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I have this homework problem which i need your help about how to start?

g is holomorphic in a simply connected domain U. show that there is a f which is holomorphic in U without zero such that g=f'/f in U.

I know that for any simply connected domain U if g is holomorphic in U then g has a primitive in U i.e. f'=g. and i should use this somehow but I could not see it how?

Thanks.

ruud
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  • Notice that $f'/f$ is the derivative of $\log f$. – Gerry Myerson Feb 24 '14 at 06:13
  • @gerry: I saw that I guess you somehow see that g is a branch of logz in U but i do not can you please give more hint. – ruud Feb 24 '14 at 06:30
  • I think the question of defining a branch of $\log f$ in a domain where $f$ is never zero has been asked on this site recently. Maybe question 3736. – Gerry Myerson Feb 24 '14 at 06:32
  • Actually, http://math.stackexchange.com/questions/74920/the-existence-of-analytical-branch-of-the-logarithm-of-a-holomorphic-function --- also http://math.stackexchange.com/questions/143609/representation-of-holomorphic-functions-by-exponential – Gerry Myerson Feb 24 '14 at 06:34
  • @gerry: firstly thanks a lot for all the links. but in my case a holomorphic function g defined on U might be vanished or not it is not said anything about it on the contrary of the other question – ruud Feb 24 '14 at 06:42
  • What's relevant is $f$ not being zero. – Gerry Myerson Feb 24 '14 at 08:03

1 Answers1

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I think it goes like this.

$g$ is holomorphic, so it has a holomorphic antiderivative, $h$.

$h$ is holomorphic, so $e^h$ is holomorphic and nonzero. Let $f=e^h$, and show it's what you want.

Gerry Myerson
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