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I have a function $f:\mathbb{C}\to\mathbb{C}$ that is analytic on $A = \{s\mid \mathfrak{R}(s)\geq1\}$ and does not vanish on $A$.

Intuitively, I would expect this to mean that $f$ has an analytic logarithm on $A$, but I am not sure if that is true.

I know that it has a holomorphic logarithm on $\{s\mid\mathfrak{R}(s)>1\}$, since that is open and simply connected.

It is also clear that there exists an open set $B \supseteq A$ in which $f$ is holomorphic and does not vanish, but I am not sure if I can assume $B$ to be simply connected, so I cannot simply conclude that $f$ has a holomorphic logarithm on $B$.

It is also clear that, at every point on the $\mathfrak{R}(s) = 1$ line, I can find some open neighbourhood in which $f$ is holomorphic and does not vanish, so I can extend the logarithm of $f$ on $A$ to that neighbourhood analytically. But then I have a ‘different’ logarithm for each point on that line and I need to somehow stitch them together, which is not clear to me.

So my question is: Is my intuition correct? Can I somehow justify the existence of an analytic logarithm of $f$ on $A$?

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Yes, because you can take $B$ to be the union of $A$ and a family of open disks with center on the imaginary axis, so that $B$ is simply connected.

Why is $B$ simply connected? Note that $z+x\in B$ for all $z\in B$ and $x>0$. So given a closed curve in $B$ you can shrink it to a point by first pushing it to the right until it lies in the open right half-plane, then shrinking it to a point there. (Or: $B$ is contractible...)

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  • $\begingroup$ That seems reasonable, but I don't know enough topology to see what theorems you are using here. Why does this imply that $B$ is simply connected? $\endgroup$ – Manuel Eberl Dec 18 '17 at 18:12
  • $\begingroup$ @ManuelEberl See edit $\endgroup$ – David C. Ullrich Dec 18 '17 at 18:19
  • $\begingroup$ Ah, interesting. That seems very reasonable to me. Thanks! $\endgroup$ – Manuel Eberl Dec 18 '17 at 18:21
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Yes, it is true. In fact, if $A\subset\mathbb C$ is open and connected, then $A$ is simply connected if and only if every analytic function $f\colon A\longrightarrow\mathbb C$ has an analytic logarithm. See, for instance, theorem 2.2 in John B. Conway's Functions of One Complex Variable I.

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  • $\begingroup$ I don't see how that relates to my problem. I don't know if my $B$ is simply connected (i.e. if I can pick it such that it is simply connected); that's the entire problem. $\endgroup$ – Manuel Eberl Dec 18 '17 at 18:02
  • $\begingroup$ @ManuelEberl Your set $A$ is simply connected and you wrote “I would expect this to mean that $f$ has an analytic logarithm on $A$, but I am not sure if that is true”. How can you think that my answer is not related to your problem? $\endgroup$ – José Carlos Santos Dec 18 '17 at 18:05
  • $\begingroup$ It is simply connected, but it is not open. $\endgroup$ – Manuel Eberl Dec 18 '17 at 18:06
  • $\begingroup$ @ManuelEberl You are right! But then I would say that your problem makes no sense, because by definition (at least, that's the usual definition) the domain of an analytic function is an open set. I'm sorry for wasting your time. $\endgroup$ – José Carlos Santos Dec 18 '17 at 18:08
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    $\begingroup$ @JoséCarlosSantos Actually "analytic" and "holomorphic" are not quite synonymous. The domain of a holomorphic function is open by definition. The usual definition is that $f$ is analytic on $E$ if for every $z\in E$ there exists $r>0$ such that $f$ is, or can be extended so as to be, holomorphic in $D(z,r)$. So the question does make sense... $\endgroup$ – David C. Ullrich Dec 18 '17 at 18:10

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