple of $2\pi ni$. If this holds for all $n$ then $\int_C df/f=0$. It follows that $f$ has a holomorphic logarithm.

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    $\begingroup$ Oh.. Nice! Thank you! $\endgroup$ – Malik Younsi Aug 31 '10 at 20:54

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