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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. $\endgroup$ – Kavi Rama Murthy Aug 17 '20 at 5:24
  • $\begingroup$ What is $\int_{\Omega}f(z)dz$? $\endgroup$ – Paul Frost Aug 17 '20 at 10:55
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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 '20 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. $\endgroup$ – Vercassivelaunos Aug 17 '20 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 '20 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 $\tilde F=\log f$. $\endgroup$ – Vercassivelaunos Aug 17 '20 at 16:25

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