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Complex Integration over a Closed Curve

by sikrut
Tags: complex, curve, integration
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sikrut
#1
Feb12-13, 07:47 PM
P: 42
(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?
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haruspex
#2
Feb12-13, 07:59 PM
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Quote Quote by sikrut View Post
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)?
sikrut
#3
Feb12-13, 08:28 PM
P: 42
"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] ???

haruspex
#4
Feb12-13, 08:39 PM
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Complex Integration over a Closed Curve

Quote Quote by sikrut View Post
"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?
sikrut
#5
Feb12-13, 09:29 PM
P: 42
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?
haruspex
#6
Feb12-13, 10:06 PM
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Quote Quote by sikrut View Post
C = 0
The integration path is a circle radius R, centred at z0. z(t) = c+reit describes a circle radius r, centred where?
sikrut
#7
Feb12-13, 10:40 PM
P: 42
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]
haruspex
#8
Feb12-13, 11:10 PM
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Quote Quote by sikrut View Post
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?
sikrut
#9
Feb12-13, 11:14 PM
P: 42
**ignore-wrong post**
sikrut
#10
Feb12-13, 11:27 PM
P: 42
Quote Quote by haruspex View Post

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]
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#11
Feb12-13, 11:34 PM
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Quote Quote by sikrut View Post
[tex] = (re^{-i\tau})^m (-rie^{-i\tau}) d\tau [/tex]
Yes. Now simplify and integrate.
sikrut
#12
Feb12-13, 11:59 PM
P: 42
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 ?
haruspex
#13
Feb13-13, 12:18 AM
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Quote Quote by sikrut View Post
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.
sikrut
#14
Feb13-13, 12:21 AM
P: 42
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


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