If $f$ analytic and $\neq0$ in a simply connected domain, show a single valued analytic branch of $\log f$ is defined on that domain

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.

• There are many definitioins of simply connected regions and the answer to this question depends on your definition. – Kavi Rama Murthy Aug 17 '20 at 5:24
• What is $\int_{\Omega}f(z)dz$? – Paul Frost Aug 17 '20 at 10:55

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.
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 $$\log f$$.
• so, we should find a specific function for $f$ to make this work, such that $\log f$ works with what is above? – User7238 Aug 17 '20 at 15:47
• I'm not sure what you mean. The above idea works for general $f$ of the type you mentioned in your question. – Vercassivelaunos Aug 17 '20 at 15:51
• Maybe I am just a little confused on what you mean by "play around with this antiderivative to find a suitable candidate for $\log f$. – User7238 Aug 17 '20 at 15:53
• 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 $\tilde F=\log f$. – Vercassivelaunos Aug 17 '20 at 16:25