# Homework Help: Complex Integration over a Closed Curve

1. Feb 12, 2013

### sikrut

(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?

2. Feb 12, 2013

### haruspex

That's a good start. And c = ? What does it give you for dz and (z-z0)?

3. Feb 12, 2013

### 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: $$\oint_\kappa z^m dz$$ ???

4. Feb 12, 2013

### haruspex

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?

5. Feb 12, 2013

### sikrut

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?

Last edited: Feb 12, 2013
6. Feb 12, 2013

### haruspex

The integration path is a circle radius R, centred at z0. z(t) = c+reit describes a circle radius r, centred where?

7. Feb 12, 2013

### sikrut

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 $(re^{i\tau} - c)^m rie^{i\tau} d\tau$

8. Feb 12, 2013

### haruspex

Yes.
No, you've not substituted correctly. If c = z0 and z = c+re-iτ, what is z-z0?

9. Feb 12, 2013

### sikrut

**ignore-wrong post**

Last edited: Feb 12, 2013
10. Feb 12, 2013

### sikrut

$$(c + re^{-i\tau} - c)^m (-rie^{-i\tau}) d\tau$$
$$= (re^{-i\tau})^m (-rie^{-i\tau}) d\tau$$

How's that?

where: $z - z_0 \equiv c + re^{-i\tau} - c$

11. Feb 12, 2013

### haruspex

Yes. Now simplify and integrate.

12. Feb 12, 2013

### sikrut

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 ?

Last edited: Feb 13, 2013
13. Feb 13, 2013

### haruspex

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. Feb 13, 2013

### sikrut

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