Complex log as integral

[Mondo]

Hello,

I read a chapter on complex logarithm integration and I am confused. [https://i2.paste.pics/8caf8d9e8129acde986cd41399893fad.png]
The author starts of by saying that in general we can say (like in real analysis) that the integral:
[math]\int_1^z{\frac{1}{z}dz} = log(z)[/math]But there is one problem namely the singularity at the origin and hence we must specify a contour K from 1 to Z before the integral above becomes well defined; Well here is my first question:
1. Why we need to take care of this singularity at origin in complex analysis and we don't need to do this in real analysis?
Next, they draw several contours and write down equation to count rotations as shown here -> https://i2.paste.pics/8caf8d9e8129acde986cd41399893fad.png
finally, they write an equation for log(Z) as:
[math]log_K(Z) = ln|Z| +i\theta_K(Z)[/math]and here is my second questions:
2. How was this equation derived? What is the log base K equal to log base e + some weird imaginary part?

Thank you.

 

[topsquark]

No problems. I wrote a reply early in the morning hours that was incomplete and decided I was too tired to fix it. So I deleted my post.

-Dan

 

[Mondo]

Hi Dan, that's fine but are you going to update it and post?

Thanks!

 

[topsquark]

Hi Dan, that's fine but are you going to update it and post?

Thanks!

I'll look at it again later tonight.

-Dan

 

[topsquark]

1. If we presume a real number z such that 1 < z then we don't need to worry about anything in the integration because the path does not pass through the origin. But if z is complex then we need to specify a path for the integral.

In complex analysis we can choose just about any path we like to do an integral but certain paths give different answers to the integral. The most important features involve closed paths about the singularities, or poles, of the integrand... in this case z = 0. There is a theorem from the Calculus of Residues that states [imath]\oint f(z) ~ dz = 2 \pi i \sum \text{(enclosed residues)}[/imath], where "residues" is a function of the poles. It is a very powerful theorem and depends only on if the integration path encloses the pole, not on the exact path itself. Many complicated real integrals can be adapted to be done in this way.

2. We don't actually need the residue theorem here for the concept. Note that any complex number z can be written in polar form: [imath]z = r e^{i \theta }[/imath]. So [imath]ln(z) = ln \left ( r e^{i \theta } \right )[/imath]. Now, [imath]e^{2 \pi n i} = 1[/imath] where n is any integer, so:
[imath]ln(z) = ln \left ( r e^{i \theta } \right ) = ln \left ( r e^{i \theta } e^{2 \pi n i} \right ) = ln \left ( r e^{i \theta + 2 \pi n i} \right ) = ln(r) + (i \theta + 2 \pi n i)[/imath]

so ln(z) is multivalued. Now, as to the number of loops, the residue theorem says that we sum over the residues so if the path circles the origin twice then we include the residue twice, if the path circles the origin three times then we include the residue three times, etc. The number of loops around the origin is a positive integer n, though the derivation I showed above can use a negative integer as well.

-Dan

 

[Mondo]

in this case z = 0

I think you wanted to say [imath]\int{f(z)} = 0[/imath]

As for the second question, in your derivation you got ln(z) on the LHS while they have [imath]log_K(z)[/imath] something doesn't add up?

Thank you for referring to residue theorem, yes is really interesting. But what surprises me more is so called "Laurent Series" - why can we represent a function near its singualrity by a Taylor series with negative powers (now called Lauren series). In other words, how these negative powers allows us to overcome singularities?

Thank you

 

[topsquark]

I think you wanted to say [imath]\int{f(z)} = 0[/imath]

As for the second question, in your derivation you got ln(z) on the LHS while they have [imath]log_K(z)[/imath] something doesn't add up?

Thank you for referring to residue theorem, yes is really interesting. But what surprises me more is so called "Laurent Series" - why can we represent a function near its singualrity by a Taylor series with negative powers (now called Lauren series). In other words, how these negative powers allows us to overcome singularities?

Thank you

No, I meant that the pole is at z = 0. The integral is not 0.

You are right about the log. I didn't look at it carefully enough. As they are labeling the curves with the index K I suspect that they are meaning that the log over the curve K is equal to the given expression. Otherwise the "ln|z|" would also have to be [imath]log_K|z|[/imath].

Here's a primer on Calculus of Residues. I can't really go through how to do the derivations due to the limitations of the Forum interface, but if you have specific questions about how to derive and apply Laurent series I'd be happy to help.

-Dan