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



On the existence of a holomorphic logarithm
I believe that your suggestion is correct:
$\newcommand{\C}{\mathbb{C}}$ 




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**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.)




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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.




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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.)


