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Question: Show that if $f(z)$ is analytic and $f(z)\neq0$ in a simply connected domain $\Omega$, then a single valued analytic branch of $\log f(z)$ can be defined in $\Omega$

My Thoughts:Since $f$ is analytic in $\Omega$ then $\int_{\Omega}f(z)dz=0$. Now, then assumption that $f(z)\neq0$ makes me think that we are going to be considering $\frac{f'(z)}{f(z)}$ at some point, because I am not sure how else that assumption would be relevant here. So would it be a good idea to try and play with something like $\int \log f(z)dz$, or something like that? Or, does the problem come down to us picking a single valued analytic branch of $\log f(z)$? Any help is greatly appreciated! Thank you.

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    $\begingroup$ There are many definitioins of simply connected regions and the answer to this question depends on your definition. $\endgroup$ Aug 17, 2020 at 5:24
  • $\begingroup$ What is $\int_{\Omega}f(z)dz$? $\endgroup$
    – Paul Frost
    Aug 17, 2020 at 10:55

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"Since $f$ is analytic in $\Omega$ then $\int_{\Omega}f(z)dz=0$." I'm not sure where this is coming from. $\Omega$ is an open set, not a contour.

You mention $\dfrac{f'(z)}{f(z)}.$ That is certainly important here, but I'm not sure how you knew that.

The main idea is suggested by an example: Suppose $f:\Omega \to \mathbb C\setminus (-\infty,0]$ is analytic. Then $\log f(z)$ is analytic on $\Omega,$ where $\log$ means the principal value logarithm. Thus

$$\tag 1 \frac{d \log f(z)}{dz} = \frac{f'(z)}{f(z)}$$

by the chain rule. So there we we see the expression you mentioned, with its connection to the logarithm.

In our problem we want to work backwards, starting with the right side of $(1).$ We are given an analytic $f$ on $\Omega$ that is never $0.$ Therefore $f'/f$ is analytic on $\Omega.$ Suppose we can find an antiderivative $g$ for $f'/f$ on $\Omega.$ We can then hope $e^g=f,$ which is the same thing as saying $g$ is an analytic logarithm of $f$ in $\Omega.$

Towards that end, consider $e^g/f.$The derivative of this is

$$\frac{fe^gg'-e^gf'}{f^2} =\frac{e^g}{f^2}(f'-f')\equiv 0.$$

Since $\Omega$ is connected, $e^g/f=c$ for some constant $c.$ Note $c\ne 0.$ This gives $(1/c)e^g = f.$ We can write $1/c = e^d$ for some constant $d.$ So we conclude $e^{d+g}=f$ in $\Omega$ and we're done.

Now, the crucial assumption that led to a solution is in boldface above. We assumed $f'/f$ has an antiderivative. Is that true? Yes: Since $\Omega$ is simply connected, every analytic function in $\Omega$ has an antiderivative.

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  • $\begingroup$ So easy to follow, and cleared up a few thoughts I was having. Thank you so much! $\endgroup$
    – User7238
    Aug 9, 2021 at 19:26
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    $\begingroup$ You are certainly welcome. $\endgroup$
    – zhw.
    Aug 10, 2021 at 0:12
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Hint: It can be shown that between any $z_1, z_2 \in \Omega$ there exists a path $\gamma : [0, 1] \to \Omega$ s.t. $\gamma(0) = z_1$, $\gamma(1) = z_2$. Assume that $\Omega$ is non-empty; take $w \in \Omega$. Define $g(z)$ to be the integral of $\frac{f'(z)}{f(z)}$ over some path $\gamma$ from $w$ to $z$. Since $\gamma$ is unique up to homotopy, it can be shown that $g(z)$ is uniquely defined since $\frac{f'(z)}{f(z)}$ has no singularities.

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It all boils down to the fact that if an analytic branch of $\log f$ exists, then we expect its derivative to be $\frac{f'}{f}$ due to the chain rule and that the derivative of the logarithm should be $\log'(z)=\frac{1}{z}$. So the next step will be to consider an antiderivative $F$ of $\frac{f'}{f}$. Such an antiderivative exists because $\frac{f'}{f}$ is analytic on a simply connected domain. Now you should play around with this antiderivative to find a suitable candidate for $\log f$.

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  • $\begingroup$ so, we should find a specific function for $f$ to make this work, such that $\log f$ works with what is above? $\endgroup$
    – User7238
    Aug 17, 2020 at 15:47
  • $\begingroup$ I'm not sure what you mean. The above idea works for general $f$ of the type you mentioned in your question. $\endgroup$ Aug 17, 2020 at 15:51
  • $\begingroup$ Maybe I am just a little confused on what you mean by "play around with this antiderivative to find a suitable candidate for $\log f$. $\endgroup$
    – User7238
    Aug 17, 2020 at 15:53
  • $\begingroup$ I meant that $F$ is already close to being a branch of $\log f$, but not quite there. A hint: if it were already a branch of $\log f$, then we would have $\exp F=f$. And though you can't prove that (since it's not true), you can prove that $\exp F=cf$ for some non-zero constant $c$. Then use that fact to find a function $\tilde F$ with $\exp \tilde F=f$. Then $\tilde F=\log f$. $\endgroup$ Aug 17, 2020 at 16:25

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