On the existence of a holomorphic logarithm Hi,
The following is probably well-known, but I couldn't find anything in the literature. Any reference would be nice.
Let $\Omega$ be a domain in the complex plane, and let $f$ be holomorphic and one-to-one in $\Omega$. Then is it true that $f'$ has a holomorphic logarithm in $\Omega$, i.e. there exists a function $g$ holomorphic in $\Omega$ such that $f'=e^g$ ?
EDIT
This is false in general, as seen by David Cohen's answer. However, I would be very interested to know under which conditions is the above true?
Thank you,
Best regards,
Malik
 A: I believe that your suggestion is correct:
$\newcommand{\C}{\mathbb{C}}$ 

THEOREM. Suppose that $U,V\subset\mathbb{C}$ are domains, and that $f:U\to V$ is a conformal isomorphism. Then $f'$ has a (single-valued) logarithm if and only if $f$ "maps the outer boundary of $U$ to the outer boundary of $V$". (I.e., if $\gamma\subset U$ is a simply closed curve parameterized in positive orientation, then $f\circ\gamma$ also has positive orientation.)

A special case is, of course, that the logarithm always exists when $U$ is simply-connected.
To prove the Theorem, note that the existence of the logarithm of $f'$ means precisely that, on any essential curve in $U$, the derivative does not wind around zero.
Claim. If $\gamma\subset\C$ is a simple closed curve, and $f$ is holomorphic and injective in a neighbourhood of $\gamma$, then the winding number of $f'\circ\gamma$ around zero is either $0$ or $-2$, according to whether $f\circ\gamma$ has the same or the opposite orientation to $\gamma$.
This proves the theorem.

The claim should be intuitively plausible. To prove it, consider the case where $\gamma$ and $f(\gamma)$ are both the unit circle. Then clearly, for $z\in\gamma$,
$$ \arg f'(z) = \arg f(z) - \arg(z) $$
if $f$ is orientation-preserving, while
$$ \arg f'(z) = -\arg f(z) - \arg(z)$$
if $f$ is orientation-reversing.
To reduce the general case to this observation, e.g. assume w.l.o.g. that the curve $\gamma$ is analytic, and change coordinates using the Riemann mapping theorem. 
(I think there should be a direct analytic proof that does not require this change of variable, but I find the above argument rather intuitive.)
A: If $f(z)=\frac{1}{z}$, then $f$ is holomorphic (and one-to-one) in $\mathbb{C}\setminus 0$.  But $f^{\prime}(z)=\frac{-1}{z^{2}}$ does not have a logarithm.  (If $e^{g}=\frac{-1}{z^{2}}$, then $e^{(i\pi-g)/2}=z$, i.e., $(i\pi-g)/2$ would be a logarithm for $z$, which is impossible.)
