Complex Integration

**InfernoZeus** · August 16th 2010, 07:33 AM

I'm stuck on a question of complex integration; there doesn't seem to be any way to integrate the function, so I imagine there must be a 'trick' somewhere, but I can't work it out

Here's the problem:

Compute

$\oint \frac{log(z)}{z^2 + 9}dz$

around $|z-4i|=3$ .

I parametrized the path as

$\gamma (t) = 4i + 3e^{it},\ 0 \leq t \leq 2\pi$

but then I end up trying to integrate

$\int_0^{2\pi} \frac{log(4i+3e^{it})3ie^{it}}{(4i+3e^{it})^2 + 9}dt$

which I can't work out. Thanks for any help

**Defunkt** · August 16th 2010, 07:46 AM

Why not use the residue theorem?

**InfernoZeus** · August 16th 2010, 07:51 AM

Originally Posted by Defunkt

Why not use the residue theorem?

I thought there have to be poles inside the contour. $f(z)$ has poles at $z=\pm 3$ , which aren't inside $\gamma (t)$ .

**Defunkt** · August 16th 2010, 07:53 AM

It actually has poles at $z= \pm 3i$

:P

EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.

**InfernoZeus** · August 16th 2010, 08:01 AM

Originally Posted by Defunkt

It actually has poles at $z= \pm 3i$

:P

EDIT: And if it had no poles then it would be analytic inside that domain, which means that the integral would've been 0.

Damn.. Such a stupid mistake

Thanks for pointing it out!

**InfernoZeus** · August 16th 2010, 08:18 AM

Alright, so how do I go about calculating the residue?

Don't I still need to integrate $f(x)$ to work out $Res(f, z_1)$ where $z_1 = 3i$ ?

**Defunkt** · August 16th 2010, 08:38 AM

No.
Since $z=3i$ is a simple pole, you know that

$Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)$

**InfernoZeus** · August 16th 2010, 09:16 AM

Originally Posted by Defunkt

No.
Since $z=3i$ is a simple pole, you know that

$Res(f, 3i) = \lim_{z \to 3i}(z-3i)f(z)$

How do we know that $z = 3i$ is a simple pole?

**Ackbeet** · August 16th 2010, 09:26 AM

How do we know that z = 3i is a simple pole?

Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.

**InfernoZeus** · August 16th 2010, 09:44 AM

Originally Posted by Ackbeet

Factor the denominator of the original integrand. The factor z - 3i appears to the first power. That implies, by definition, pretty much, that it's a simple pole.

Man, I wish the definition I have was that simple... Why can't my lecturer's explain it like that?! Thanks everyone

**Ackbeet** · August 16th 2010, 09:55 AM

Well, you gotta make sure the numerator doesn't also go to zero or do something else equally strange at the same location. The precision necessary to correctly define a simple pole means you're probably going to be doing something similar to what your lecturers have. However, this is an intuitive way to think of poles, and it works in quite a few cases.

**InfernoZeus** · August 16th 2010, 12:04 PM

In a slightly related question, compute

$\oint \frac{\sin (2z)}{z^{15}}$

around $|z| = 2$ .

There's obviously a pole at $z=0$ but is that a simple pole, as the numerator will be zero at that point...

**Ackbeet** · August 16th 2010, 12:08 PM

The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?

**InfernoZeus** · August 16th 2010, 12:36 PM

Originally Posted by Ackbeet

The numerator will be zero at that point, correct. However, the denominator has a 15th order zero at the origin. You can show that the sin function has a first-order zero at the origin. Therefore, the integrand has a 14th order pole at the origin. Make sense?

Cool, that makes sense. So I can work out the residue at $z=0$ , by using

$b_{1} = \frac{1}{2\pi i} \int_\gamma f(\zeta )d\zeta$

right? I did that and after a bit of manipulation, arrived at

$\frac{1}{2^{15} \pi } \int_{0}^{2\pi }\frac{\sin{4e^{it}}}{e^{14it}}$

which I can't integrate. I've got no idea how to do it..

**Ackbeet** · August 16th 2010, 12:53 PM

I would just evaluate the residue using this formula.

Thread: Complex Integration

LinkBack

Thread Tools

Display