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in the book The Prime Numbers and Their Distribution from Tenenbaum is a note about the existence of a complex continuation of the logarithm:

Let $\alpha>0$ and an analytic function $f(s)$ with no zeros for $Re(s)>\alpha$ and $f(s)$ is real and positive for $s \in (\alpha,\infty)$. then $\log(f(s))$ is defined for $Re(s)>\alpha$ and $Re(\log(f(s))=\log(|f(s)|)$. But there is no proof in book.

I would like to proof it, I know:

Let $f:D \longrightarrow \mathbb{C}$ is analytic and $D$ is simple connected domain, $f(s) \neq 0$ for all $s \in D$. Then there is an analytic funtion $h: D \longrightarrow \mathbb{C}$ with the property

$f(z)=\exp(h(z))$

so $h$ is an analytic branch of logarithm of $f$. But how I can obtain the property $Re( \log(f(s))= \log(|f(s)|)$?

The motivation is to get a logarithm of the riemann $\zeta$ function which is real for all $s>1$.

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    $\begingroup$ if $f(z) = \exp(h(z))$ then $\log |f(z)| = Re(h(z))$ $\endgroup$ – reuns Apr 24 '16 at 7:07
  • $\begingroup$ The condition $\alpha>0$ is of course irrelevant here $\endgroup$ – Hagen von Eitzen Apr 24 '16 at 7:19
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from $\frac{1}{1-z}= \sum_{n=0}^\infty z^n$ for any $|z| < 1$ you know that whenever $|z| < 1$: $$\log(1-z) = 2 i \pi k-\sum_{n=1}^\infty \frac{z^n}{n}$$ hence if $0 < |f(z)| < 1$ then $$\log(f(z)) = \log(1-(1-f(z))) = 2 i k \pi - \sum_{\nu = 1}^\infty \frac{(1-f(z))^\nu}{\nu}$$ for some integer $k$.

because $\log(a f(z)) = \log(a) + \log(f(z))+2 i k \pi $ for some $k$, you get that in any region where $0 < |f(z)| < C$ : $$\log(f(z)) = \log(f(z)/C)+ \log(C) + 2 i k \pi = \log(1-(1-f(z)/C)) +2 i k \pi + \log(C)$$ $$= \log(C) + 2 i m \pi -\sum_{\nu = 1}^\infty \frac{(1-f(z)/C)^\nu}{\nu}$$

is analytic on that region.

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  • $\begingroup$ Thanks for the detailed answer, do you mean $log(f(z))= log(|f(z)/C|)+f(C)+2 i \pi k$ ? $\endgroup$ – niemannski Apr 24 '16 at 7:20
  • $\begingroup$ @niemannski : no why do you add an absolute value ? the first line is for any $z$ complex $|z| < 1$ $\endgroup$ – reuns Apr 24 '16 at 7:24
  • $\begingroup$ @niemannski : I just used the fact that $\log(1-z)$ is analytic on the open unit disk $|z| < 1$ : the logarithm is locally analytic $\endgroup$ – reuns Apr 24 '16 at 7:26
  • $\begingroup$ I was confused by $ \log(f(z)C/C)=\log(f(z)/C)+log(C)+ 2 \pi k$, but it is now clear thanks a lot! $\endgroup$ – niemannski Apr 24 '16 at 7:32

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