$\Omega$ 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 $\Omega$ ?
The existence of analytical branch of the logarithm of a holomorphic function
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0No problem. However, I made your change in the question clearly visible, in order for my answer to still make sense. – 2011-10-23
2 Answers
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 $H^1(X,\mathbb Z)=0$.
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0@GeorgesElencwajg, are you available to answer some questions about your answer? I am in doubt if I should consider $df = \partial f + \bar{\partial} f$ and how could I use this to show that $exp(log(f)) = f$. – 2014-05-13
Edit: I misread the question and only address the case $n=1$.
If you allow the function $f$ to have roots, then the answer is clearly no. There is no analytic branch of the logarithm of $z$ in any neighborhood of $0$.
If you exclude roots, the answer is yes.
More generally:
Let $f: G \to \mathbb{C}$ be holomorphic with $0 \notin f(G)$ and assume that $G$ is a simply connected domain (in particular $G$ convex suffices). Then $f$ has a holomorphic logarithm $g: G \to \mathbb{C}$, that is $f(z) = e^{g(z)}$ for all $z \in G$.
Fix $z_0 \in G$ arbitrarily, choose $g(z_0)$ in such a way that $f(z_0) = \exp{g(z_0)}$ and put g(z) = g(z_0) + \int_{z_0}^{z} \frac{f'(w)}{f(w)}\,dw. Check that $g$ is well-defined, holomorphic, and that $\frac{d}{dz} [f \cdot e^{-g}] = 0$, hence $f = e^g$.
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0Here is how I came about this question: consider the set of all the symmetric complex $n \times n$ matrix and its subset $H$ consisting of all symmetric nonsingular complex $n \times n$ matrix $B$ such that ReB > 0 (positive definite), then function $f:B \to {(\det (B))^{\frac{1}{2}}}$ is well-defined and analytic on $H$. – 2011-10-23