Complex Logarithm: Analytic Function on $1<|z|<2$

[Fermat1]

Is the complex logarithm an example of a function analytic on the domain $1<|z|<2$ such that it's derivative on the domain is $\frac{1}{z}$?

Thanks

 

[Prove It]

Is the complex logarithm an example of a function analytic on the domain $1<|z|<2$ such that it's derivative on the domain is $\frac{1}{z}$?

Thanks

No, the complex logarithm is never analytic on the negative real axis...

 

[chisigma]

Is the complex logarithm an example of a function analytic on the domain $1<|z|<2$ such that it's derivative on the domain is $\frac{1}{z}$?

Thanks

Yes, it is!...

A detailed demonstrationj of the fact that $\ln z$ is analytic everywhere in the complex plane with the exception of the point z=0 is in...

http://mathhelpboards.com/analysis-50/evaluating-log-imaginary-number-11064.html#post51453

... and...

http://mathhelpboards.com/analysis-50/evaluating-log-imaginary-number-11064.html#post51462

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

No, the complex logarithm is never analytic on the negative real axis...

By choosing a suitable branch, the complex logarithm is analytic on the negative real axis.

 

[Opalg]

Is the complex logarithm an example of a function analytic on the domain $1<|z|<2$ such that it's derivative on the domain is $\frac{1}{z}$?

No. The complex logarithm can be defined as an analytic function locally in the neighbourhood of any nonzero complex number. But it cannot be defined globally as an analytic function on the annulus $1<|z|<2$. In fact, it cannot even be defined as a continuous function on that annulus.

The complex logarithm $\log z$ satisfies $\log z = \ln|z| + i\arg z$, where $\arg z$ is some value of the argument of $z$. Suppose that $z$ goes round the circle $|z| = 1.5$, starting at the point $1.5$ on the real axis, and that we take $\arg z = 0$ at that starting point. As $z$ goes round the circle with $\log z$ varying continuously, $\arg z$ will increase until it approaches the value $2\pi$ as $z$ completes the circuit. But if $\log z$ is to be continuous on the annulus, $\arg z$ would have to go back to $0$ when $z$ reaches the real axis after going round the circle. So somewhere along the route, there has to be a discontinuity. It could be on the negative real axis, as Prove It wants, or on the positive real axis, or at some other point. But there has to be a discontinuity somewhere, and consequently a point where $\log z$ fails to be analytic.

 

[chisigma]

No. The complex logarithm can be defined as an analytic function locally in the neighbourhood of any nonzero complex number. But it cannot be defined globally as an analytic function on the annulus $1<|z|<2$. In fact, it cannot even be defined as a continuous function on that annulus.

To be fair to say that the function f (x) = ln z is locally analytic at each point of the complex plane with the exception of z = 0 and at the same time to say that in a region of the complex plane not including the point z = 0 it is not globally analytic seems a little contradictory ...

To fill this gap and to my my exclusive personal interest I would be very happy to learn a script where the mathematical concepts of local analyticity and global analyticity are described in an understandable way ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

No. The complex logarithm can be defined as an analytic function locally in the neighbourhood of any nonzero complex number. But it cannot be defined globally as an analytic function on the annulus $1<|z|<2$. In fact, it cannot even be defined as a continuous function on that annulus...

... somewhere along the route, there has to be a discontinuity. It could be on the negative real axis, as Prove It wants, or on the positive real axis, or at some other point. But there has to be a discontinuity somewhere, and consequently a point where $\log z$ fails to be analytic...

