Complex Integration over a Closed Curve

[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?

 

[haruspex]

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)?

 

[sikrut]

"c" is the complex number around which K_m 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 ?

 

[haruspex]

"c" is the complex number around which K_m 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?

 

[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?

 

[haruspex]

C = 0

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

 

[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

 

[haruspex]

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?

 

[sikrut]

**ignore-wrong post**

 

[sikrut]

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

 

[haruspex]

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

Yes. Now simplify and integrate.

 

[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 ?

 

[haruspex]

how do I prove the case K_m = -2\pi i 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 x^m.dx, is only valid when m is not -1. So put m=-1 there and see how you would integrate what results.

 

[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