Complex Integration over a Closed Curve

In summary, you can use Cauchy's integral theorem without proof to parameterize K_m. If you want to set up the equation for z(\tau) using z(\tau) = c+re^{i\tau}, you need to set up c and z0 correctly.
  • #1
sikrut
49
1
(a) Suppose [itex]\kappa[/itex] is a clockwise circle of radius [itex]R[/itex] centered at a complex number [itex]\mathcal{z}[/itex]0. Evaluate: [tex]K_m := \oint_{\kappa}{dz(z-z_0)^m} [/tex]
for any integer [itex] m = 0, \pm{1},\pm{2}, ,... [/itex]Show that

[itex] K_m = -2\pi i[/itex] if [itex] m = -2;[/itex] else :[itex] K_m = 0 [/itex] if [itex] m = 0, \pm{1}, \pm{2}, \pm{3},...[/itex]

Note the minus sign here: [itex]\kappa[/itex] is clockwise.



I am not allowed to use or assume the validity of the residue theorem, but I can use Cauchy's integral theorem without proof.

I was trying to parameterize [itex]K_m[/itex] using

[itex]z(\tau) = c + re^{i\tau} , \tau \in [a,b][/itex] with [itex] a \equiv {\theta_a}[/itex] and [itex] b \equiv \theta_b,[/itex] if [itex] \theta_a < \theta_b, [/itex]

But I'm just stuck on how to set this up at this point. Any ideas?
 
Physics news on Phys.org
  • #2
sikrut said:
I was trying to parameterize [itex]K_m[/itex] using

[itex]z(\tau) = c + re^{i\tau} , \tau \in [a,b][/itex] with [itex] a \equiv {\theta_a}[/itex] and [itex] b \equiv \theta_b,[/itex] if [itex] \theta_a < \theta_b, [/itex]
That's a good start. And c = ? What does it give you for dz and (z-z0)?
 
  • #3
"c" is the complex number around which Km would be centered in the form of [itex]re^{i\tau}[/itex]

So could I assume a clockwise circle centered at 0, making both c and z0 = 0, which makes would make my equation: [tex] \oint_\kappa z^m dz [/tex] ?
 
  • #4
sikrut said:
"c" is the complex number around which Km would be centered in the form of [itex]re^{i\tau}[/itex]

So could I assume a clockwise circle centered at 0, making both c and z0 = 0, which makes would make my equation: [tex] \oint_\kappa z^m dz [/tex] ?
You don't need to make them both 0. z0 is a given, so you can't decide what it is. But you can choose c. You need to use the form [itex]z\left(\tau\right)=c+re^{i\tau}[/itex] to substitute for z in the integral. This is supposed to represent the curve of interest (circle of radius R centered at z0) as tau varies and c remains constant. What value of c will give you that?
What does that substitution give you for dz and for z-z0?
 
  • #5
C = 0

Tau in the exponent is negative because of the clockwise direction.

So [itex] z(\tau) = re^{-i\tau} [/itex], [itex]\tau \in [0, 2\pi] [/itex]

I would plug this in for Z and its respective derivative to replace dz with d[itex]\tau[/itex]
But what would I do with z_0?
 
Last edited:
  • #6
sikrut said:
C = 0
The integration path is a circle radius R, centred at z0. z(t) = c+reit describes a circle radius r, centred where?
 
  • #7
Oh.. z0 is c, which is just some arbitrary constant that I can leave in the equation, where my [itex] (z - z_0)^m dz [/itex] becomes [itex] (re^{i\tau} - c)^m rie^{i\tau} d\tau [/itex]
 
  • #8
sikrut said:
Oh.. z0 is c,
Yes.
which is just some arbitrary constant that I can leave in the equation, where my [itex] (z - z_0)^m dz [/itex] becomes [itex] (re^{i\tau} - c)^m rie^{i\tau} d\tau [/itex]
No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?
 
  • #9
**ignore-wrong post**
 
Last edited:
  • #10
haruspex said:
No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?

[tex] (c + re^{-i\tau} - c)^m (-rie^{-i\tau}) d\tau [/tex]
[tex] = (re^{-i\tau})^m (-rie^{-i\tau}) d\tau [/tex]

How's that?where: [itex] z - z_0 \equiv c + re^{-i\tau} - c [/itex]
 
  • #11
sikrut said:
[tex] = (re^{-i\tau})^m (-rie^{-i\tau}) d\tau [/tex]
Yes. Now simplify and integrate.
 
  • #12
So here's my integration, which I'm pretty sure is correct. I just need help with one more thing.
[tex] (re^{-i\tau})^m(-rie^{-i\tau})d\tau = -ir^{m+1}e^{-(m+1)i\tau} d\tau [/tex]
[tex] \int_{0}^{2\pi} -ir^{m+1}e^{-(m+1)i\tau} d\tau= \frac{r^{m+1}e^{-(m+1)i\tau}}{m+1} |_{0}^{2\pi}[/tex]

And so for all m [itex]\neq[/itex] -1, this result is always 0. Now, how do I prove the case [itex] K_m = -2\pi i[/itex] if m = -1 ?
 
Last edited:
  • #13
sikrut said:
how do I prove the case [itex] K_m = -2\pi i[/itex] if m = -1 ?
Go back to the last point before you did the integration. The integration step you performed, using the usual rule for integrating xm.dx, is only valid when m is not -1. So put m=-1 there and see how you would integrate what results.
 
  • #14
ahhh, I see I see. Ok thank you for all your help hauspex! This one is done.
sorry it took so long for that one part to click. it's been a long day.I might need some more help tomorrow with another :D
 

What is complex integration over a closed curve?

Complex integration over a closed curve is a mathematical concept used to calculate the integral of a complex-valued function over a closed curve in the complex plane. It is a generalization of the concept of line integration in multivariable calculus.

What is the significance of closed curves in complex integration?

Closed curves are important in complex integration because they allow us to define a path over which we can integrate a function in the complex plane. This path forms a closed loop, meaning that it starts and ends at the same point, allowing us to calculate the integral using the fundamental theorem of calculus.

What is the difference between contour integration and complex integration over a closed curve?

Contour integration is a special case of complex integration over a closed curve, where the curve is a smooth path in the complex plane. Complex integration over a closed curve, on the other hand, allows for more general paths, including self-intersecting and discontinuous curves.

How is complex integration over a closed curve calculated?

To calculate complex integration over a closed curve, we use the Cauchy integral theorem, which states that the integral of a complex-valued function over a closed curve is equal to the sum of its residues inside the curve. This allows us to reduce the integral to a simpler form that can be easily evaluated.

What are some real-world applications of complex integration over a closed curve?

Complex integration over a closed curve has many applications in physics and engineering, such as in calculating electric and magnetic field strengths, fluid flow, and heat transfer. It is also commonly used in the study of complex analysis and in the development of mathematical models for various systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
813
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
813
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
119
  • Calculus and Beyond Homework Help
Replies
12
Views
936
  • Calculus and Beyond Homework Help
Replies
6
Views
971
Back
Top