Let $f$ be analytic with continuous derivative on open $U\subset \Bbb{C}$ such that $|f(z)-1|<1$ on $U$. The question continues by asking to show that $\int_{\gamma}{f'\over f}dz=0$ on any closed path $\gamma$, but I am fixated on why is it so clear that there exists a continuous $\log(f)$ in that area. I know that the origin isn't in $U$ which makes it workable.

I cannot quite refer to the claim that there simply exists a branch because it doesn't say much to me.

In another source (a question in this site) I saw a claim according to which $\int_{[z_0,z]}{f'(z)\over f(z)}dz+f(z_0)$ is holomorphic as ${f'\over f}$ is analytic. Why does it obviously follow?; When is the integral of an analytic function, holomorphic? What are the ground rules by which the existence of such branch, ($\log (f)$)is promised?

I would appreciate a more authoritative perspective on those cases.

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    $\begingroup$ If $f$ is holomorphic, then also $f'$ is holomorphic. If $f(z)\ne0$ on the whole domain, then also $f'/f$ is holomorphic. Every holomorphic function has an antiderivative. $\endgroup$ – egreg Dec 28 '16 at 22:21

$g(z) = \log z$ is holomorphic on $U = |z-1| < 1$ so if $f(z)$ is a holomorphic function $V \to U$ then $g(f(z))$ is holomorphic on $V$ (this is no more than the chain rule).

And if $h(z)$ is holomorphic on $W$ a simply connected open then $H(z) = \int_a^z h(s)ds$ is holomorphic on $W$. This follows from the Cauchy integral theorem $\int_\gamma h(s)ds = 0$ whenever $\gamma$ is a closed contour, so that $\int_a^z h(s)ds$ is well-defined (it doesn't depend on the path $a \to z$) and it is obviously complex-differentiable.

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  • $\begingroup$ So $f$ is holomorphic as well? In particular? $\endgroup$ – Meitar Dec 28 '16 at 22:04
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    $\begingroup$ @Meitar It is trivial that (complex) analytic $\implies $ holomorphic. The Cauchy integral formula proves the converse is true. $\endgroup$ – reuns Dec 28 '16 at 22:08
  • $\begingroup$ What I am confused about is that for $g(f)$ to be holomorphic, $g'(f(z))f'(z)$ has to be continuous, but all my uncertainty is because I can't tell whether or not $f$ is continuous. $\endgroup$ – Meitar Dec 28 '16 at 22:14
  • $\begingroup$ Another thing that stuck me: I wasn't sure there was actually a branch of $\log z$ in $D_1(1,0)$. Now I thought of $\{z:\arg(z)\in(-{\pi\over 2},{\pi\over 2})\}$. Is that admissible? $\endgroup$ – Meitar Dec 28 '16 at 22:31
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    $\begingroup$ @Meitar Yes, or $\log(z) = \int_1^z \frac{ds}{s} = -\int_0^{z-1} \frac{ds}{1-s}=-\sum_{k=1}^\infty \frac{(1-z)^k}{k}$ works fine. I don't see what you mean for $g(f(z))$. Using the Taylor expansion $f(z) = f(z_0)+ (z-z_0)f'(z_0)+o(|z-z_0|)$ and $g(s) = g(s_0)+(s-s_0)g'(s_0)+o(|s-s_0|)$ so that $g(f(z)) = g(f(z_0))+ (z-z_0) g'(f(z_0))f'(z_0)+ o(|z-z_0|)$ (the chain rule) and hence it is complex differentiable. $\endgroup$ – reuns Dec 28 '16 at 22:52

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