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Complex Integration over a Closed Curve 
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#1
Feb1213, 07:47 PM

P: 47

(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(zz_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? 


#2
Feb1213, 07:59 PM

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#3
Feb1213, 08:28 PM

P: 47

"c" is the complex number around which K_{m} 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 z_{0} = 0, which makes would make my equation: [tex] \oint_\kappa z^m dz [/tex] ??? 


#4
Feb1213, 08:39 PM

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Complex Integration over a Closed Curve
What does that substitution give you for dz and for zz_{0}? 


#5
Feb1213, 09:29 PM

P: 47

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? 


#6
Feb1213, 10:06 PM

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#7
Feb1213, 10:40 PM

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Oh.. z_{0} 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
Feb1213, 11:10 PM

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#9
Feb1213, 11:14 PM

P: 47

**ignorewrong post**



#10
Feb1213, 11:27 PM

P: 47

[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
Feb1213, 11:34 PM

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#12
Feb1213, 11:59 PM

P: 47

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 ? 


#13
Feb1313, 12:18 AM

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#14
Feb1313, 12:21 AM

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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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