Analytic Function I: Proving f(z)=log z Not Analytic on Domain D w/ γ

[asqw121]

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]

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]

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]

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...

http://www.123homepage.it/u/i71462702._szw380h285_.jpg.jfif

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

$\chi$ $\sigma$

 

[blue_raver22]

for your question. I am happy to provide a response to your content.

First, let's define what it means for a function to be analytic on a domain. A function f(z) is said to be analytic on a domain D if it is differentiable at every point in D. This means that the limit of the function as z approaches any point in D exists and is unique.

Now, let's consider the function f(z) = log z. This function is not defined at z = 0, as the logarithm of 0 is undefined. However, we can define a branch of the logarithm function that is continuous and differentiable on a domain D that does not include the origin.

Now, let's assume that there exists a domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin. This means that every point on the curve γ is contained within the domain D.

Since the function f(z) = log z is not defined at z = 0, it cannot be differentiable at any point on the curve γ, including the points on the curve that are closest to the origin. This is because the limit of the function as z approaches these points does not exist, as the function is undefined at z = 0.

Therefore, we can conclude that the function f(z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin. This is because there are points on the curve γ where the function is not differentiable, which violates the definition of analyticity.

In conclusion, we have proven that the function f(z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin. This further supports the fact that the function is not analytic on the entire complex plane, as it is not defined at z = 0.