Very well!... it is clear that, since the function ln z analytic in every point of the real positive axis, on it the function is continuous and differentiable. If, therefore, there must be 'somewhere' discontinuity, we can take into account the positive imaginary axis, the negative real axis (as Prove It wants), and finally the negative imaginary axis... there seems no other reasonable choices. Now let's pick a point in the region indicated by Fermat where ln z is 'locally analytic', the point z = -1 + i. Let us look at the following figure...
View attachment 2778In z= -1 + i ln z is analytic, i.e. it is expressed as Taylor series that converges inside a circle and everywhere inside this circle the function is analytic. Çlearly on the segment joining the point z = -1 + i and the point z = 0, the function is analytic, so that the circle of convergence is the one shown in the figure. But within the circle there are parts of the positive imaginary axis and parts of the negative real axis , so that both can not be the 'boundaries of discontinuities' we are looking for. If we develop ln z around the point z = -1-i we come to also exclude the negative imaginary axis, so that this 'border of discontinuity' seems to be a little like the the 'Araba Fenice' ... everyone says that it is but nobody knows where it is (Thinking)...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Very well!... it is clear that, since the function ln z analytic in every point of the real positive axis, on it the function is continuous and differentiable. If, therefore, there must be 'somewhere' discontinuity, we can take into account the positive imaginary axis, the negative real axis (as Prove It wants), and finally the negative imaginary axis... there seems no other reasonable choices. Now let's pick a point in the region indicated by Fermat where ln z is 'locally analytic', the point z = -1 + i. Let us look at the following figure...
https://www.physicsforums.com/attachments/2778In z= -1 + i ln z is analytic, i.e. it is expressed as Taylor series that converges inside a circle and everywhere inside this circle the function is analytic. Çlearly on the segment joining the point z = -1 + i and the point z = 0, the function is analytic, so that the circle of convergence is the one shown in the figure. But within the circle there are parts of the positive imaginary axis and parts of the negative real axis , so that both can not be the 'boundaries of discontinuities' we are looking for. If we develop ln z around the point z = -1-i we come to also exclude the negative imaginary axis, so that this 'border of discontinuity' seems to be a little like the the 'Araba Fenice' ... everyone says that it is but nobody knows where it is (Thinking)...

Kind regards

$\chi$ $\sigma$

The basic difficulty here is that "$\log z$" is not really a function at all, in the strict sense of the word. A function by definition must be single-valued. But $\log z = \ln|z| + i\arg z$, and $\arg z$ is not single-valued: it is only defined modulo a multiple of $2\pi$. So if you want to define $\log z$ as an analytic function in some domain, you need to specify which branch of the $\arg$ function to use throughout that domain. In a simply-connected domain like the disk of radius $\sqrt2$ centred at $z=-1+i$, that can be done (in that case, by choosing the value of $\arg z$ that lies between $0$ and $2\pi$), but in a domain that includes a circuit of the origin, it is not possible to give a choice of $\arg z$ that is continuous throughout the domain.

 

[chisigma]

The basic difficulty here is that "$\log z$" is not really a function at all, in the strict sense of the word. A function by definition must be single-valued. But $\log z = \ln|z| + i\arg z$, and $\arg z$ is not single-valued: it is only defined modulo a multiple of $2\pi$. So if you want to define $\log z$ as an analytic function in some domain, you need to specify which branch of the $\arg$ function to use throughout that domain. In a simply-connected domain like the disk of radius $\sqrt2$ centred at $z=-1+i$, that can be done (in that case, by choosing the value of $\arg z$ that lies between $0$ and $2\pi$), but in a domain that includes a circuit of the origin, it is not possible to give a choice of $\arg z$ that is continuous throughout the domain.

True to the saying of Dante Alighieri I have chosen as signature I decide not to waste any more time on this topic. Sooner or later I will open a math note that validates incontrovertibly the concept I said (Wave)...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

True to the saying of Dante Alighieri I have chosen as signature I decide not to waste any more time on this topic. Sooner or later I will open a math note that validates incontrovertibly the concept I said (Wave)...

Kind regards

$\chi$ $\sigma$

I'll be looking forward to this incontrovertible validation!
(Sorry. I just felt an irresistible need to post.)

 

[chisigma]

I'll be looking forward to this incontrovertible validation!
(Sorry. I just felt an irresistible need to post.)

When will be I will decide!... What is certain is that in any case, your contribution will be negligible ... (sorry but I felt the need to post) (Bigsmile)...

Kind regards

$\chi$ $\sigma$