# Complex Integration over a Closed Curve

by sikrut
Tags: complex, curve, integration
 P: 37 (a) Suppose $\kappa$ is a clockwise circle of radius $R$ centered at a complex number $\mathcal{z}$0. Evaluate: $$K_m := \oint_{\kappa}{dz(z-z_0)^m}$$ for any integer $m = 0, \pm{1},\pm{2}, ,...$Show that $K_m = -2\pi i$ if $m = -2;$ else :$K_m = 0$ if $m = 0, \pm{1}, \pm{2}, \pm{3},....$ Note the minus sign here: $\kappa$ 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 $K_m$ using $z(\tau) = c + re^{i\tau} , \tau \in [a,b]$ with $a \equiv {\theta_a}$ and $b \equiv \theta_b,$ if $\theta_a < \theta_b,$ But I'm just stuck on how to set this up at this point. Any ideas?
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P: 7,952
 Quote by sikrut I was trying to parameterize $K_m$ using $z(\tau) = c + re^{i\tau} , \tau \in [a,b]$ with $a \equiv {\theta_a}$ and $b \equiv \theta_b,$ if $\theta_a < \theta_b,$
That's a good start. And c = ? What does it give you for dz and (z-z0)?
 P: 37 "c" is the complex number around which Km would be centered in the form of $re^{i\tau}$ So could I assume a clockwise circle centered at 0, making both c and z0 = 0, which makes would make my equation: $$\oint_\kappa z^m dz$$ ???
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P: 7,952

## Complex Integration over a Closed Curve

 Quote by sikrut "c" is the complex number around which Km would be centered in the form of $re^{i\tau}$ So could I assume a clockwise circle centered at 0, making both c and z0 = 0, which makes would make my equation: $$\oint_\kappa z^m dz$$ ???
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 $z\left(\tau\right)=c+re^{i\tau}$ 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?
 P: 37 C = 0 Tau in the exponent is negative because of the clockwise direction. So $z(\tau) = re^{-i\tau}$, $\tau \in [0, 2\pi]$ I would plug this in for Z and its respective derivative to replace dz with d$\tau$ But what would I do with z_0?
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P: 7,952
 Quote by sikrut C = 0
The integration path is a circle radius R, centred at z0. z(t) = c+reit describes a circle radius r, centred where?
 P: 37 Oh.. z0 is c, which is just some arbitrary constant that I can leave in the equation, where my $(z - z_0)^m dz$ becomes $(re^{i\tau} - c)^m rie^{i\tau} d\tau$
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P: 7,952
 Quote by sikrut Oh.. z0 is c,
Yes.
 which is just some arbitrary constant that I can leave in the equation, where my $(z - z_0)^m dz$ becomes $(re^{i\tau} - c)^m rie^{i\tau} d\tau$
No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?
 P: 37 **ignore-wrong post**
P: 37
 Quote by haruspex No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?
$$(c + re^{-i\tau} - c)^m (-rie^{-i\tau}) d\tau$$
$$= (re^{-i\tau})^m (-rie^{-i\tau}) d\tau$$

How's that?

where: $z - z_0 \equiv c + re^{-i\tau} - c$
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P: 7,952
 Quote by sikrut $$= (re^{-i\tau})^m (-rie^{-i\tau}) d\tau$$
Yes. Now simplify and integrate.
 P: 37 So here's my integration, which I'm pretty sure is correct. I just need help with one more thing. $$(re^{-i\tau})^m(-rie^{-i\tau})d\tau = -ir^{m+1}e^{-(m+1)i\tau} d\tau$$ $$\int_{0}^{2\pi} -ir^{m+1}e^{-(m+1)i\tau} d\tau= \frac{r^{m+1}e^{-(m+1)i\tau}}{m+1} |_{0}^{2\pi}$$ And so for all m $\neq$ -1, this result is always 0. Now, how do I prove the case $K_m = -2\pi i$ if m = -1 ?
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 Quote by sikrut how do I prove the case $K_m = -2\pi i$ if m = -1 ?