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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

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It seems to me that a minor modification of David Cohen's answer provides counterexamples whenever $\Omega$ is not simply connected. Of course, whenever $\Omega$ is simply connected then every non-vanishing function admits a logarithm. – Andreas Blass May 31 '13 at 19:17
I agree. If $\Omega$ is not simply connected then $\mathbb C\setminus\Omega$ has at least one bounded component, say $K$. If $a$ is any point of $K$, then $z-a$ has no holomorphic logarithm on $\Omega$, and it follows that $f(z)=(z-a)^2$ is a counterexample. – Etienne May 31 '13 at 21:31
@Etienne Matheron : In general, $f(z)=(z-a)^2$ is not necessarily one-to-one on $\Omega$. You probably mean that $f(z)=1/(z-a)$ is a counterexample. – Kalim May 31 '13 at 23:47
@Andreas Blass : Yes indeed, but I meant under which conditions on $\Omega$ and $f$ does the above holds. In particular, I think it is true if $f$ does not reverse orientation of curves inside $\Omega$. – Kalim May 31 '13 at 23:48
@Malik. Yes, $f(z)=\frac1{z-a}$! – Etienne Jun 1 '13 at 4:12
up vote 5 down vote accepted

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.)

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Yes, this is the proof I had in mind, thank you. However, I was wondering : do you have any idea where I could look to find a reference for this result? It is quite natural, so I am pretty sure it appears somewhere in the literature, but I couldn't find anything. – Kalim Jun 5 '13 at 14:57
@Malik: For the reference, try MR1343250 Fulton, William Algebraic topology. A first course. Graduate Texts in Mathematics, 153. Springer-Verlag, New York, 1995. xviii+430 pp. ISBN: 0-387-94326-9; 0-387-94327-7. Although I am not sure if you can find this precise argument there, Chapter 3 treats winding numbers. There are applications given in Chapter 4, and Chapter 6 has a section on winding numbers and homology. – Margaret Friedland Jun 5 '13 at 15:27
@Margaret Friedland : Thank you for the reference. It seems like a good book, but I didn't find what I'm looking for. – Kalim Jun 6 '13 at 12:35

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.)

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We have $e^g=-1/z^2$, and thus $e^{-g}=-z^2$. How do you obtain $e^{-2g}=z$? But I believe this gives a counter-example because -z^2 does not have a holomorphic logarithm on $\mathbb{C} \setminus \{0\}$. Clearly I overlooked something this morning... Thank you – Kalim May 31 '13 at 18:01

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