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Analytic function I

asqw121

New member
Jun 11, 2013
5
Prove f((z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin?
Thanks
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Prove f((z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin?
Thanks
Isn't it because log(z) is undefined when z = 0 + 0i?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Prove f((z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin?
Thanks
The question is controversial and depends from the definition of complex logarithm. From De Moivre's writing $z= r\ e^{i\ \theta}$, where $r$ is the 'modulus' and $\theta$ is the 'argument', we derive $\ln z = \ln r + i\ \theta$. The point of controversial is the precise definition of $\theta$. If we accept the approach of the German mathematician Berhard Riemann, then the function $\ln z$ can be analitycally extended to the whole complex plane with the only exception of the point z=0. For details see...


Wolfram Demonstrations Project

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The question is controversial and depends from the definition of complex logarithm. From De Moivre's writing $z= r\ e^{i\ \theta}$, where $r$ is the 'modulus' and $\theta$ is the 'argument', we derive $\ln z = \ln r + i\ \theta$. The point of controversial is the precise definition of $\theta$. If we accept the approach of the German mathematician Berhard Riemann, then the function $\ln z$ can be analitycally extended to the whole complex plane with the only exception of the point z=0. For details see...


Wolfram Demonstrations Project
An illustrative example of Riemann's analytic extension of the function $\ln z$ is based on the following pitcure...



The function $\ln z$ is analytic in $z= s_{0}=1$, so that we can develop it in Taylor's series...

$$\ln z = 2\ \pi\ k\ i + (z-1) - \frac{(z-1)^{2}}{2} + \frac{(z-1)^{3}}{3}-...\ (1)$$

The (1) converges inside a disc of radious 1 centered in $s_{0}=1$ and that means that in any point inside the disc the (1) permits the computation of $\ln z$ and all its derivatives. The knowledge of the derivatives in $z=s_{1} = e^{i \frac{\pi}{4}}$, which is inside the disc, permits to write the Taylor's expansion of $\ln z$ around $s_{1}$ and this series converges in a disc of radious 1 centered in $s_{1}$, and that means that we have extended the region of the complex plane where $\ln z$ is analytic. Proceeding along this way we can compute the function and its derivatives in $s_{2}= e^{i\ \frac{\pi}{2}}$, $s_{3}= e^{i\ \frac{3\ \pi}{4}}$, $s_{4}= e^{i\ \pi}$ and so one. When we return to $s_{0}$ we obtain the (1) with an 'extra term' $2\ \pi\ i$ and, very important detail, we have found a region of the complex plane surrounding the point $z=0$ where $\ln z$ is analytic...

Kind regards

$\chi$ $\sigma$