TITLE: The existence of analytical branch of the logarithm of a holomorphic function
QUESTION [9 upvotes]: $\Omega$ is a convex open set in $\mathbb {C}^n$ and $f$ is an analytical function Edit: without zero point on $\Omega$, then can we define an analytical branch of $\ln {f}$ on $\Omega$ ?

REPLY [17 votes]: I am happy to report that, yes, you can define $\log f$ even in much greater generality than in your situation.
Namely, if you have a real differentiable manifold $M$ and a function $f\in \mathcal C^\infty (M)$ never zero on $M$, then $f$ has a $\mathcal C^\infty $ logarithm 
as soon as the first De Rham cohomology group of $M$ vanishes: $H^1_{DR}(M, \mathbb R)=0$.   
The definition of the logarithm is straightforward: fix a point $x_0\in M$ and define  $$(\log f)(x)= \int_\gamma \frac {df}{f}    $$
where $\gamma$ is a differentiable path joining $x_0$ to $x$, along which we can integrate the closed  $1$-form $\frac {dg}{g}$.
The vanishing cohomology hypothesis ensures that the value of $\log f$ at $x$ does not depend of the path $\gamma$ chosen.     
If $M$ happens to be a complex holomorphic manifold and if $f\in \mathcal O(M)$ is holomorphic, then the logarithm of $f$ will automatically be holomorphic: $\log f\in \mathcal O(M)$.
This applies to your case since a simply connected manifold -and a fortiori a convex set in a vector space- has zero first De Rham cohomology group.
Finally, just for old times' sake, let me sum this up in the language of classical physics :
$$\text{Every conservative vector field has a potential}$$
A variant
 Specialists in complex manifolds are addicted to the exact sequence of sheaves on the complex manifold $X$:
$$0\to 2i\pi \mathbb Z\to \mathcal O_X \stackrel {exp}{\to} \mathcal O^\ast_X  \to 0  $$
A portion of the associated cohomology long exact sequence  is 
$$ \Gamma(X,\mathcal O_X) \stackrel {exp}{\to} \Gamma(X,\mathcal O^\ast_X) \to H^1(X,\mathbb Z)  $$
which shows again that every nowhere vanishing holomorphic function on $X$ is an exponential (in other words: has a logarithm) as soon as $H^1(X,\mathbb Z)=0$.